Ans: (b) Distance of the point A(-3, -4) from x-axis = absolute value of the ordinate = 4 units.
Q3: AOBC is a rectangle whose three vertices are A(0, 2), B(0, 0) and 8(4, 0). The square of the length of its diagonal is equal to: (a) 36 (b) 20 (c) 16 (d) 4
Ans: (b) In rectangle AOBC, AB is the diagonal, where A(0, 2) and B(4, 0). ∴ By distance formula,
Q4: The coordinates of the centre of a circle are (2a, a – 7). Find the value(s) of ‘a’ if the circle passes through the point (11, -9) and has diameter 10√2 units.
Ans: Let the point P be (x, y). Since, P is equidistant from A(7, 1) and 8(3, 5), we get PA = PB
Hence, abscissa of point P is 2 more than its ordinate.
Q6: If the mid-point of the line segment joining the points (a, 4) and (2, 2b) is (2, 6), then the value of (a + b) is given by: (a) 6 (b) 7 (c) 8 (d) 16
Ans: (a) Let the coordinates be A(a, 4), B(2, 2b) and C(2, 6) respectively. On comparing the coordinates, we get ∴ a + b = 2 + 4 = 6
Q7: Two of the vertices of ΔPQR are P(-1, 5) and Q(5, 2). The coordinates of a point which divides PQ in the ratio 2 : 1 are: (a) (3, -3) (b) (5, 5) (c) (3,3) (d) (5, 1)
Ans: (c) Let (x, y) be the coordinates of a point that divides P(-1, 5) and Q(5, 2) in the ratio 2:1. By section formula, ∴ Required coordinates are (3, 3).
Q8: The line represented by intersects x-axis and y-axis respectively at P and Q. The coordinates of the mid-point of line segment PQ are: (a) (2, 3) (b) (3, 2) (c) (2, 0) (d) (0, 3)
Ans: (a) ⇒ 3x + 2y = 12 … (i) Put x = 0 in (i), we have 3(0) + 2y = 12 ⇒ y = 12/2 = 6 Coordinates of Q are (0, 6). Put y = 0 in (i), we have 3x + 2(0) = 12 ∴ Coordinates of P are (4, 0) ∴ Coordinates of mid point of PQ
Q9: The mid-point of the line segment joining the points P(-4, 5) and Q(4, 6) lies on: (a) x-axis (b) y-axis (c) origin (d) neither x-axis nor y-axis
Ans: Let P, Q and R divides the line joining AB into four equal parts. Now, point P divides the line segment AB in 1 : 3. ∴ The coordinates of P are given by
∴ The coordinates of P are Point Q divides the line segment in 1:1 i.e., Q is mid point of AB. ∴ The coordinates of Q are (0, 5). Point R divides AB in 3 : 1 The coordinates of R are given by ∴ The coordinates of R are
Q12: Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the point of intersection.
Ans: Let the point P(0, y) on y-axis divides the line segment joining the points A(5, -6) and B(-1, -4) in the ratio k : 1. ∴ By section formula, we have, ⇒ -k + 5 = 0 ⇒ k = 5 … (i) Hence, the required point is and the required ratio is 5 : 1.
Q13: If the points A(6, 1), B(p, 2), C(9, 4) and D(7, q) are the vertices of a parallelogram ABCD, then find the values of p and q. Hence, check whether ABCD is a rectangle or not.
Ans: Since, ABCD be a parallelogram. ∴ Mid-point of AC = Mid-point of BO Thus, it is not a rectangle.
Previous Year Questions 2024
Q1: Assertion (A): The point which divides the line segment joining the points A (1, 2) and B (–1, 1) internally in the ratio 1 : 2 is Reason (R): The coordinates of the point which divides the line segment joining the points A(x1, y1) and B(x2, y2) in the ratio m1 : m2 are (1 Mark) (CBSE 2024)
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Q3: Points A(–1, y) and B(5, 7) lie on a circle with centre O(2, –3y) such that AB is a diameter of the circle. Find the value of y. Also, find the radius of the circle. (3 Marks) (CBSE 2024)
Ans: A (– 1, y); B(5, 7) Since, AB is a diameter of circle and O is the centre of the circle. OA = OB i.e., O divides AB in 1 : 1 So m1 : m2 = 1 : 1 So ⇒ ⇒ – 6y = y + 7 ⇒ – 7y = 7 ⇒ y = – 1 Point O = (2, 3), A = (–1, – 1) Now, So, radius of circles = 5 units
Q4: Find the ratio in which the line segment joining the points (5, 3) and (–1, 6) is divided by Y-axis. (3 Marks) (CBSE 2024)
Ans: If y-axis divides points (5, 3) and (–1, 6) then coordinate of that point will be (0, y). Let P(0, y) divides A(5, 3) and B(–1, 6) in k : 1. m1 : m2 = k : 1 ⇒ 0 × (k + 1) = – k + 5 ⇒ 0 = – k + 5 ⇒ k = 5 So, m1 : m2 = 5 : 1
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Previous Year Questions 2023
Q1: The distance of the point (-1, 7) from the x-axis is (1 Mark) (2023) (a) -1 (b) 7 (c) 6 (d) √50 [2023, 1 Mark]
Ans: (b) Distance from x-axis = y-coordinate of point = 7 units
Q2: Assertion (A): Point P(0, 2) is the point of intersection of the y-axis with the line 3x + 2y = 4. (1 Mark) (2023) Reason (R): The distance of point P(0, 2) from the x-axis is 2 units. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false but Reason (R) is true.
Assertion (A):To find the intersection of the y-axis with the line 3x + 2y = 4, set x = 0:3(0) + 2y = 4 ⇒ y = 2. So, the point of intersection is P(0,2).Assertion (A) is true.
Reason (R):The distance of point P(0,2) from the x-axis is indeed 2 units.Reason (R) is true.
However, Reason (R) does not explain Assertion (A); it is just a separate true statement. (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Q3: The distance of the point (-6, 8) from origin is (1 Mark) (2023) (a) 6 (b) -6 (c) 8 (d) 10
Ans: (d) Distance of the point (-6, 8) from origin (0, 0)
= 10 Units
Q4: The points (-4, 0), (4, 0) and (0, 3) are the vertices of a (1 Mark) (CBSE 2023) (a) right triangle (b) isosceles triangle (c) equilateral triangle (d) scalene triangle
Ans: (b) The points be A(-4, 0), B(4, 0) and C(0, 3). Using distance formula
= 8 units
= 5 units
= 5 units And, AB2 ≠ BC2 + CA2 [∵ BC = CA] ∴ ΔABC is an isosceles triangle.
Q5: The centre of a circle is (2a, a – 7). Find the values of ‘a’ if the circle passes through the point (11, -9). Radius of the circle is 5√2 cm. (3 Marks) (2023)
Radius of the circle is 5√2 cm. ∴ Distance between centre (2a, a – 7) and (11, – 9 ) = radius of circle.
Q6: In what ratio, does the x-axis divide the line segment joining the points A(3, 6) and B(-12, -3) ? (1 Mark) (2023) (a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) 2 : 1
Ans: (d) Let the point on the x-axis be (x, 0) which divides the line segment joining the points A(3, 6) and B(-12, -3) in the ratio k : 1
Using section formula, we have
(x, 0) = (-12)k + 3(1)k + 1 , (-3)k + 6(1)k + 1
⇒ -3k + 6k + 1 = 0
⇒ -3k + 6 = 0
⇒ k = 2
Hence, the required ratio is 2 : 1.
Q7: Case Study: Jagdish has a Field which is in the shape of a right angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O. (4/5/6 Marks) (CBSE 2023)
Based on the above information, answer the following questions: (i) Taking O as origin, coordinates of P are (-200, 0) and of Q are (200, 0). PQRS being a square, what are the coordinates of R and S? (ii) (a) What is the area of square PQRS? OR (b) What is the length of diagonal PR in square PQRS? (iii) If S divides CA in the ratio K: 1, what is the value of K, where point A is (200, 800)?
The coordinates of R and S are (200, 400) and (-200, 400). (ii) (a) The length PQ = 200 + 200 = 400 units. Area of square PQRS = 400 x 400 = 160000 sq. units.
OR
(b) Length of diagonal PR = √2 x length of side = 400√2 units. (iii) Here, Using section formula, we have ∴ Kx2 + x1K + 1 , Ky2 + y1K + 1 = (-200, 400)
Ans: (a) Given, the equation of line is 4x- 3y = 9. Putting x = 0, we get 4x(0) – 3y = 9 ⇒ y = -3 So, the line 4x – 3y = 9 intersects the y-axis at (0, -3).
Q2: The point on x-axis equidistant from the points P(5, 0) and Q(-1, 0) is (2022) (a) (2, 0) (b) (-2, 0) (c) (3, 0) (d) (2, 2)
Ans: (a) Let coordinates of the point on the x-axis be R (x, 0).
√((x – 5)² + (0 – 0)²) = √((x + 1)² + (0 – 0)²)
Simplify:
(x – 5)² = (x + 1)²
Expand:
x² – 10x + 25 = x² + 2x + 1
Solve for x:
-10x + 25 = 2x + 1
⇒ -12x = -24
⇒ x = 2
So, the point is (2, 0).
Q3: The x-coordinate of a point P is twice its y-coordinate. If P is equidistant front Q(2, -5) and R(-3, 6), then the coordinates of P are (2022) (a) (8, 16) (b) (10, 20) (c) (20, 10) (d) (16, 8)
Ans: (d) Let coordinate of point P= t So, .(x-coordinate of point P = 2t ∴ Point is P (2t, t). Given, PQ = RP ⇒ PQ2 = RP2 ⇒ (2t – 2)2 + (t + 5)2 = (2t + 3)2 + (t – 6)2 [By distance formula] ⇒ 4t2 – 8t + 4 + t2 + 10t + 25 = 4t2+ 12t + 9 + t2– 12t + 36 ⇒ 2t = 16 t = 8 P = 2t = 2 x 8 = 16 Coordinates of P are (16, 8).
Q4: The ratio in which the point (-4, 6) divides the line segment joining the points A(-6, 10) and B(3, -8) is (2022) (a) 2 : 5 (b) 7 : 2 (c) 2 : 7 (d) 5 : 2
Ans: (c) Let point P(-4, 6) divides the line segment AB in the ratio m1: m2.
By section formula, we have
(-4,6) = 3m1 – 6m2m1 + m2 , -8m1 + 10m2m1 + m2
Now, -4 = 3m1 – 6m2m1 + m2
⇒ 3m1 – 6m2 = -4m1 – 4m2
⇒ 7m1 = 2m2 ∴ m1 : m2 = 2:7
Putting the value of m1 : m2 in the y-coordinate, we get
-8m1 +10m1 + m2 = -8 × 27 + 10
-167 + 10 = 6
Hence, the required ratio is 2:7.
Q5: Case Study: Shivani is an interior decorator. To design her own living room, she designed wail shelves. The graph of intersecting wail shelves is given below: (2022)
Based on the above information, answer the following questions: (i) If O is the origin, then what are the coordinates of S? (a) (-6, -4) (b) (6, 4) (c) (-6, 4) (d) (6, -4)
Ans: (d) Coordinates of A are (-2, 4) and coordinates of C are (4, -4). Let (x, 0) divides the line segment joining the points A and C in the ratio m1 : m2 By section formula, we have
(x, 0) = 4m1 – 2m2m1 + m2 , -4m1 + 4m2m1 + m2
Now, 0 = -4m1 + 4m2m1 + m2
⇒ -4m1 + 4m2 = 0
⇒ m1 : m2 = 1:1
(iv) The distance between the points P and G is (a) 16 units (b) 3√74 units (c) 2√74 units (d) √74 units
Ans: (b) Coordinates of vertices of rectangle IJKL are respectively I(2, -2), J(2, -6), K(8, -6),L(8, -2).
Previous Year Questions 2021
Q1: Case Study : Students of a school are standing in rows and columns in their school playground to celebrate their annual sports day. A, B, C and D are the positions of four students as shown in the figure. (2021)
Based on the above, answer the following questions: (i) The figure formed by the four points A, B, C and D is a (a) square (b) parallelogram (c) rhombus (d) quadrilateral
Ans: Let the given points be A(7, 10), B(-2, 5) and C(3, – 4]. Using distance Formula, we have Also, AB2 + BC2 = 106 + 106 = 212 = AC2 So. ABC is an isosceles right angled triangle with ∠B = 90°.
Q5: The point on the x-axis which is equidistant from (-4, 0) and (10, 0) is (2020) (a) (7, 0) (b) (5, 0) (c) (0, 0) (d) (3, 0)
To find the point on the x-axis equidistant from (-4, 0) and (10, 0), let the point be (x, 0).
Using the distance formula:
√( (x + 4)² ) = √( (x – 10)² )
Square both sides:
(x + 4)² = (x – 10)²
Expand:
x² + 8x + 16 = x² – 20x + 100
Cancel x² and simplify:
8x + 16 = -20x + 100
28x = 84
x = 3
Final Answer:
The point is (3,0).
Q6: If the point P(k, 0) divides the line segment joining the points A(2, -2) and B(-7, 4) in the ratio 1:2 then the value of k is (2020) (a) 1 (b) 2 (c) -2 (d) -1
Ans: Let coordinates of the point A be (x, y) and O is the mid point of AB. By using mid-point formula, we have ⇒ -4 = x + 3 and 4 = y + 4 ⇒ x = -7 and y = 0 ∴ Coordinates of A are (-7, 0).
Q4: In what ratio is the line segment joining the points P(3, -6) and Q(5, 3) divided by x-axis? (2019)
Ans: Let the point R(x, 0) on x-axis divides the line segment PQ in the ratio k: 1. ∴ By section formula, we have ∴ Required ratio is 2 : 1 .
Q5: Find the ratio in which the segment joining the points (1, -3) and (4, 5) is divided by x-axis? Also find the coordinates of this point on x-axis. (2019)
Ans: Let the coordinates of A be (x, y). Here, 0(3, – 1] is the mid point of AB. By using mid point formula, we have⇒ x = 4, y = – 8∴ Coordinates of A are (4, – 8).
Q8: The line segment joining the points A(2, 1) and B(5, -8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x – y + k = 0, find the value of k. (2019)
Ans: Let P(x1, y1) and Q(x2, y2) are the points of trisection of line segment AB. ∴ AP = PQ = QB Now. point P divides AB internally in the ratio 1 : 2 ∴ By section formula, we have Since, point P(3, -2) lies on the line 2x – y + k = 0 ⇒ 6 + 2 + k = 0 ⇒ k = – 8
Q9: Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection. (2019)
Ans: Let point P(x1, y1) divides the line segment joining the points A(-2, -5) and B(6, 3) in the ratio k: 1 ∴ Coordinates of P are The point P lies on line x – 3y = 0 ∴ Required ratio is 13 : 3.Now, coordinates of P are
Q10: In what ratio does the point P(-4, y) divide the line segment joining the points A(-6, 10) and B(3, -8) Hence find the value of y. (2019)
Ans: Let the point P(-4, y) divides the line segment joining the points A and B in the ration k: 1 ∴ By section formula, coordinates of P are ∴ Required ratio is 2: 7.
Now,
Also read: Important Definitions & Formulas: Coordinate Geometry
Previous Year Questions 2017
Q1: If two adjacent vertices of a parallelogram are (3, 2) and (–1, 0) and the diagonals intersect at (2, –5), then find the coordinates of the other two vertices. (CBSE 2017)
Ans: The given line segment is A(3, – 2) and B(–3, –4). Here, C(x, y) and C'(x’, y’) are the points of trisection of AB. Then, AC : CB = 1 : 2 and AC’ : C’B = 2 : 1
Ans: Given an equilateral triangle ABC of side 3 units. Also, coordinates of vertex A are (2, 0). Let the coordinates of B = (x, 0) and C = (x’, y’). Then, using the distance formula,
On squaring both sides, we get 9 = (x – 2)2 x2 + 4 – 4x = 9 x2 – 4x – 5 = 0 x2 – 5x + x – 5 = 0 (x – 5) (x + 1) = 0 x = 5, – 1 But x = – 1 (Neglected , since it lies on positive x-axis) Coordinates of B are (5, 0). Now, AC = BC [Q Sides of an equilateral triangle are equal] or AC2 = BC2 By using the distance formula, √(x’ – 2)2 + (y’ – 0)2 = √(x’ – 5)2 + (y’ – 0)2 x’ 2 + 4 – 4x’ + y’ 2 = x’ 2 + 25 – 10x’ + y’ 2 6x’ = 21 x’ = 7 / 2
Also, AC = 3 units. (given)
√((x – 2)² + (y’ – 0)²) = 3
On squaring both sides:
x² + 4 – 4x + y’² = 9
494 + 4 – 4 × 72 + y’² = 9 [∵ x’ = 72]
⇒ y’² = 9 – 94 = 274
⇒ y’ = √274 = ± 3√32
But, C lies in the first quadrant.
y’ = 3√32
Coordinates of C are ( 72 , 3√32 )
Hence, the coordinates of B and C are (5, 0) and ( 72 , 3√32 ) respectively.
Q4: Show that ∆ABC, where A(–2, 0), B(2, 0), C(0, 2) and ∆PQR where P(–4, 0), Q(4, 0), R(0, 4) are similar triangles. (CBSE 2017)
Ans: Let ABC be an isosceles triangle with length of base BC = a ⇒ 16a = 32 × 6 ⇒ a = 12 cm ∴ BC = 12 cm, AB = 10 cm, AC= 10 cm
Draw AD ⊥ BC so that BD = DC = 12/2 = 6 cm ln Δ ABD, AD2 = AB2 – BD2 (By Pythagoras theorem) ⇒ AD2 = 102 – 62 = 100 – 36 ⇒ AD2 = 64 ⇒ AD = 8 cm So, height of ΔABD, h = 8 cm
Q2: The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the lengths of other two sides of the triangle.
Ans: Let ABC be a right angled triangle. Since, the perimeter the right triangle is 60 cm, AB + BC+ CA = 60 ⇒ AB + BC+ 25 = 60 {∵ AC = 25 cm} ⇒ AB + BC = 35 … (i) ln ΔABC, AC2 = AB2 + BC2 {By using Pythagoras theorem} Note that (AB + BC)2 = AB2 + 2 (AB) (BC) + BC2. ⇒ (25)2 = (AB + BC)2 – 2(AB)(BC) ⇒ 2(AB)(BC) = (35)2 – (25)2 {∵ Using(i)} ⇒ (AB)(BC) = 300 … (ii) Now, (AB – BC)2 = (AB + BC)2 – 4(AB)(BC) ⇒ (AB – BC)2 = (35)2 – 4 × 300 {∵ Using(i)} ⇒ AB – BC = √25 = 5 …(iii)
Adding eqn (i) & (iii), we get 2AB = 40 ⇒ AB = 20 cm From (i), AB+ BC= 35 ⇒ BC = 35 – 20 = 15 cm Hence, lengths of other two sides of the triangle are 20 cm and 15cm.
Q3: In the adjoining figure, PQ||XY||BC, AP = 2 cm, PX = 1.5 cm and BX = 4 cm. If QY = 0.75 cm, then AQ + CY = (a) 6 cm (b) 4.5 cm (c) 3 cm (d) 5.25 cm
Ans: (c) In ΔABC, DE||BC By Basic Proportionality Theorem,
Q6: If a line drawn parallel to one side of triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to third side. State and prove the converse of the above statement.
Ans: Statement: If a line is drawn parallel to one side of a triangle, then it divides the other two sides in the same ratio. Given: In ΔABC, ∵ IIBC, which intersects AB and AC at ‘D’ and ‘E’ respectively. To prove: AD/DB = AE/EC Construction: Draw OM ⊥ AC and EN ⊥ AB. Also, join B to E and C to D.
Q7: If in two triangles ΔDEF and ΔPQR, ∠D = Q and ∠R = E, then which of the following is not true? (a) DE/QR = DF/PQ (b) EF/PR = DF/PQ (c) EF/RP = DE/QR (d) DE/PQ = EF/RP
Ans: (c) In ΔABC and ΔDAC ∠BAC = ∠ADC = 90° ∠C = ∠C (Common) So, ΔABC ∼ ΔADC (By AA similarity) ⇒ x2 = 16 × 4 = 64 ⇒ x = 8 cm
Q10: In triangles ABC and DEF, ∠B = ∠E , ∠F = ∠C and AB = 3DE. Then, the two triangles are: (a) congruent but not similar (b) congruent as well as similar (c) neither congruent nor similar (d) similar but not congruent
Ans: In ΔAPC and ΔBQC, we have ∠PAC = ∠QBC (Each 90°) ∠ACP = ∠BCQ (common) ∴ ΔAPC ~ ΔBQC (By AA similarity) ⇒ AP/BQ = AC/BC ⇒ x/y = AC/BC …(i) Again, in ΔACR and ΔABQ, we have ∠ACR = ∠ABQ (Each 90°) ∠CAR = ∠BAQ (Common) ∴ ΔACR ~ ΔABQ (By AA similarity) ⇒ AC/AB = CR/BQ ⇒ AC/AB = z/y …(ii) From (i) and (ii), we have
Hence, proved.
Q15: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR. Show that ΔABC ~ ΔPQR.
Ans: Given, ΔABC and ΔPQR in which AD and PM are the medians such that ∴ ΔABD ~ ΔPQM [By SSS similarity] ⇒ ∠B = ∠Q Now, in ΔABC and ΔPQR, AB/PQ = BC/QR [Given] ∠B = ∠Q (Proved above) ∴ ΔABC ~ ΔPQR (By SAS similarity)
Q16: The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that DF × EF = FB × FA.
Given, In ΔABC, AD ⊥ BC and AD² = BD × DC ⇒ AD × AD = BD × DC ⇒ AD/BD = DC/AD In ΔABD and ΔCAD, AD/BD = DC/AD and ∠ADB = ∠ADC = 90° ⇒ ΔABD ~ ΔCAD ⇒ ∠BAD = ∠ACD and ∠ABD = ∠CAD Now, in ΔABD, ∠BAD + ∠CAD = 90° ⇒ ∠BAD + ∠BAD = 90° [∵ ∠ABD = ∠CAD] ⇒ ∠BAC = 90°
Q18: In the adjoining figure, ΔCAB is a right triangle, right angled at A and AD ⊥ BC. Prove that ΔADB ~ ΔCDA. Further, if BC = 10 cm and CD = 2 cm, find the length of AD.
In ΔADB and ΔCDA, ∠ADB = ∠CDA [Each 90°] ∠DAB = ∠DCA [Each (90° – B)] ∴ ΔADB ~ ΔCDA [By AA similarity] Also, AD/CD = BD/AD ⇒ AD² = BD × CD ⇒ AD² = 8 × 2 [∵ BO = BC – CD] ⇒ AD = 4 cm
Previous Year Questions 2024
Q1: In ΔABC, DE || BC (as shown in the figure). If AD = 2 cm, BD = 3 cm, BC = 7.5 cm, then the length of DE (in cm) is: (CBSE 2024) (a) 2.5 (b) 3 (c) 5 (d) 6
Ans: Let the measures of the two angles be x° and y° (x > y). Given: x + y = 180° …….. (i) (Sum of Supplementary angles are 180°) Also, x – y = 18° …….. (ii) (Given)
Adding both the equations, we get: 2x = 198° x = 99° By putting value of x in equation (i) we get, 99° + y = 180° y = 180° – 99° = 81° Hence, y = 81° and x = 99°.
Q4: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio. (2024)
∠AED = ∠ACB (Corresponding angles) ∠ADE = ∠ABC (Corresponding angles) ∠EAD is common to both the triangles
∴ ΔAED ~ ΔACB by AAA similarity
⇒ AC / AE = AB / AD ( ∴ corresponding sides of similar triangles are proportional) ……………(i) ⇒ Also, AC = AE + EC and AB = AD + BD Putting these values in (i), we get
⇒ (AE + EC) / AE = (AD + BD) / AD
⇒ EC / AE = BD / AD
Hence proved.
Q5: Sides AB and BC and median AD of a ΔABC are respectively proportional to sides PQ and QR and median PM of ΔPQR. Show that ΔABC ∼ ΔPQR. (2024)
Q7: In the given figure, PA, QB, and RC are each perpendicular to AC. If AP = x, BQ = y, and CR = z, then prove that (1/x) + (1/z) = (1/y). (CBSE 2024)
Given: AZ = 3 cm, ZC = 2 cm, BM = 3 cm, and MC = 5 cm.
Proof:
We can calculate the lengths of AC and BC:
AC = AZ + ZC = 3 + 2 = 5 cm,
BC = BM + MC = 3 + 5 = 8 cm.
Next, in triangles AXY and ABM, we know that:
∠AXY = ∠ABM (corresponding angles since XZ is parallel to BC),
∠XAY = ∠BAM (common angle).
∴ By the AA similarity criterion (Angle-Angle similarity): ΔAXY ~ ΔABM.
Since these triangles are similar, their corresponding sides are proportional: AX / AB = XY / BM = AZ / AC.
From the proportionality, we substitute the known values: AX / AB = AZ / AC, which simplifies to: XY / BM = AZ / AC.
Substituting the known values of AZ = 3 cm and AC = 5 cm, we get: XY / 3 = 3 / 5.
Now, solving for XY, we multiply both sides by 3: XY = (3 × 3) / 5 = 9 / 5 = 1.8 cm.
Thus, the length of XY is 1.8 cm.
Q3: Assertion (A): The perimeter of ΔABC is a rational number. Reason (R): The sum of the squares of two rational numbers is always rational. (2023)
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false but Reason (R) is true.
Q3: A vertical pole of length 19 m casts a shadow 57 m long on the ground and at the same time a tower casts a shadow 51m long. The height of the tower is (2022) (a) 171m (b) 13 m (c) 17 m (d) 117 m
Ans: 13 m Let Aman starts from A point and continues 5 m towards west and readied at B point, from which he goes 12 m towards North reached at C point finally. In ΔABC, we have
AC2 = AB2 + BC2………… [By Pythagoras theorem] AC2 = 52 + 122 AC2 = 25 + 144 = 169 AC = 13m So, Aman is 13 m away from his starting point.
Ans: Given, PQIIBC PQ = 3 cm, BC = 9 cm and AC = 7.5 cm Since. PQ || BC ∴ ∠APQ = ∠ABC (Corresponding angles are equal) Now, in ΔAPQ and ΔABC ∠APQ =∠ABC (Corresponding angles are equal) ∠A = ∠A (Common) ΔAPQ ∼ ΔABC (AA similarity)
∴ APAB = AQAC = PQBC
⇒ AQAC = 39
⇒ AQ7.5 = 13
⇒ AQ = 7.53 = 2.5 cm
Q3: In the given figure, EA/EC = EB/ED, prove that ΔEAB ~ ΔECD. (CBSE 2020)
Acc to Converse of basic proportionality theorem, It states that if any line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
It is given that
Since, EA/EC = EB/ED ∠1 = ∠2 [Vertically opposite angles] So, by SAS similarity criterion ΔEAB ~ ΔECD Hence, proved.
Practice Test: Triangles
Start Test
Previous Year Questions 2019
Q1: In the figure, GC||BD and GE||BF. If AC = 3cm and CD = 7 cm, then find the value of AE / AF. (2019)
Ans: (b) Let the terms in AP be a – d, a and a + d. ∴ a – d + a + a + d = 30 ⇒ 3a = 30 ⇒ a = 10 So, middle term is 10.
Q2: Case Study: A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.
Based on the given information answer the following questions: (i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. (ii) Determine the distance of the 8th round. (iii) (a) Find the total distance run after completing all 10 rounds. OR (iii) (b) If a runner completes only the first 6 rounds, what is the total distance run by the runner?
Ans: (i) The Arithmetic progression so formed will be 300, 350, 400, 450, ….. up to 10 terms. ∴ a4 = 450; a5 = 500; a6 = 550 (ii) Here, a = 300, d = 50, n = 8 ∴ an = a + (n – 1)d ⇒ a8 = 300 + (8 – 1)50 = 300 + 7(50) = 300 + 350 = 650 (iii)(a) We have, a = 300, d = 50, n = 10, then total distance run after completing all 10 rounds is given by
= 5(600 + 450) = 5(1050) = 5250 metres (iii) (b) We have, a = 300, d = 50, n = 6 If a runner completes only first 6 rounds, then total distance run is given by n
= 3(850) = 2550 metres.
Q3: Case Study: In order to organise Annual Sports Day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below: The length of innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane. Based on given information, answer the following questions, using concept of Arithmetic Progression. (i) What is the length of the 6th lane? (ii) How long is the 8th lane than that of 4th lane? (iii) (a) While practising for a race, a student took one round each in first six lanes. Find the total distance covered by the student. OR (iii) (b) A student took one round each in lane 4 to lane 8. Find the total distance covered by the student.
Ans: (i) We have, a= 400 m, d = 7.6 m ∴ Length of 6th lane(a6) = a + (6 – 1)d =a + 5d = 400 + 5(7.6) = 438 m (ii) Length of 8th lane, a8 = a + 7d = 400 + 7(7.6) = 453.2 m Length of 4th lane, a4 = a+ 3d = 400 + 3(7.6) = 422.8 m ∴ Difference = (453.2 – 422. 8) m = 30.4 m (iii) (a) Total distance covered by a student in first six lanes, is given by, 6
[∵ a = 400, d = 7.6] = 3(800 + 38) = 2514 m OR (iii) (b) Total distance covered by a student from lane 4 to lane 8, is, S8 – S3
Q4: An AP consists of ‘n ‘ terms whose nth term is 4 and the common difference is 2. If the sum of ‘n’ terms of AP is -14, then find ‘n’. Also, find the sum of the first 20 terms.
Ans: Let a be the first term and d be the common difference and nth term be an of A.P. ∴ an = a + [n – 1] d ⇒ 4 = a + (n – 1)2 {∵ an = 4, d = 2} ⇒ a + 2n = 6 …(i) Now, sum of ‘n’ terms of A.P., Sn = – 14
Q5: The sum of the first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is 1 : 3. Calculate the first and the thirteenth terms of the AP.
Ans: Given, S6 = 42 and a10/a30 = 1/3 As, an = a + (n – 1)d, where a is the first term, d is the common difference, n is the number of terms. ⇒ 3a + 27d =a + 29d ⇒ 2a = 2d ⇒ a = d … (i) Now, sum of first six terms of an A.P. (S6) = 42
⇒ 3[2a + 5d] = 42 [Using (i)] ⇒ 3 × 7a = 42 ⇒ a = 2 So, a = d = 2 ∴ a13 = a + (13 – 1)d = 2 + 12 x 2 = 26 Hence, first term is 2 and thirteenth term is 26.
Q6: The sum of the third term and the seventh term of an AP is 6 and their product is 8. Find the Q sum of the first sixteen terms of the AP.
Ans: Let a be the first term and d be the common difference of an A.P. Given, a3 + a7 = 6 ⇒ a + 2d + a + 6d = 6 ⇒ 2a + 8d = 6 ⇒ a + 4d = 3 …(i) Also, a3 a7 = 8 ⇒ (a + 2d) (a + 6d) = 8 ⇒ ( 3 – 4d + 2d) (3 – 4d + 6d) = 8 (using (i)) ⇒ ( 3 – 2d) (3 + 2d) = 8 ⇒ 9 – 4d2 = 8
Q7: The minimum age of children eligible to participate in a painting competition is 8 years. It is observed that the age of the youngest boy was 8 years and the ages of the participants, when seated in order of age, have a common difference of 4 months. If the sum of the ages of all the participants is 168 years, find the age of the eldest participant in the painting competition.
Ans: Given, first term, a = 8 years Common difference, d = 4 months = 1/3 year Let there are ‘n’ number of participants. ∵ Sn = 168 years
Previous Year Questions 2024
Q1: The ratio of the 10th term to its 30th term of an A.P. is 1 : 3 and the sum of its first six terms is 42. Find the first term and the common difference of A.P. (CBSE 2024)
Ans: We have First term, a1 = – 14 Fifth term, a5 = 2 Last term, an = 62 Let d be the common difference and n be the number of terms. ∵ a5 = 2 ⇒ -14 +(5 – 1)d = 2 ⇒ 4d = 16 ⇒ d =4 Now, an = 62 ⇒ -14 + (n – 1)4 = 62 ⇒ 4n – 4 = 76 ⇒ 4n = 80 ⇒ n= 20 There are 20 terms in A.P.
Q4: Which term of the A.P. : 65, 61, 57, 53, _____ is the first negative term? (CBSE 2023)
Ans: Given, A.P. is 65, 61, 57, 53,….. Here, first term a = 65 and common difference, d = -4 Let the nth term is negative. Last term, an = a + (n – 1) = 65 + (n – 1)(-4) = 65 – 4n + 4 = 69 – 4n, which will be negative when n = 18 So, 18th term is the first negative term.
Q5: Assertion: a, b, c are in AP if and only if 2b = a + c Reason: The sum of the first n odd natural numbers is n2. (CBSE 2023) (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A). (c) Assertion (A) is true but Reason (R) is false. (d) Assertion (A) is false but Reason (R) is true.
Ans: (b) Sol: Since, a, b ,c are in A.P. then b – a = c -b ⇒ 2b = a + c First n odd natural number be 1, 3, 5 ….. (2n – 1). which form an A.P. with a = 1 and d = 2 Sum of first n odd natural number = n/2[2a + (n -1)d] = n/2 [2 + (n – 1)2] = n2 Hence, assertion and reason are true but reason is not the correct explanation of assertion.
Q6: The sum of the first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term. (2023)
Ans: Here, a = 15 and S15 = 750 ∵ Sn = n/2[2a + (n -1)d] ∴ S15 = 15/2 [2 x 15 + (15 -1)d] = 750 ⇒ 15(15 + 7d) = 750 ⇒ 15 + 7d = 50 ⇒ 7d = 35 ⇒ d = 5 Now, 20th term = a + (n – 1)d = 15 + (20 – 1) 5 = 15 + 95 = 110
Q7: Rohan repays his total loan of Rs. 1,18,000 by paying every month starting with the first installment of Rs. 1,000. If he increases the installment by Rs. 100 every month. what amount will be paid by him in the 30th installment? What amount of loan has he paid after the 30th installment? (2023)
Ans: Total amount of loan Rohan takes = Rs. 1,18,000 First installment paid by Rohan = Rs. 1000 Second instalment paid by Rohan = 1000 + 100 = Rs. 1100 Third installment paid by Rohan = 1100 + 100 = Rs. 1200 and so on. Let its 30th installment be n. Thus, we have 1000,1100,1200, which forms an A.P. with first term (a) = 1000 and common difference (d) = 1100 – 1000 = 100 nth term of an A.P. an= a + (n – 1)d For 30th installment, a30 = a + (30 – 1)d = 1000 + (29) 100 = 1000 + 2900 = 3900 So Rs. 3900 will be paid by Rohan in the 30th installment. Now, we have a = 1000, last term (l)= 3900 Sum of 30 installment, S30 = 30/2[a + 1] ⇒ S30 = 15(1000 + 3900) = Rs. 73500 Total amount he still have to pay after the 30th installment = (Amount of loan) – (Sum of 30 installments) = Rs. 1,18,000 – Rs. 73,500 = Rs. 44,500 Hence, Rs. 44,500 still has to be paid after the 30th installment.
Q8: The ratio of the 11th term to the 18th term of an AP is 2:3. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms. (2023)
Ans: Let a and d be the first term and common difference of an AP. Given that, a11 : a18 = 2 : 3
⇒ a + 10da + 17d = 23
⇒ 3a + 30d = 2a + 34d ⇒ a = 4d …(i) Now, a5 = a + 4d = 4d + 4d = 8d [from Eq.(i)] And a21 = a + 20d = 4d + 20d = 24d [from Eq. (i)] a5 : a21 = 8d : 24d = 1 : 3 Now, sum of the first five terms, S5 = 5/2 [2a + (5−1)d] = 5/2 [2(4d) + 4d] [from Eq.(i)] = 5/2 (8d + 4d) = 5/2 × 12d = 30d And, sum of the first 21 terms, S21 = 21/2 [2a + (21−1)d] = 21/2 [2(4d) + 20d]= 21/2 × 28 d = 294 d from Eq..(i)] So, ratio of the sum of the first five terms to the sum of the first 21 terms is, S5 : S21 = 30d : 294d = 5 : 49
Q9: 250 logs are stacked in the following manner: 22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row?
Ans: Let there be n rows to pile of 250 logs Here, the bottom row has 22 logs and in next row, 1 log reduces It means, we get an AP 22, 21, 20, 19, ………………… n with first term or a = 22 and d = -1 Now, we know that total logs are 250 or we can say that, Sn =250 Since sum of n terms of an A.P. Sn = n/2 (2a + (n-1) d) = 250 Therefore, n/2 (2 x 22 + (n-1) x (-1)) or 500 = n (44 – (n-1)) 500 = n (45- n) n2 – 45 n + 500 = 0 n2 – 25n – 20n + 500 = 0 n(n – 25) – 20(n – 25) = 0 By solving this, we get (n-20) (n-25) = 0 Since, there are 22 logs in first row and in next row, 1 log reduces, then we can not have more than 22 terms ∴ n ≠ 25 and n = 20 Means, 20th row is the top row of the pile Now let’s find out number of logs in 20th row We know that value of nth term of an A.P. = a + (n-1) d n20 = [22 + (20-1) (-1)] =(22-19) = 3 Therefore, there are 3 logs in the top row.
Q10: The next term of the A.P.: √7, √28, √63 is: (a) √70 (b) √80 (c) √97 (d) √112 (CBSE 2023)
Ans: Since a, 7, b, 23 are in A.P. ∴ Common difference is same. ∴ 7 – a = b -7 = 23 – b Taking second and third terms, we get b – 7 = 23 – b ⇒ 2b = 30 ⇒ b = 15 Taking first and second terms, we get ⇒ 7 – a = b – 7 ⇒ 7 – a = 15 – 7 ⇒ 7 – a = 8 ⇒ a = -1 Hence, a = -1, b = 15.
Q2: Find the number of terms of the A.P. : 293, 235, 177,….., 53 (2022)
Ans: Let the first term and common difference of an A.P. be a and d, respectively. Given a3= 5 and a7 = 9 a + (3 – 1 ) d = 5 and a + (7 – 1)d = 9 a + 2d = 5 ————–(i) and a + 6d = 9————–(ii) On subtracting (i) from (ii), we get ⇒ 4d = 4 ⇒ d = 1 From (i), a + 2(1) = 5 ⇒ a + 2 = 5 ⇒ a = 3 So. required A.P. is a, a + d, a + 2d, a + 3d…… i.e. 3, 3 +1, 3 + 2(1), 3 + 3(1), ….., i.e.. 3, 4, 5, 6, …..
Previous Year Questions 2020
Q1: If -5/7, a, 2 are consecutive terms in an Arithmetic Progression, then the value of a’ is (2020) (a) 9/7 (b) 9/14 (c) 19/7 (d) 19/14
Ans: Let a1 = (a – b)2, a2 = (a2 + b2) and a3= (a + b)2 Now. a2 – a1 = (a2 + b2) – (a – b)2 = a2 + b2 – (a2 + b2 – 2ab) = a2 + b2– a2– b2 + 2ab = 2ab Again a3 – a2 = (a + b)2 – (a2 + b2) = a2 + b2 + 2ab – a2 – b2 = 2ab ∴ a2 – a1 = a3 – a2 So, (a – b)2, (a2 + b2) and (a + b)2 are in A.P.
Q5: The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the numbers. (2020)
Ans: Let the four consecutive numbers be (a – 3d), (a – d), (a + d), (a + 3d).
Sum of four numbers = 32 (Given) ⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32 ⇒ 4a = 32 ⇒ a = 8 Also, (a – 3d)(a + 3d)(a – d)(a + d) = 715
⇒ a2 – 9d2a2 – d2 = 715
⇒ 15a2 – 135d2 = 7a2 – 7d2
⇒ 8a2 = 128d2
⇒ d2 = 8a2128 = 8 × 64128
⇒ d = ± 2
If d = 2. then the numbers are (8 – 6), (8 – 2), (8 + 2) and (8 + 6) i.e., 2,6, 10, 14 . If d = -2. then the numbers are (8 + 6), (8 + 2), (8 – 2). (8 – 6) i.e.,14 ,10 ,6 ,2 . Hence, the numbers are 2, 6, 10, 14 or 14, 10, 6, 2.
Q6: Find the sum of the first 100 natural numbers. (CBSE 2020)
Ans: Given, a4 = -15 and a9 = -30 a + 3d = – 15 (i) a + 8d = -30 (ii) On subtracting (ii)from (i), we have -5d = 15 ⇒ d = – 3 Put d = – 3 in (i), we have a + 3(-3)= – 15 ⇒ a – 9 = – 15 ⇒ a = – 6 Now, Sn = n/2 [2a + (n – 1)d] ⇒ S16 = 16/2 [2(-6) + (16 – 1) (-3)] = 8 [2(-6) + (15) (-3)] = 8 [-12 – 45] = -456
Q8: In an A.P. given that the first term (a) = 54. the common difference (d) = -3 and the nth term (an) = 0. Find n and the sum of the first n terms (Sn) of the A.P. (2020)
Ans: (−5) + (−8) + (−11) + … + (−230) . Common difference of the A.P. (d) = a2 – a1 = -8-(-5) = -8+5 = -3 So here, First term (a) = −5 Last term (l) = −230 Common difference (d) = −3 So, here the first step is to find the total number of terms. Let us take the number of terms as n. Now, as we know, an = a + (n-1) d So, for the last term, – 230 = -5 + ( n-1) (-3) – 230 = -5 – 3n + 3 -230 + 2 = -3n – 228/-3 = n n = 76 Now, using the formula for the sum of n terms, we get Sn = 76/2 [2(-5) + (76-1)(-3)] = 38 [-10 + (75)(-3)] =38 (-10-225) = 38(-235) = -8930 Therefore, the sum of the A.P is Sn = -8930
Q10: Show that the sum of all terms of an A.P. whose first term is a, the second term is b and the last term is c is equal to (a + c)(b + c – 2a)2(b – a) . (CBSE 2020)
Ans: Given: first term a1 = a, second term, a2 = b and last term, l = c. So, common difference, d = a2 – a1 = b – a Let this A.P. contains n terms. Then, l = a + (n – 1)d ⇒ c = a + (n – 1) (b – a) ⇒ c – a = (n – 1)(b – a)
Ans: Two-digit numbers which are divisible by 3 are 12, 15, 13….. 99. which forms an A.P. with first term (a) = 12, common difference (d) = 15 – 12 = 3 and last term (l) or nth term = 99 a + (n – 1)d = 99 ⇒ 12 + (n – 1)3 = 99 ⇒ 3n = 99 – 9 ⇒ n = 90/3 ⇒ n = 30
Q5: If the 9th term of an AR is zero, then show that its 29th term is double its 19th term. (2019, 2 Marks)
Ans: We have, first term, a = 3, common difference, d = 15 – 3 =12 nth term of an A.P. is given by an = a + (n – 1)d ∴ a21 = 3 + (20) x 12 = 3 + 240 = 243 Let the rth term of the AP. be 120 more than the 21st term. ⇒ a + (r – 1) d = 243 + 120 ⇒ 3 + (r – 1) 12 = 363 ⇒ (r – 1) 12 = 360 ⇒ r – 1= 30 ⇒ r = 31
Q7: If the 17th term of an A.P. exceeds its 10th term by 7, find the common difference. (2019)
Ans: According to question, a17 – a10 = 7 i.e. a + 16d- (a + 9d) = 7 where a = first term d = common difference ⇒ 7d = 7 ∴ d = 1
Q8: Ramkali would require ₹ 5000 to get her daughter admitted to a school after a year. She saved ₹ 150 in the first month and increased her monthly savings by ₹ 50 every month. Find out if she will be able to arrange the required money after 12 months. Which value is reflected in her efforts? (CBSE 2019, 15)
Ans: The saving in first month is₹ 150. The saving in second month is ₹(150 + 50) = ₹ 200 Similarly, saving goes on increasing every month by ₹ 50. Savings = ₹ 150, ₹ 200, ₹ 250, ₹300,….. Savings forms an A.P. in which first term (a) = 150 and common difference, (d) = 50 Then, total savings for 12 months
S12 = n2 [2a + (n – 1)d]
⇒ S12 = 122 [2 × 150 + (12 – 1) × 50]
⇒ S12 = 6 [300 + 550]
⇒ S12 = 6 × 850 = ₹ 5100
Then, Ramkali would be able to save ₹ 5,100 in 12 months and she needs ₹5,000 to send her daughter to school. Hence, Ramkali would be able to send her daughter to school. Values: Putting efforts to send her daughter to school shows her awareness regarding girls education and educating a child.
Also read: Flashcards: Arithmetic Progressions
Previous Year Questions 2017
Q1: A sum of ₹ 4,250 is to be used to give 10 cash prizes to students of a school for their overall academic performance. If each prize is ₹ 50 less than its preceding prize, find the value of each of the prizes. (CBSE 2017)
Ans: Let the value of first most expensive prize be ₹ a. Then, according to the given condition, prizes are a, a – 50, a – 100, a – 150 ……. The given series forms an A.P., with a common difference of (– 50). Here, first term = a Common difference d = – 50 Number of terms, n = 10 and, sum of 10 terms, S10 = ₹ 4,250
By formula, Sn = n2 [2a + (n – 1)d]
⇒ S10 = 102 [2 × a + (10 – 1) × (-50)]
⇒ 4250 = 5(2a – 450)
⇒ 850 = 2a – 450
⇒ a = 13002 = ₹ 650
Hence, the value of the prizes are: ₹ 650, ₹ 600, ₹ 550, ₹ 500, ₹ 450, ₹ 400, ₹ 350, ₹ 300, ₹ 250, ₹ 200.
Q2: A child puts one five-rupee coin of her savings in the piggy bank on the first day. She increases her savings by one five-rupee coin daily. If the piggy bank can hold 190 coins of five rupees in all, find the number of days she can continue to put the five-rupee coins into it and find the total money she saved. Write your views on the habit of saving. (CBSE 2017)
Ans: Since, child puts ₹ 5 on 1st day, ₹ 10 (2 × 5) on 2nd day, ₹ 15(3 × 5) on 3rd day and so on. Total savings = 190 coins = 190 × 5 = ₹ 950 So, the series of her daily savings is, ₹ 5, ₹ 10, ₹ 15, ….. Clearly, this series is an A.P. So, first term, a = 5 Common difference, d = 5 Sum of total savings, Sn = 950 Let, n be the last day when piggy bank becomes full.
∴ Sn = n2 [2a + (n – 1)d]
⇒ 950 = n2 [2 × 5 + (n – 1) × 5]
⇒ 1900 = n[10 + 5n – 5]
⇒ 1900 = n[5n + 5]
⇒ 5n2 + 5n – 1900 = 0
⇒ n2 + n – 380 = 0
⇒ n2 + 20n – 19n – 380 = 0 (on splitting the middle term)
n(n + 20) – 19(n + 20) = 0 (n – 19) (n + 20) = 0 n = 19 or – 20 But ‘n‘ cannot be negative, hence n = 19. Hence, she continuous the savings for 19 days and saves ₹ 950. Views on habit of saving: (1) Child is developing a very good habit of savings. (2) Consistent saving can create a wonder.
Q3: Write the nth term of the A.P. 1m , 1 + mm , 1 + 2mm, ……. (CBSE 2017)
Q2: Case Study: A garden designer is planning a rectangular lawn that is to be surrounded by a uniform walkway. The total area of the lawn and the walkway is 360 square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are 12 metres by 10 metres. Based on the information given above, answer the following questions: (i) Formulate the quadratic equation representing the total area of the lawn and the walkway, taking width of walkway= x m. (ii) (a) Solve the quadratic equation to find the width of the walkway ‘x’. OR (ii) (b) If the cost of paving the walkway at the rate of ₹ 50 per square metre is ₹ 12,000, calculate the area of the walkway. (iii) Find the perimeter of the lawn.
Ans: (i) Length of rectangular lawn, l = (12 + 2x) m Breadth of rectangular lawn, b = (10 + 2x) m ∴ Area of rectangular lawn = 360 m2 (given) ⇒ (12 + 2x)(10 + 2x) = 360 ⇒ 120 + 24x + 20x + 4x2 = 360 ⇒ 4x2 + 44x + 120 = 360 ⇒ 4x2 + 44x – 240 = 0 ⇒ x2 + 11x – 60 = 0
(ii) (a) Quadratic equation is x2 + 11x – 60 = 0 ⇒ x2 + 15x – 4x – 60 = 0 ⇒ x(x + 15) – 4(x + 15) = 0 ⇒ (x + 15)(x – 4) = 0 ⇒ x = -15 or 4 ⇒ x = 4 ( x can’t be negative) ∴ Width of the walkway= 4 m.
OR (ii) (b) (iii) Length of lawn , l= 12 m, Breadth of lawn, b = 10 m ∴ Perimeter of lawn= 2(l + b) = 2(12 + 10) = 2 x 22 = 44 m
Q3: A student scored a total of 32 marks in class tests in Mathematics and Science. Had he scored 2 marks less in Science and 4 marks more in Mathematics, the product of his marks would have been 253. Find his marks in the two subjects.
Ans: Let the marks scored by a student in Mathematics be x and Science be y. According to question, x + y = 32 ⇒ y = 32 – x …(i) Also, (x + 4)(y – 2) = 253 ⇒ xy – 2x + 4y – 8 = 253 ⇒ xy – 2x + 4y = 261 ⇒ x(32 – x) – 2x + 4(32 – x) = 261 [from(i)] ⇒ 32x – x2 – 2x + 128 – 4x = 261 ⇒ 26x -x2 + 128 = 261 ⇒ x2 – 26x + 133 = 0 ⇒ x2 – 19x – 7x + 133 = 0 ⇒ x(x – 19) – 7(x – 19) = 0 ⇒ (x – 19)(x – 7) = 0 ⇒ x = 7, 19 From (i) when get y = 25, 13 respectively if x = 7, then y = 25 and if x = 19, then y = 13. ∴ Marks in mathematics and science are 7 and 25 or 19 and 13 respectively.
Q4: There is a circular park of diameter 65 mas shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.
Ans: (d) We have, bx2 + ax + c = 0 On comparing with Ax2 + Bx + C = 0, we get A = b, B = a and C = c ∴ D = B2 – 4AC = a2 – 4bc [∵ Discriminant = B2 – 4AC]
Q6: Which of the following quadratic equations has real and equal roots? (a) (x + 1)2 = 2x + 1 (b) x2 +x= 0 (c) x2 – 4 = 0 (d) x2+ x + 1= 0
Ans: Given quadratic equation 4x2 + kx + 1 = 0 has real and equal roots. Comparing with ax² + bx + c = 0, we have a = 4, b = k, c = 1 ∴ D = b2 – 4ac = k2 – 4(4)(1) = 0 ⇒ k2 – 16 = 0 ⇒ k² = 16 ⇒ k = ±4
Q8: Find the smallest value of p for which the quadratic equation x2 – 2(p + 1)x + p2 = 0 has real roots. Hence, find the roots of the equation so obtained.
Ans: Given, x2 – 2(p + 1)x + p2 = 0 has real roots. Compare it with ax2 + bx + c = 0, we get a = 1, b = -2(p + 1), c = p2 ∴ For real roots, we have D ≥ 0 (-2(p + 1))2 – 4 × 1 × p² ≥ 0 ⇒ 4(p2 + 1 + 2p) – 4p2 ≥ 0 ⇒ 4p2 + 4 + 8p – 4p2 ≥ 0 ⇒ 8p + 4 ≥ 0 => p ≥ -1/2 ∴ Smallest value of p is -1/2. Put value of ‘p’ in given equation, So, 1/2, 1/2 are the roots.
Previous Year Questions 2024
Q1: In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100km/h, and by doing so, the time of flight is increased by 30 minutes. Find the original duration of the flight. (CBSE 2024)
Ans: Let the speed of aircraft be x km/hr. Time taken to cover 2800 km by speed of x km/hr = 2800/x hrs. New speed is (x – 100) km/hr so time taken to cover 2800 km at the speed of (x – 100) km/hr = 2800x – 100 hrs
Q2: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is found, the fraction. (CBSE 2024)
Ans: (b) 25/16 Given that, Roots of quadratic equation 4x2 − 5x + k = 0 are real and equal On comparing with ax2 + bx + c = 0, We get, a = 4, b = −5 and c = k For real and equal roots, ⇒ b2 − 4ac = 0 ⇒ (−5)2 − 4(4)(k) = 0 ⇒ 16k = 25 ⇒ k = 25/16
Q4: A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor. (CBSE 2024) (i) Assuming the original length of each side of a tile is x units, make a quadratic equation from the above information. (ii) Write the corresponding quadratic equation in standard form. (iii) (a) Find the value of x, the length of the side of a tile by factorisation. OR (b) Solve the quadratic equation for x using the quadratic formula.
Ans: (i) Let the original side length of each tile be x units. The area of the rectangular floor using 200 tiles = 200 x2 unit2 The area with increased side length (each side increased by 1 unit) using 128 tiles = 128 (x + 1)2 unit2 So, the required quadratic equation is: 200x2 = 128 (x + 1 )2 (ii) We have, 200x2 = 128 (x + 1)2 ⇒ 200x2 = 128 (x2 + 2x + 1) ⇒ 200x2 = 128x2 + 256x + 128 ⇒ 72x2 − 256x − 128 = 0 which is the quadratic equation is standard form. (iii) (a) We have, 72x2 − 256x − 128 = 0 or, 9x2 − 32x − 16 = 0 or, 9x2 − 36x + 4x − 16 = 0 or, 9x (x − 4) + 4 (x − 4) = 0 or, (x − 4) (9x + 4) = 0
Since the side cannot be negative, thus x = 4 units. OR (b) We have 9x2 − 32x − 16 = 0 On comparing with ax2 + bx + c = 0, we get a = 9, b = −32 and c = −16 Using quadratic formula,
Q5: If the roots of equation ax2 + bx + c = 0, a ≠ 0 are real and equal, then which of the following relations is true? (CBSE 2024) (a) a = b2/c (b) b2 = ac (c) ac = b2/4 (d) c = b2/c
Ans: (c) ac = b2/4 If the discriminant is equal to zero, i.e., b2 − 4ac = 0 where a, b, c are real numbers and a ≠ 0, then roots of the quadratic equation ax2 + bx + c = 0, are real and equal. Thus, b2 – 4ac = 0 or ac = b2/4
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Previous Year Questions 2023
Q6: Find the sum and product of the roots of the quadratic equation 2x2 – 9x + 4 = 0. (CBSE 2023)
Ans: Let α and β be the roots of given quadratic equation 2x2 – 9x + 4 = 0. Sum of roots = α + β = -b/a = (-9)/2 = 9/2 and Product of roots, αβ = c/a = 4/2 = 2
Q7: Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other. (CBSE 2023)
Ans: Let the first root be α, then the second root will be 6a Sum of roots = -b/a ⇒ a + 6a = 14/p ⇒ 7a = 14/p ⇒ a = 2/p Product of roots = c/a ⇒ a x 6a = 8/p ⇒ 6a2 = 8/p
⇒ 6(2p)² = 8p
⇒ 6 x 4p² = 8p
⇒ p = 6 x 4/8 ⇒ p = 3 Hence, the value of p is 3.
Q8: The least positive value of k for which the quadratic equation 2x2 + kx + 4 = 0 has rational roots is (2023) (a) ±2√2 (b) 4√2 (c) ±2 (d) √2
Ans: Given quadratic equation is 4x2 – 5 = 0 Discriminant, D = b2 – 4ac = 02 – 4(4)(-5) = 80 > 0 Hence, the roots of the given quadratic equation are real and distinct.
Q10: Find the value of ‘p’ for which the quadratic equation px(x – 2) + 6 = 0 has two equal real roots. (2023)
Ans: The given quadratic equation is px(x – 2) + 6 = 0 ⇒ px2 – 2xp +6 = 0 On comparing with ax2+ bx + c = 0, we get a = p, b = -2p and c = 6 Since, the quadratic equations has two equal real roots.
∴ Discriminant D = 0 ⇒ b2 – 4ac = 0 ⇒ (-2p)2 – 4 x p x 6 = 0 ⇒ 4p2 – 24p = 0 ⇒ p2 – 6p = 0 ⇒ p(p – 6) = 0 ⇒ p = 0 or p = 6 But p ≠ 0 as it does not satisfy equation Hence, the value of p is 6.
Q11: Case Study: While designing the school yearbook, a teacher asked the student that the length and width of a particular photo be increased by n units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide. Based on the above information. Answer the following Questions: (i) Write an algebraic equation depicting the above information. (ii) Write the corresponding quadratic equation in standard form. (iii) What should be the new dimensions of the enlarged photo? OR
Can any rational value of x make the new area equal to 220 cm2? (2023)
Ans: Area = 18 x 12 cm = 216 cm2 Length (l) is increased by x cm So, new length =(18 + x ) cm New width = (12 + x) cm
(i) Area of photo after increasing the length and width = (18 + x)(12 + x) = 2 x 18 x 12 i.e., (18 + x) (12 + x) = 432 is the required algebraic equation. (ii)From part (i) we get, (18 + x) (12 + x) = 432 ⇒ 216 + 18x + 12x + x2 = 432 ⇒ x2 + 30x – 216 = 0 (iii) x2 + 30x – 216 = 0 ⇒ x2+ 36x – 6x – 216 = 0 ⇒ x(x+ 36) – 6 (x+ 36) = 0 ⇒ x = 6, -36 -36 is not possible. So, new length = (18 + 6) cm = 24 cm New width = (12 + 6) cm = 18cm So. new dimension = 24cm x 18 cmOR
According to question (18 + x) (12 + x) = 220 ⇒ 216 + 30x + x2 = 220 ⇒ x2 + 30x + 216 – 220 = 0 ⇒ x2 + 30x – 4 = 0 For rational value of x. discriminant (D) must be perfect square. So, D = b2 – 4ac = (30)2 – 4(1)(-4) = 900 + 16 = 916 ∴ 916 is not a perfect square. So, no rational value of x is possible.
Q12: The roots of the equation x2 + 3x – 10 = 0 are: (a)2,–5 (b)–2,5 (c)2,5 (d)–2,–5 (CBSE 2023)
Ans: (a) To find the roots of the quadratic equation x2 + 3x − 10 = 0, we can use the quadratic formula:
For the equation x2 + 3x − 10 = 0: a = 1 b = 3 c = −10 Substitute these values into the formula:
Now, calculate the two roots:
(i) For x = (-3 + 7) / 2 = 4/2 = 2 (ii) For x = (-3 – 7)/2 = -10/2 = -5
The roots of the equation are 2 and -5. So, the correct answer is: (a) 2,−5
Also read: Short Answer Questions: Quadratic Equations
Previous Year Questions 2022
Q13: If the sum of the roots of the quadratic equation ky2 – 11y + (k – 23) = 0 is 13/21 more than the product of the roots, then find the value of k. (2022)
Ans: Given, quadratic equation is ky2 – 11y + (k – 23) = 0 Let the roots of the above quadratic equation be α and β. Now, Sum of roots, α + β = -(-11)/k = 11/k …(i) and Product of roots, αβ = (k-23)/k …(ii) According to the question, α + β = αβ + 13/21
∴ 11k = k – 23k + 1321 … [From equations (i) and (ii)]
Ans: x2 – 2ax – (4b2 – a2) = 0 ⇒ x2 + (2b – a)x – (2b + a)x – (4b2 – a2) = 0 ⇒ x(x + 2b – a) – (2b + a)(x + 2b – a) = 0 ⇒ (x + 2b – a)(x – 2b – a) = 0 ⇒ (x + 2b – a) = 0, (x – 2b – a) = 0 ∴ x = a − 2b, a + 2b
Q15: In the picture given below, one can see a rectangular in-ground swimming pool installed by a family In their backyard. There is a concrete sidewalk around the pool that is width x m. The outside edges of the sidewalk measure 7 m and 12 m. The area of the pool is 36 sq. m.
Based on the information given above, form a quadratic equation in terms of x Find the width of the sidewalk around the pool. (2022)
Ans: Given, width of the sidewalk = x m. Area of the pool = 36 sq.m ∴ Inner length of the pool = (12 – 2x)m Inner width of the pool = (7 – 2x)m ∴ Area of the pool = A = l x b ⇒ 36 = (12 – 2x) x (7 – 2x) ⇒ 36 = 84 – 24x – 14x +4x2 ⇒ 4x2 – 38x + 48 = 0 ⇒ 2x2 – 19x + 24 = 0, is the required quadratic equation. Area of the pool given by quadratic equation is 2x2 – 19x + 24 = 0 ⇒ 2x2 – 16x – 3x + 24 = 0 ⇒ 2x(x – 8) – 3(x – 8) = 0 ⇒ (x – 8)(2x – 3) = 0 ⇒ x = 8 (not possible) Hence, x = 3/2 = 1.5 Width of the sidewalk =1.5m
Q16: The sum of two numbers is 34. If 3 Is subtracted from one number and 2 is added to another, the product of these two numbers becomes 260. Find the numbers. (2022)
Ans: Let one number be x and another number be y. Since, x + y = 34 ⇒ y = 34 – x (i) Now. according to the question. (x – 3) (y + 2) = 260 (ii) Putting the value or y from (i) in (ii), we get ⇒ (x – 3)(34 – x + 2) = 260 ⇒ (x – 3)(36 – x) = 260 ⇒ 36x – x2 – 108 + 3x = 260 ⇒ x2 – 39x +368 = 0 ⇒ x2– 23x – 16x + 368 = 0 ⇒ x(x – 23) – 16(x – 23) = 0 ⇒ (x – 23)(x – 16) =0 ⇒ x = 23 or 16 Hence; when x = 23 from (i), y = 34 – 23 = 11 When x = 16. then y = 34 – 16 = 18 Hence the required numbers are 23 and 11 or 16 and 18.
Q17: The hypotenuse (in cm) of a right-angled triangle is 6 cm more than twice the length of the shortest side. If the length of the third side is 6 cm less than thrice the length of the shortest side, then find the dimensions of the triangle. (2022)
Let ΔABC be the right angle triangle, right angled at B, as shown in the figure. Also, let AB = c cm, BC = a cm and AC = b cm Then, according to the given information, we have b = 6 + 2a …..(i) (Let a be the shortest side) and c = 3a – 6 …(ii) We know that, b2 = c2 + a2 …..(Pythagoras Theorem) Putting the values of b and c from (i) and (ii) in above equation ⇒ (6 + 2a)2 = (3a – 6)2 + a2 …[Using (i) and (ii)] ⇒ 36 + 4a2 + 24a = 9a2 + 36 – 36a + a2 ⇒ 60a = 6a2 ⇒ 6a = 60 …[∵ a cannot be zero] ⇒ a = 10 cm Now, putting value of a in equation (i), we get b = 6 + 2 × 10 = 26 and putting value of a in equation (ii), we get c = 3 × 10 – 6 = 24 Thus, the dimensions of the triangle are 10 cm, 24 cm and 26 cm.
Q18: Solve the quadratic equation: x2 – 2ax + (a2 – b2) = 0 for x. (2022)
Ans: Roots of quadratic equation are given as 2 and – 5. Sum of roots = 2 + (-5) = -3 Product of roots = 2 (-5) = -10 Quadratic equation can he written as x2 – (sum of roots)x + Product of roots = 0 ⇒ x2 + 3x – 10 = 0
Ans: Let the sides of the two squares be x m and y m, where ; x > y. Then, their areas are x2 and y2 and their perimeters are 4x and 4y respectively. By the given condition, x2 + y2 = 544 ——–(i) and 4x – 4y = 32 ⇒ x – y = 8 ⇒ x = y + 8 ———— (ii) Substituting the value of x from (ii) in (i) we get ⇒ (y + 8)2 + y2 = 544 ⇒ y2 + 64 + 16y + y2 = 544 ⇒ 2y2 + 16y – 480 = 0 ⇒ y2 + 8y – 240 = 0 ⇒ y2 + 20y – 12y – 240 = 0 ⇒ y(y + 20) – 12(y + 20) = 0 ⇒ (y – 12) (y + 20) = 0 ⇒ y = 12 (∵ y ≠ – 20 as length cannot be negative) From (ii), x = 12 + 8 = 20 Thus, the sides of the two squares are 20 m and 12 m.
Q23: A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. (2020)
Q26: A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more to cover the same distance. Find the original speed of the train. (CBSE 2020)
Ans: Let the original speed of the train be x km/h. Then, time taken to cover the journey of 480 km , t = 480 / x hours Time taken to cover the journey of 480 km with speed of (x – 8) km/h = 480 / x − 8 hours Now, according to question,
Ans: Let the length of the side of one square be x m and the length of the side of another square be y m. Given, x2 + y2= 157 _(i) and 4x + 4y=68 _(ii) x + y = 17 y = 17 – x _(iii) On putting the value of y in (i), we get x2 + (17 – x)2 = 157 ⇒ x2 + 289 + x2 – 34x = 157 => 2x2 – 34x +132 = 0 ⇒ x2 – 17x + 66 = 0 ⇒ x2 – 11x – 6x + 66 =0 ⇒ x(x – 11) – 6(x – 11) = 0 ⇒ (x – 11) (x – 6) = 0 ⇒ x = 6 or x = 11 On putting the value of x in (iii), we get y = 17 – 6 = 11 or y = 17 – 11 = 6 Hence, the sides of the squares be 11 m and 6 m.
Q1: A system of two linear equations in two variables is inconsistent, if the lines in the graph are: (a) coincident (b) parallel (c) intersecting at one point (d) intersecting at right angles
Ans: We have, x + 3y = 6 3y- 2x = -12 On comparing with general equation, we get a1 = 1, b1 = 3, c1 = -6 a2 = – 2, b2 = 3, c2 = 12 Hence, the given pair of equations is consistent.
Hence, (6, 0) is the solution of given system of equations.
Q3: Solve the following pair of equations algebraically: 101x + 102y = 304 102x + 101y = 305
Ans: We have, 101x + 102y = 304 …(i) 102x + 101y = 305 …(ii) Adding equations (i) and (ii), we get 101x + 102y + 102x + 101y = 304 + 305 ⇒ 203x + 203y = 609 ⇒ x + y = 3 …(iii) Subtracting equation (ii) from (i), we get 101x + 102y – 102x – 101y = 304 – 305 ⇒ y – x = -1 …(iv) Adding equations (iii) and (iv), we get x + y + y – x = 3 + (-1) ⇒ 2y = 2 ⇒ y = 1 Substitute of y’ in (iv), we get 1 – x = -1 ⇒ x=1 + 1 = 2 Thus, the solution is x = 2 and y = 1.
Q4: In a pair of supplementary angles, the greater angle exceeds the smaller by 50°. Express the given situation as a system of linear equations in two variables and hence obtain the measure of each angle.
Ans: Let the greater angle be x and smaller angle be y. Using statement, we have x + y = 180° …(i) x – y = 50° …(ii) Adding equations (i) and (ii), we get x + y + x – y = 180° + 50° ⇒ 2x = 230° ⇒ x = 115° Using equation (i), we get 115° + y = 180° ⇒ y = 180° – 115° = 65° So, greater angle is 115° and smaller angle is 65°.
Q5: A man lent a part of his money at 10% p.a. and the rest at 15% p.a. His income at the end of the year is ₹1,900. If he had interchanged the rate of interest on the two sums, he would have earned ₹200 more. Find the amount lent in both cases.
Ans: Let the amount lent at 10% p.a. = ₹x Let the amount lent at 15% p.a. = ₹y According to question, Adding (i) and (ii), we get 25x + 25y = 400000 ⇒ x + y = 16000 …(iii) Subtracting eqn. (i) from (ii), we get 5x – 5y = 2000 ⇒ x – y = 4000 …(iv) adding eqn. (iii) and (iv), we get 2x = 20000 ⇒ x = 10000 Put value of x in eqn. (iii), we get y = 6000 Hence, amount lent at 10% p.a. is ₹10000 and amount lent at 15% p.a. is ₹6000.
Q6: Vijay invested certain amounts of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. He received ₹1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received ₹ 20 more as annual interest. How much money did he invest in each scheme?
Ans: Let the amount invested in scheme A be ₹x and in scheme B be ₹y respectively. According to question,
⇒ 8x + 9y = 186000 ….(i)
⇒ 9x + 8y = 188000 ….(ii) On solving equations (i) and (ii), we get 17y = 170000 ⇒ y = 10000 Substituting the value of y in equation (i), we get 8x + 9(10000) = 186000 ⇒ 8x = 186000 – 90000 ⇒ 8x = 96000 ⇒ x = 96000/8 ⇒ x = 12000 ∴ Vijay invested ₹12000 in scheme A and ₹10000 in scheme B.
Q7: A two-digit number is such that the product of its digits is 12. When 36 is added to this number, the digits interchange their places. Find the number.
Ans: Let the tens digit of a number be a and ones digit be b, then two digit number can be written as 10a + b Given, ab = 12 …(i) Also, 10a + b + 36 = 10b + a ⇒ 9a – 9b + 36 = 0 ⇒ a – b + 4 = 0 ⇒ a – b = -4 …(ii) Squaring on both sides, we get (a – b)2 = 16 ⇒ (a + b)2 – 4ab = 16 ⇒ (a + b)2 = 16 + 4(12) ⇒ a + b = ±8 [∴ Using (i)] When a + b = 8, then a = 2, b = 6 [Using (iii)] When a + b = -8, then a = -6, b = -2 [Rejected] Number = 10a + b = 10(2) + 6 = 26
Q8: If x = 1 and y = 2 is a solution of the pair of linear equations 2x – 3y + a= 0 and 2x + 3y – b = 0, then: (a) a = 2b (b) 2a = b (c) a + 2b = 0 (d) 2a + b = 0
Ans: (b) We have, 2x – 3y + a = 0 ..(i) Put x = 1 and y = 2 in (i), we get ∴ 2 – 6 + a = 0 ⇒ a = 4 Put x = 1 and y = 2 in (ii), we get Also, given 2x + 3y – b = 0 ∴ 2 + 6 – b = 0 ⇒ b = 8 ⇒ b = 2 * 4 ⇒ b = 2a [From (i)]
Q9: The value of ‘k’ for which the system of linear equations 6x + y = 3k and 36x + 6y = 3 have infinitely many solution is: (a) 6 (b) 1/6 (c) 1/2 (d) 1/3
Ans: (a) We have, 2x + 1 = 0 and 3y – 5 = 0 ⇒ x = -1/2 and y = 5/3 ∴ Given system of equations has a unique solution.
Previous Year Questions 2024
Q1: The pair of linear equations x + 2y + 5 = 0 and – 3x = 6y – 1 has. (CBSE 2024) (a) unique solution (b) exactly two solutions (c) infinitely many solutions (d) no solutions
Ans: 2x + y = 13 …(i) 4x – y = 17 …(ii) On adding eqn.(i) and eqn.(ii) 6x = 30 x = 5 Put the value of x in eqn.(i) 2 × 5 + y = 13 ⇒10 + y = 13 ∴ y = 3 So, x – y = 5 – 3 = 2
Q3: The value of k for which the pair of linear equations 5x + 2y − 7 = 0 and 2x + ky + 1 = 0 do not have a solution is ______. (CBSE 2024) (a) 5 (b) 4/5 (c) 5/4 (d) 5/2
Ans: Given the pair of linear equations as, ⇒ x + 2y = 9 …(i) ⇒ y − 2x = 2 ⇒ −2x + y = 2 …(ii) Multiplying eqn (i) by 2 and adding to eqn (ii), we get ⇒ (−2x + y) + ( 2x + 4y) = 2 + 18 ⇒ 5y = 20 ⇒ y = 4 Putting in eqn (i), ⇒ x + 2(4) = 9 ⇒ x = 9 − 8 ⇒ x = 1 So, the required solution is x = 1 and y = 4
Q5: Check whether the point (−4, 3) lies on both the lines represented by the linear equations x + y + 1 = 0 and x − y = 1. (CBSE 2024)
Ans: Given the equations of line are ⇒ x + y = −1 …(i) ⇒ x − y = 1 …(ii) The intersection point of both the lines will be the point that lies on both the lines, So, adding eqn (i) and (ii), ⇒ (x + y) + (x − y) = −1 + 1 ⇒ 2x = 0 ⇒ x = 0 Putting in eqn (i), ⇒ 0 + y = −1 ⇒ y = −1 So, the point will be (x, y) = (0, −1) Hence, the point (−4, 3) does not lie on both the lines.
Q6: In the given figure, graphs of two linear equations are shown. The pair of these linear equations is: (CBSE 2024)
(a) consistent with unique solutions. (b) consistent with infinitely many solutions. (c) inconsistent. (d) inconsistent but can be made consistent by extending these lines.
Ans: Given system of linear equations are 7x − 2y = 5 …(i) 8x + 7y = 15 …(ii) By multiplying eq. (i) by 7 and eq. (ii) by 2, we get 49x − 14y = 35 16x + 14y = 30 65x = 65 ∴ x = 1 Substituting the value of x is eq (i), we get 7(1) − 2y = 5 or, 7 − 2y = 5 or −2y = −2, or y = 1 Therefore, x = 1 and y = 1
Q8: Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now? (CBSE 2024)
Ans: Let the age of Rashmi = x years and the age of Nazma = y years Three years ago, Rashmi’s age = (x − 3) years Nazma’s age = (y − 3) years According to the question, (x − 3) = 3(y − 3) ⇒ x − 3 = 3y − 9 ⇒ x = 3y − 6 …(i) Ten years later, Rashmi’s age = x + 10 Nazma’s age= y + 10 According to the question, (x + 10) = 2(y + 10) x + 10 = 2y + 20 x = 2y + 10 …(ii) From eq. (i) and (ii), we get 3y − 6 = 2y + 10 y = 16 Substituting the value of y in eq. (i), we get x = 3 × 16 − 6 = 48 − 6 = 42 Thus, Rashmi is 42 years old, and Nazma is 16 years old.
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Previous Year Questions 2023
Q9: The pair of linear equations 2x = 5y + 6 and 15y = 6x – 18 represents two lines which are (2023) (a) intersecting (b) parallel (c) coincident (d) either intersecting or parallel
Ans: (c) Sol: x + ky = 5 At x = 2, y = 1 2 + k(1) = 5 ∴ k = 3
Q11: The pair of linear equations x + 2y + 5 = 0 and -3x – 6y + 1 = 0 has (2023) (a) A unique solution (b) Exactly two solutions (c) Infinitely many solutions (d) No solution
x = 5 —————(i) y = 7 —————(ii) Draw the line x = 5 parallel to the y-axis and y= 7 parallel to the x-axis. ∴ The graph of equation (i) and (ii) is as follows The lines x = 5 and y = 7 intersect each other at (5, 7).
Q13: Using the graphical method, find whether a pair of equations x = 0 and y = -3 is consistent or not. (2023)
Ans: Let x and y be two numbers such that x> y According to the question, and x + 2y = 13 —- (ii) Subtracting (i) from (ii), we get 3y = 9 ⇒ y = 3 Substitute y = 3 in (i) we get x – 3 = 4 ⇒ x = 7
Q15: (A) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1 It becomes 1/2 if we only add 1 to the denominator. What is the fraction? OR (B) For which value of ‘k’ will the following pair of linear equations have no solution? (2023) 3x + y = 1 (2k – 1)x + (k – 1)y = 2k + 1
Ans: (A) Let required fraction be x/y According to question,
x + 1y – 1 = 1
⇒ x + 1 = y – 1
⇒ x = y – 2 … (i)
Also, xy + 1 = 12
⇒ 2x = y + 1 …(ii) From equations (i) and (ii), we get 2y — 4 = y + 1 y = 5 ∴ x = 3 Required fraction x/y is 3/5 OR (B) 3x + y = 1 (2k – 1 )x + (k – 1 )y = 2k + 1
For no solution;
⇒ 32k – 1 = 1k – 1 ≠ 12k + 1
2k – 1 = 3k – 3
⇒ k = 2
Also, 1k – 1 ≠ 12k + 1
2k + 1 ≠ k – 1
⇒ k ≠ -2
Q16: Two schools ‘P’ and ‘Q’ decided to award prizes to their students for two games of Hockey Rs. x per student and Cricket Rs. y per student. School ‘P’ decided to award a total of Rs. 9,500 for the two games to 5 and 4 students respectively, while school ‘Q’ decided to award Rs. 7,370 for the two games to 4 and 3 students respectively.
Based on the given information, answer the following questions. (i) Represent the following information algebraically (in terms of x and y). (ii) (a) What is the prize amount for hockey?
OR
(b) Prize amount on which game is more and by how much? (iii) What will be the total prize amount if there are 2 students each from two games? (CBSE 2023)
Ans: (i) For Hockey, the amount given to per student = x For cricket, the amount given to per student = y From the question, 5x + 4y =9500 (i) 4x + 3y = 7370 (ii)
(ii) (a) Multiply (1) by 3 and (2) by 4 and then subtracting, we get
15x + 12y- (16x + 12y) = 28500 – 29480 ⇒ – x = – 980 ⇒ x = ₹980 The prize amount given for hockey is Rs. 980 per student (b) Multiply (1) by 4 and (2) by 5 and then subtracting, we get 20x + 16y- 20x – 15y = 38000 – 36850 ⇒ y = 1150 The prize amount given for cricket is more than hockey by (1150 – 980) = 170. (iii) Total prize amount = 2 x 980 + 2 x 1150 = Rs. (1960 + 2300) = Rs. 4260
Also read: Facts that Matter: Pair of Linear Equations in Two Variables
Previous Year Questions 2022
Q17: The pair of lines represented by the linear equations 3x + 2y = 7 and 4x + 8y -11 = 0 are (2022) (a) perpendicular (b) parallel (c) intersecting (d) coincident
Ans: (d) Sol: Given equations are, y = 2 and y = – 3.
Clearly, from the graph, we can see that both equations are parallel to each other. So, there will be no solution.
Q19: A father is three times as old as his son. In 12 years time, he will be twice as old as his son. The sum of the present ages of the father and the son is (2022) (a) 36 years (b) 48 years (c) 60 years (d) 42 years
Ans: (b) Sol: Let age of father be ‘x’ years and age of son be ‘y’ years. According to the question, x = 3y ..(i) and x + 12 = 2 (y + 12) ⇒ x – 2y = 12 ..(ii) From (i) and (ii), we get x = 36, y = 12 ∴ x + y = 48 years
Q20: If 17x – 19y = 53 and 19x – 17y = 55, then the value of (x + y) is (2022) (a) 1 (b) -1 (c) 3 (d) -3
The general form of a linear equation is ax + by + c = 0. So, comparing terms:For the first equation, a1 = 1, b1 = 1, c1 = −4. For the second equation, a2 = 2, b2 = k, c2= −3.
For the lines to be parallel (and hence have no solution), we need:
a1a2 = b1b2 ≠ c1c2
So, 1/2 = 1/k
Cross-multiplying gives:
k = 2
Now, let’s check the condition for the
c1c2 = -4-3 = 43
Since 1/2 = 1/k when k = 2 but 1/2 ≠ 4/3, the condition for no solution is satisfied.
Thus, the value of kk for which the equations have no solution is: 2 So, the correct answer is (b) 2.
Q22: The solution of the pair of linear equations x = -5 and y = 6 is (2021) (a) (-5, 6) (b) (-5, 0) (c) (0, 6) (d) (0, 0)
Ans: (a) Sol: (-5, 6) is the solution of x = -5 and y = 6.
Q23: The value of k for which the pair of linear equations 3x + 5y = 8 and kx + 15y = 24 has infinitely many solutions, is (2021) (a) 3 (b) 9 (c) 5 (d) 15
Ans: (a) Sol: 32x + 33y = 34 …(i) 33x + 32y = 31 …(ii) Adding equation (i) and (ii) and subtracting equation (ii) from (i), we get 65x + 65y = 65 or x + y = 1 …(iii) and – x + y = 3 …(iv) Adding equation (iii) and (iv), we get y = 2 Substituting the value of y in equation (iii), x = -1
Q25: Two lines are given to be parallel. The equation of one of the lines is 3x – 2y = 5. The equation of the second line can be (2021) (a) 9x + 8 y = 7 (b) – 12 x – 8 y = 7 (c) – 12 x + 8y = 7 (d) 12x + 8y = 7
Ans: (c) Sol: If two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel, then
a1a2 = b1b2 ≠ c1c2 It can only possible between 3x – 2y = 5 and -12x + 8y = 7.
Q26: The sum of the numerator and the denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to 1/3. Find the fraction. (2021)
Ans: 5/13 Let the numerator be x and the denominator be y of the fractions. Then, the fraction = x /y. Given , x + y = 18 – (i) and ⇒ 3x – y = 2 . . (ii) Adding (i) and (ii), we get 4x = 20 ⇒ x = 5 Put the value of x in (i), we get 5+ y= 18 ⇒ y = 13 ∴ The required fraction is 5/13
Q27: Find the value of K for which the system of equations x + 2y = 5 and 3x + ky + 15 = 0 has no solution. (2021)
x + 2y = 5 3k + ky = – 15 has no solution. ∴ For K = 6 the given system of equations has no solution.
Q28:Case study-based questions are compulsory. A bookstore shopkeeper gives books on rent for reading. He has a variety of books in his store related to fiction, stories, quizzes etc. He takes a fixed charge for the first two days and an additional charge for subsequent days Amruta paid ₹22 for a book and kept it for 6 days: while Radhika paid ₹16 for keeping the book for 4 days. Assume that the fixed charge is ₹x and the additional charge (per day) is ₹y. Based on the above information, answer any four of the following questions. (i) The situation of the amount paid by Radhika. is algebraically represented by (2021) (a) x – 4 y = 16 (b) x + 4 y = 16 (c) x – 2 y = 16 (d) x + 2 y = 16
Ans: (d) Sol: For Amruta, x + (6 – 2)y = 22 i. e., x + 4y = 22 …(i) For Radhika, x + (4 – 2)y = 16 i.e.,x + 2y = 16 …(ii) Solving equation (i) and (ii). we get x = 10 and y = 3 i.e., Fixed charges (x) = 10 …(iii) and additional charges per subsequent day (y) = ₹ 3 …(iv) x + 2 y = 16 [From equation (ii)]
(ii) The situation of the amount paid by Amruta. is algebraically represented by (2021) (a) x – 2y = 11 (b) x – 2y = 22 (c) x + 4 y = 22 (d) x – 4 y = 11
Ans: (c) Sol: For Amruta, x + (6 – 2)y = 22 i. e., x + 4y = 22 …(i) For Radhika, x + (4 – 2)y = 16 i.e.,x + 2y = 16 …(ii) Solving equation (i) and (ii). we get x = 10 and y = 3 i.e., Fixed charges (x) = 10 …(iii) and additional charges per subsequent day (y) = ₹ 3 …(iv) x + 4 y = 22 [From equation (i)]
(iii) What are the fixed charges for a book? (2021) (a) ₹ 9 (b) ₹ 10 (c) ₹ 13 (d) ₹ 15
Ans: (b) Sol: For Amruta, x + (6 – 2)y = 22 i. e., x + 4y = 22 …(i) For Radhika, x + (4 – 2)y = 16 i.e.,x + 2y = 16 …(ii) Solving equation (i) and (ii). we get x = 10 and y = 3 i.e., Fixed charges (x) = 10 …(iii) and additional charges per subsequent day y = ₹ 3 …(iv) x = ₹ 10 [From equation (iii)]
(iv) What are the additional charges for each subsequent day for a book? (2021) (a) ₹ 6 (b) ₹ 5 (c) ₹ 4 (d) ₹ 3
Ans: (d) Sol: For Amruta, x + (6 – 2)y = 22 i. e., x + 4y = 22 …(i) For Radhika, x + (4 – 2)y = 16 i.e.,x + 2y = 16 …(ii) Solving equation (i) and (ii). we get x = 10 and y = 3 i.e., Fixed charges (x) = 10 …(iii) and additional charges per subsequent day y = ₹ 3 …(iv) y = ₹ 3 [From equation (iv)]
(v) What is the total amount paid by both, if both of them have kept the book for 2 more days? (2021) (a) ₹ 35 (b) ₹ 52 (c) ₹ 50 (d) ₹ 58
Ans: (c) For Amruta, x + (6 – 2)y = 22 i. e., x + 4y = 22 …(i) For Radhika, x + (4 – 2)y = 16 i.e.,x + 2y = 16 …(ii) Solving equation (i) and (ii). we get x = 10 and y = 3 i.e., Fixed charges (x) = 10 …(iii) and additional charges per subsequent day y = ₹ 3 …(iv) Total amount paid for 2 more days by both = (x + 4 y) + 2 y + (x + 2y ) + 2 y = 2 x + 10y = 2 x 10 + 10 x 3 = ₹ 50
Q29: The pair of equations x = a and y = b graphically represent lines which are (2020) (a) Intersecting at (a, b) (b) Intersecting at (b, a) (c) Coincident (d) Parallel
Ans: (a) Sol: The pair of equations x = a and y = b graphically represent lines which are parallel to the y-axis and x-axis respectively. The lines will intersect each other at (a, b).
Q30: If the equations kx – 2y = 3 and 3x + y = 5 represent two intersecting lines at unique points, then the value of k is _________. (2020)
Ans: For any real number except k = -6 kx – 2y = 3 and 3x + y = 5 represent lines intersecting at a unique point. ⇒ k3 ≠ -21 ⇒ k ≠ -6 For any real number except k ≠ -6 The given equation represent two intersecting lines at unique point.
Q31: The value of k for which the system of equations x + y – 4 = 0 and 2x + ky = 3 has no solution. is (2020) (a) -2 (b) ≠2 (c) 3 (d) 2
Q32: Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are given by 2y – x = 8, 5y – x = 14, and y – 2x = 1. (2020)
2y – x = 8 ..(i) 5y – x = 14 …(ii) and y – 2x = 1 …(iii) are given below:
From the graph of lines represented by given equations, we observe that Lines (i) and (iii) intersect each other at C(2, 5), Lines (ii) and {iii) intersect each other at B(1, 3) and Lines (i) and (ii) intersect each other at 4(-4, 2). Coordinates of the vertices of the triangle are A(-4, 2), B(1, 3) and C(2, 5).
Q33: Solve the equations x + 2y = 6 and 2x – 5y = 12 graphically. (2020)
From the graph, the two lines intersect each other at point (6, 0) ∴ x = 6 and y = 0
Q34: A fraction becomes 1/3 when 1 is subtracted from the numerator, and it becomes 1/4 when 8 is added to its denominator. Find the fraction. (CBSE 2020)
From (ii), 4x = y +8 so, 4x – y – 8 = 0 … (iv) Subtracting (iii) from (iv), we get x = 5 Substituting the value of x in (iii), we get y = 12 Thus, the required fraction is 5/12
Q35: The present age of a father is three years more than three times the age of his son. Three years hence, the father’s age will be 10 years more than twice the age of the son. Determine their present ages. (2020)
Ans:Let the present age of son be x years and that of father be y years.
According to question, we have y = 3x+ 3 ⇒ 3x – y + 3 = 0 (i) And y + 3 = 2(x + 3) + 10 ⇒ y + 3 = 2x + 6 +10 ⇒ 2x – y + 13 = 0 (ii) Subtracting (ii) from (i), we get x = 10 Substituting the value of x in (ii). we get y = 33 So. the present age of the son is 10 years and that of the father is 33 years.
Ans: Given lines are 2x + 3y = 2 and x – 2y = 8 2x + 3y = 2 and x – 2y = 8 ∴ We will plot the points (1, 0), (-2, 2) and (4, – 2 ) and join them to get the graph of 2x + 3y = 2 and we will plot the points (0, -4), (8, 0) and (2, -3) and join them to get the graph of x – 2y = 8
Q37: A train covered a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken 4 hours less than the scheduled time and if the train were slower by 6 km/hr, it would have taken 6 hours more than the scheduled time. Find the length of the journey. (CBSE 2020)
Ans: Let the original uniform speed of the train be x km/hr and the total length of journey be l km. Then, scheduled time taken by the train to cover a distance of l km = l/x hours Now,
lx + 6 = lx – 4
⇒ lx – lx + 6 = 4
⇒ x + 6 – xx(x + 6) = 4
⇒ 6lx(x + 6) = 4
⇒ l = 2x(x + 6)3 … (i)
Also,
lx – 6 = lx + 6
⇒ lx – 6 – lx = 6
⇒ x – x + 6(x – 6)x = 6
⇒ 6l(x – 6)x = 6
⇒ l = x(x – 6) … (ii)
From equations (i) and (ii), we have
2x(x + 6)3 = x(x – 6)
⇒ 2x + 12 = 3x – 18
⇒ x = 30
Putting the value of x in eq. (ii), we get l = 30(30 – 6) = 30 × 24 = 720 Hence, the length of the journey is 720 km.
Also read: Facts that Matter: Pair of Linear Equations in Two Variables
Previous Year Questions 2019
Q38: Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Using this graph, find the values of x and y which satisfy both the equations. (2019)
Ans: Let the larger angle be x° and the smaller angle be y°. We know that the sum of two supplementary pairs of angles is always 180°.
We have x° + y° = 180° (i) and x° – y° = 18° (ii) [Given] By (1), we have x° = 180° – y° _(iii) Put the value of x° in (ii), we get 180° – y° – y° = 18° ⇒ 162° = 2y° ⇒ y = 81 From (3), we have x° = 180° – 81° = 99° The angles are 99° and 81°
Q40: Solve the following pair of linear equations: 3x – 5y =4, 2y+ 7 = 9x. (2019)
3x – 5y = 4, ……. (i) 2y+ 7 = 9x 9x – 2y = 7 …….. (ii) Multiply (i) by 3 and subtract from (ii), as
⇒ 9x – 2y – 9x + 15y = -5 → 13y = -5
⇒ y = -513
Put y = -513 in (i), we get
3x – 5 -513 = 4
⇒ 3x + 2513 = 4
⇒ 3x = 4 – 2513
⇒ x = 2713 × 3 = 913
Hence, x = 9/13 and y = -5/13
Q41: A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father. (2019)
Ans: Let the ages of two children be x and y respectively.
Father’s present age = 3(x +y) After 5 years, sum of ages of children = x + 5 + y + 5 = x + y + 10 and age of father = 3(x + y) + 5 According to the question, 3(x + y) + 5 = 2(x + y+ 10) 3x + 3y + 5 = 2x + 2y + 20 ⇒ x + y = 15 Hence, present age of father = 3(x + y) = 3 x 15 = 45 years
Q42: A fraction becomes 1/3 when 2 is subtracted from the numerator and will becomes 1/2 when 1 is subtracted from the denominator. Find the fraction. (2019)
x – 4y + p = 0 (i) and 2x + y – q – 2 = 0 (ii) It is given that x = 3 and y = 1 is the solution of (i) and (ii) ∴ 3 – 4 x 1+ p = 0 ⇒ p = 1 and 2 x 3 + 1 – q – 2 = 0 ⇒ q = 5 ∴ q = 5p
Q45: For what value of k, does the system of linear equations 2x + 3y=7 and (k – 1)x + (k + 2) y = 3k have an infinite number of solutions? (CBSE 2019)
Ans: (c) Since both polynomials cut the x-axis at two distinct points each, the total number of distinct zeroes of both the polynomials combined is 2.
Q5: If α and β are the zeroes of the polynomial p(x) = x2 – ax – b, then the value of (α + β + αβ) is equal to: (2025) (a) a + b (c) a- b (b) -a – b (d) -a + b
Ans: (c) Given polynomial is p(x) = x2 – ax – b, and α and β are zeroes of p(x) ∴ Sum of zeroes = α + β =a Product of zeroes = αβ = – b Now, α + β + αβ = a – b Concept Applied If α and β are the zeroes of quadratic polynomial p(x) = ax2 + bx + c, then α + β = -b/a, αβ = c/a.
Q6: If α and β are zeroes of the polynomial p(x) = kx2 – 30x + 45k and α +β = αβ, then the value of ‘k’ is: (2025) (a) (b) (c) 3/2 (d) 2/3
Ans: The given polynomial is p(x) = (p + 1)x2 + (2p + 3)x + (3p + 4) Let α and β are zeroes of given polynomial ⇒ p + 1 = 2p + 3 ⇒ p = -2 Hence, the value of p is – 2.
Q10: If α and β are zeroes of the polynomial p(x) = x2 – 2x – 1, then find the value of (2025)
Ans: Let the zeroes of the polynomial x2 + ax + b be 3x and 4x. ⇒ 49b = 12a2. Hence proved.
Q13: Find the zeroes of the polynomial p(x) = 3x2 – 4x – 4. Hence, write a polynomial whose each of the zeroes is 2 more than the zeroes of p(x). (2025)
Ans: (b) Let, f(x) = x2 – 5x + 4 Let p should be added to f(x) then 3 becomes zero of polynomial. So, f(3) + p = 0 ⇒ (3)2 – 5 × (3) + 4 + p = 0 ⇒ 9 + 4 – 15 + p = 0 ⇒ – 2 + p = 0 ⇒ p = 2
So, 2 should be added.
Q2: Find the zeroes of the quadratic polynomial x2 – 15 and verify the relationship between the zeroes and the coefficients of the polynomial. (2024)
Ans: x2 – 15 = 0 x2 = 15 x = ± √15 Zeroes will be α = √15 , β = – √15 Verification: Given polynomial is x2 – 15 On comparing above polynomial with ax2 + bx + c, we have a = 1, b = 0, c = –15 sum of zeros = α + β Product of zeros = αβ Hence, verified.
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Previous Year Questions 2023
Q1: The graph of y = p(x) is given, for a polynomial p(x). The number of zeroes of p(x) from the graph is (2023)
Ans: (b) The distance between point C and G is 6 units.
Q5: The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6. (2022) (a) x2 + 5x + 6 (b) x2 – 5x + 6 (c) x2 – 5 x – 6 (d) – x2 + 5x + 6
Ans: (a) Let α, β be the zeroes of required polynomial p(x). Given, α + β=-5 and α.β=6 p(x) = x2 – (Sum of zeros)x + (Product of zeros) ∴ p(x)=k[x2 – (-5)x + 6] = k[x2 + 5x + 6] Thus, one of the polynomial which satisfy the given condition is x2+ 5x + 6
Previous Year Questions 2021
Q1: If one zero of the quadratic polynomial x2 + 3x + k is 2 then find the value of k. (2021)
Ans: (b) Given, 2 is a zero of the polynomial p(x) = x2 + 3x + k ∴ p (2) = 0 ⇒ (2)2 + 3(2) + k = 0 ⇒ 4 + 6 + k = 0 ⇒ 10 + k = 0 ⇒ k= -10
Q3: The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6 is ________. (2020) (a) x2 + 5x + 6 (b) x2 – 5x + 6 (c) x2– 5x – 6 (d) -x2 + 5x + 6
Ans: (a) Let α, β be the zeroes of required polynomial p(x) Given, α+ β = -5 and αβ = 6 p(x) = k[x2 – (- 5)x + 6] = k[x2 + 5x + 6] Thus, one of the polynomial which satisfy the given condition is x2 + 5x + 6.
Q4: Form a quadratic polynomial, the sum and product of whose zeroes are (-3) and 2 respectively. (CBSE 2020)
Ans: Let α, β be the zeroes of required polynomial Given, α + β = -3 and αβ = 2 ∴ p(x) = k[x2 – (-3)x + 2] = k(x2 + 3x + 2) For k = 1 , p (x) = x2 + 3x + 2 Hence, one of the polynomial which satisfy the given condition is x2 + 3x + 2.
Q5: The zeroes of the polynomial x2 – 3x – m(m + 3) are: (a) m, m + 3 (b) –m, m + 3 (c) m, – (m + 3) (d) –m, – (m + 3) (CBSE 2020)
x2 − 3x − m(m + 3) = 0 Let’s find the zeroes by applying the quadratic formula:
Substitute into the formula:
Simplify under the square root:
Taking the square root:
So, the zeroes are –m and m + 3. Thus, the correct answer is (b) –m, m + 3.
Practice Test: Polynomials
Start Test
Previous Year Questions 2019
Q1: Find the value of k such that the polynomial x2 – (k + 6)x + 2(2k – 1) has the sum of its zeroes equal to half of its product. [Year 2019, 3 Marks]
Ans: 7 The given polynomial is x2 -(k + 6)x + 2(2k – 1) According to the question Sum of zeroes = 1/2(Product of Zeroes ): ⇒ k + 6 = 1/2 x 2 (2k – 1) ⇒ k + 6 = 2k – 1 ⇒ k = 7
Ans: (c) 8n = (2 × 2 × 2)n Since, factors of 8n do not contain 5 in it. So, 8n can’t ends with 0. ⇒ 0 can’t be the unit digit of 8n.
Q3: If x is the LCM of 4, 6, 8 and y is the LCM of 3, 5, 7 and p is the LCM of x and y, then which of the following is true? (2025) (a) p = 35x (b) p = 4y (c) p = Bx (d) p = 16y
Ans: (a) We have to find the H.C.F of 70 – 5 = 65 and 125 – 8 = 117. ∴ 65 = 5 × 13, 117 = 32 × 13 ⇒ HCF (65, 117) = 13 ∴ Required greatest number = 13.
Q6: Assertion (A): For any two prime numbers p and q, their HCF is 1 and LCM is p + q. Reason (R): For any two natural numbers, HCF x LCM = product of numbers. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) (ii) is true. (2025)
Ans: The smallest number divisible by both 644 and 462 is LCM of 644 and 462. 644 = 2 × 2 × 7 × 23 462 = 2 × 3 × 7 × 11 ∴ LCM (644,462) = 2 × 2 × 3 × 7 × 11 × 23 = 21252
Q8: Two numbers are in the ratio 4: 5 and their HCF is 11. Find the LCM of these numbers. (2025)
Hide Answer Ans: Let two numbers be 4x and 5x. HCF (4x, 5x) = 11 ∴ Numbers are 4 × 11 = 44 and 5 × 11 = 55 Since, product of two numbers= HCF × LCM ⇒ 44 × 55 = 11 × LCM
Q9: Three sets of Physics, Chemistry and Mathematics books have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. The number of Physics books is 144, the number of Chemistry books is 180 and the number of Mathematics books is 192. Assuming that the books are of same thickness,determine the number of stacks of Physics, Chemistry and Mathematics books. (2025)
Ans: Number of Physics books = 144 Number of Chemistry books = 180 Number of Mathematics books = 192 Since, 180 = 2 × 2 × 3 × 3 × 5 144 = 2 × 2 × 2 × 2 × 3 × 3 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 ∴ HCF(180, 144, 192) = 2 x 2 x 3 = 12 ∴ Number of stacks of Physics books = 144/12 = 12 Number of stacks of Chemistry books = 180/12 = 15 Number of stacks of Mathematics books = 192/12 = 16
Q10: Let p, q and r be three distinct prime numbers. Check whether p·q·r + q is a composite number or not. Further, give an example for 3 distinct primes p, q, r such that (i) p·q·r + 1 is a composite number. (ii) p·q·r + 1 is a prime number. (2025)
Ans: Given: p, q, r be distinct prime numbers. We have, pqr + q = q(pr + 1) Here, q is a prime number and pr+ 1 > 1, Thus, pqr + q has factors 1, q, (pr+ 1) and q(pr + 1). Hence, pqr + q is a composite number. …(i) (i) Take p = 3, q = 5 and r = 7, we have pqr + 1 = 3 × 5 × 7 + 1 = 105 + 1 = 106 Thus, pqr + 1 is a composite number for p = 3, q = 5 and r = 7. (ii) Take p = 2, q = 3 and r = 5, we have pqr + 1 = 2 × 3 × 5 + 1 = 30 + 1 = 31 Thus, pqr + 1 is a prime number for p = 2, q = 3 and r = 5.
Q11:is a/an (2025) (a) natural number (b) integer (c) rational number (d) irrational number
Ans: It is given that, √3 is an irrational number. We have to prove that is an irrational number. 3
Let us assume be a rational number. Then, where b ≠ 0 and a, b are co-prime integers. which is a rational number. ⇒ √3 is a rational number. But √3 is an irrational number. So, our assumption is wrong. Hence, is an irrational number.
Q14: Prove thatis an irrational number given that √2 is an irrational number. (2025)
Ans: It is given that, √2 is an irrational number. We have to prove that is an irrational number.
Let us assume be a rational number. ⇒√2 is a rational number. But √2 is an irrational number. So, our assumption is wrong. Hence,is an irrational number.
Previous Year Questions 2024
Q1: The smallest irrational number by which √20 should be multipled so as to get a rational number, is: (CBSE 2024) (a) √20 (b) √2 (c) 5 (d) √5
Ans:(a) HCF(2520, 6600) = 40 LCM(2520, 6600) = 252 × k ∴ HCF × LCM = Ist No. × IInd No. ∴ 40 × 252 × k = 2520 × 6600 ⇒ k = 2520 x 660040 x 252 ⇒ k = 1650
Q5: Teaching Mathematics through activities is a powerful approach that enhances students’ understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class 5 students. She announced the number 2 in her class and asked the first student to multiply it by a prime number and then pass it to the second student. The second student also multiplied it by a prime number and passed it to the third student. In this way by multiplying by a prime number, the last student got 173250. Now, Mukta asked some questions as given below to the students: (CBSE 2024) (A) What is the least prime number used by students? (B) How many students are in the class? OR What is the highest prime number used by students? (C) Which prime number has been used maximum times?
Ans: (A) So least prime no. used by students = 3(because 2 is announced by the teacher, so the least number used by the students is 3) (B)As the last student got 173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11 there are 7 factors other than 2, which is announced by teacher. So, Number of student = 7 OR Highest prime number used by student = 11 (C)Prime number 5 is used maximum times i.e., 3 times.
Q6: LCM (850, 500) is: (CBSE 2024) (a) 850 × 50 (b) 17 x 500 (c) 17 x 52 x 22 (d) 17 × 53 × 2
Ans: Let us assume that 6 – 4√5 be a rational number Let …[b ≠ 0; a and b are integers] We know that, is a rational number. But this contradicts the fact that √5 is an irrational number. So, our assumption is wrong. Therefore, 6 – 4√5 is an irrational number.
Q8: Show that 11 × 19 × 23 + 3 × 11 is not a prime number. (CBSE 2024)
Ans: We have 11 × 19 × 23 + 3 × 11 ⇒ 11(19 × 23 + 3) ⇒ 11(437 + 3) ⇒ 11(440) ⇒ 11(2 × 2 × 2 × 5 × 11) ⇒ 2 × 2 × 2 × 5 × 11 × 11 As it can be represented as a product of more than two primes (1 and number itself). So, it is not a prime number.
Q9: If two positive integers p and q can be expressed as p = 18 a²b¹ and q=20 a³b², where a and b are prime numbers, then LCM (p, q) is: (CBSE 2024) (a) 2 a²b² (b) 180 a²b² (c) 12 a²b² (d) 180 a³b²
Ans: Assuming 5 – 2√3 to be a rational number. Here RHS is rational but LHS is irrational. Therefore our assumption is wrong. Hence, 5 – 2√3 is an irrational number.
Q11: Show that the number 5 × 11 × 17 + 3 × 11 is a composite number. (CBSE 2024)
Ans: The numbers are prime numbers and composite numbers. Prime numbers can be divided by 1 and itself. A composite number has factors other than 1 and itself. (5 × 11 × 17) + (3 × 11) = (85 × 11) + (3 × 11) = 11 × (85 + 3) = 11 × 88 = 11 × 11 × 8 = 2 × 2 × 2 × 11 × 11 The given expression has 2 and 11 as its factors. Therefore, it is a composite number.
Q12: In a teachers’ workshop, the number of teachers teaching French, Hindi and English are 48, 80 and 144 respectively. Find the minimum number of rooms required if in each room the same number of teachers are seated and all of them are of the same subject. (CBSE 2024)
Ans: The number of rooms will be minimum if each room accommodates a maximum number of teachers. Since in each room the same number of teachers are to be seated and all of them must be of the same subject. Therefore, the number of teachers in each room must be HCF of 48, 80, and 144. The prime factorisations of 48, 80 and 144 are as under 48 = 24 × 31 80 = 24 × 51 144 = 24 × 32 ∴ HCF of 48, 80 and 144 = 16 Therefore, in each room 16 teachers can be seated.
Q13: Directions: Assertion (A) is followed by a statement of Reason (R). Select the correct option from the following options: (a) Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely. (b) Both, Assertion (A) and Reason (R) are true. Reason (R) does not explain Assertion (A). (c) Assertion (A) is true but Reason (R) is false. (d) Assertion (A) is false but Reason (R) is true. Assertion (A): If the graph of a polynomial touches x-axis at only one point, then the polynomial cannot be a quadratic polynomial. Reason (R): A polynomial of degree n(n >1) can have at most n zeroes. (CBSE 2024)
Ans: Assertion (A): “If the graph of a polynomial touches the x-axis at only one point, then the polynomial cannot be a quadratic polynomial.” A quadratic polynomial is of the form ax2 + bx + c. It can have two real roots, one real root (if the discriminant is zero), or no real roots (if the discriminant is negative). If the graph of a quadratic polynomial touches the x-axis at exactly one point, this means it has a repeated real root (a double root), which is possible for quadratic polynomials. Hence, the assertion is false. Reason (R): “A polynomial of degree n (n > 1) can have at most n zeroes.” A polynomial of degree n can have at most n real or complex zeroes. This is a true statement, so the reason is true. Given the above analysis, the correct answer is: (d) Assertion (A) is false but Reason (R) is true.
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Previous Year Questions 2023
Q1: The ratio of HCF to LCM of the least composite number and the least prime number is (2023) (a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) 1 : 3
Then √3 = a/b; where a and b ( ≠ 0) are co-prime positive integers. Squaring on both sides, we get 3 = a2/b2 ⇒ a2 = 3b2 ⇒ 3 divides a2 ⇒ 3 divides a _________(i) = a = 3c, where c is an integer Again, squaring on both sides, we get a2 = 9c2 ⇒3b2 = 9c2 ⇒b2 = 3c2 ⇒ 3 divides b2 ⇒ 3 divides b _________(ii) From (i) and (ii), we get 3 divides both a and b. ⇒ a and b are not co- prime integers. This contradicts the fact that a and b are co-primes. Hence, √3 is an irrational number.
Also read: Short Answer Questions: Real Numbers – 1
Previous Year Questions 2022
Q1: Two positive numbers have their HCF as 12 and their product as 6336. The number of pairs possible for the numbers is (2022) (a) 2 (b) 3 (c) 4 (d) 1
Ans:(a) Sol: Given, HCF = 12 Let two numbers be 12a and 12b So. 12a x 12b = 6336 ⇒ ab = 44 We can write 44 as product of two numbers in these ways: ab = 1 x 44 = 2 x 22 = 4x 11 Here, we will take a = 1 and b = 44 ; a = 4 and b = 11. We do not take ab = 2 x 22 because 2 and 22 are not co-prime to each other.
For a = 1 and b = 44, 1st no. = 12a = 12, 2nd no. = 12b = 528 For a = 4 and b = 11, 1st no. = 12a = 48, 2nd no. = 12b = 132 Hence, we get two pairs of numbers, (12, 528) and (48, 132).
Q2: If ‘n’ is any natural number, then (12)n cannot end with the digit (2022) (a) 2 (b) 4 (c) 8 (d) 0
Ans: We have, 2 x 3 x 5 + 5 and 5 x 7 x 11 + 7 x 5. We can write these numbers as: 2 x 3 x 5 + 5 = 5(2 x 3 + 1) = 1 x 5 x 7 and 5 x 7 x 11 + 7 x 5 = 5 x 7(11 + 1) = 5 x 7 x 12 = 1 x 5 x 7 x 12 Since, on simplifying. we find that both the numbers have more than two factors. So. these are composite numbers.
Ans: We know that LCM × HCF = Product of two numbers ∴ LCM (135, 225) = Product of 135 and 225 / HCF(135, 225) = 135 x 225 / 45 = 675 So, LCM (135, 225) = 675
Also read: Short Answer Questions: Real Numbers – 1
Ans: Using the formula: HCF (a, b) x LCM (a, b) = a x b ∴ HCF (336, 54) x LCM (336, 54) = 336 x 54 ⇒ 6 x LCM(336, 54) = 18144 ⇒ LCM (336, 54) = 18144 / 6 = 3024
Q2: The HCF of two numbers a and b is 5 and their LCM is 200. Find the product of ab. (2019)
Ans: Let us assume that √5 is a rational number. Then √5 = a/b where a and b (≠ 0} are co-prime integers, if Squaring on both sides, we get 5 = a2b2 ⇒ a2 = 5b2 ⇒ 5 divides a2 ⇒ 5 divides a ———-(i) ⇒ a = 5c, where c is an integer Again, squaring on both sides, we get a2 = 25c2 ⇒ 5b2 = 25c2 ⇒ b2 = 5c2 ⇒ 5 divides b2 ———-(ii) ⇒ 5 divides b From (i) and {ii), we get 5 divides both a and b. ⇒ a and b are not co-prime integers. Hence, our supposition is wrong. Thus, √5 is an irrational number.
Ans: Let us assume √2 be a rational number. Then, √2 = p/q where p, q (q ≠ 0) are integers and co-prime. ; On squaring both sides. we get 2 = p2q2 ⇒ p2 = 2q2 ————(i) ⇒ 2 divides p2 ⇒ 2 divides p ———–(ii) So, p = 2a, where a is some integer. Again squaring on both sides, we get p2 = 4a2 ⇒ 2q2 = 4a2 (using (i)) ⇒ q2 = 2a2 ⇒ 2 divides q2 ⇒ 2 divides q ———–(iii) From (ii) and (iii), we get 2 divides both p and q. ∴ p and q are not co-prime integers. Hence, our assumption is wrong. Thus √2 is an irrational number.
Q7: Prove that 2 + 5√3 is an irrational number given that √3 is an irrational number. (2019)
Ans: Suppose 2 + 5√3 is a rational number. We can find two integers a, b (b ≠ 0) such that 2 + 5√3 = a/b, where a and b are co -prime integers. 5√3 = ab – 2 ⇒ √3 = 15 [ a b – 2] ⇒ √3 is a rational number.
[ ∵ a, b are integers, so 15 [ a b – 2] is a rational number] But this contradicts the fact that √3 is an irrational number. Hence, our assumption is wrong. Thus, 2 + 5√3 is an irrational number.
Q8: Write the smallest number which is divisible by both 306 and 657. (CBSE 2019)
Ans: Given numbers are 306 and 657. The smallest number divisible by 306 and 657 = LCM(306, 657) Prime factors of 306 = 2 × 3 × 3 × 17 Prime factors of 657 = 3 × 3 × 73 LCM of (306, 657) = 2 × 3 × 3 × 17 × 73 = 22338 Hence, the smallest number divisible by 306 and 657 is 22,338.
Q1: Which of the following groups do not constitute a food chain? (1 Mark) (i) Wolf, rabbit, grass, lion (ii) Plankton, man, grasshopper, fish (iii) Hawk, grass, snake, grasshopper, frog (iv) Grass, snake, wolf, tiger (a) (i) and (iv) (b) (i) and (iii) (c) (ii) and (iii) (d) (ii) and (iv)
Ans: (c) A food chain shows a sequence of organisms where each is eaten by the next one. In options (ii) and (iii), the order of organisms is incorrect — they do not represent a proper feeding relationship (for example, man does not eat grasshopper or fish directly, and grass cannot be eaten by a hawk). Hence, (ii) and (iii) do not constitute a food chain.
Q2: The percentage of solar energy which is not converted into food energy by the leaves of green plants in a terrestrial ecosystem is about: (1 Mark)
Ans: (d) Only about 1% of the sunlight that falls on green plants is converted into food energy, so 99% of solar energy remains unutilised.
Q3: Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below
Assertion (A): The amount of ozone in the atmosphere began to drop sharply in the 1980s. Reason (R): The oxygen atoms combine with molecular oxygen to form ozone. (1 Mark)
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Ans: (b)Ozone levels dropped sharply in the 1980s due to synthetic chemicals like CFCs, not because oxygen atoms combine with molecular oxygen (which actually forms ozone).
Q4: “Excessive use of chemicals and pesticides in agriculture adversely affects the environment.” Justify this statement. (2 Marks)
Ans: Excessive use of pesticides and chemicals in agriculture leads to their accumulation in soil and water. These chemicals enter the food chain through plants and aquatic organisms and are not degradable. They get progressively accumulated at each trophic level, a process called biological magnification, causing harmful effects on living organisms and the environment.
Q5: Identify from the following a group containing all non-biodegradable substances. (1 Mark)
Ans: (c) DDT, polyester, and glass cannot be broken down by biological processes; hence, they are non-biodegradable substances.
Q6: Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below (1 Mark)
Assertion (A): Animals will not get energy if they eat (consume) coal as food. Reason (R): Specific enzymes are needed for the breakdown of a particular food.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Animals cannot get energy from coal because enzymes are specific and only break down particular biological substances. Since coal cannot be broken down by these enzymes, it cannot provide energy.
Q7: Some harmful chemicals get accumulated in human bodies through the food chain. Name this phenomenon. Explain the reason for the maximum concentration of these chemicals found in our bodies. (3 Marks)
Ans: This phenomenon is called biological magnification.
Harmful chemicals such as pesticides and fertilizers enter the food chain through soil and water. These chemicals are not degradable and thus accumulate in the tissues of organisms.At each successive trophic level, their concentration increases because they are passed on through food. Since human beings occupy the top level in the food chain, the maximum concentration of these chemicals accumulates in our bodies.
Q8: Other than the abiotic components, which of the given biotic components are not required to make an aquarium with small herbivorous fishes a self-sustaining system? (1 Mark) (i) Aquatic plants and aquatic animals (ii) Terrestrial plants and terrestrial animals (iii) Decomposers as bacteria and fungi (iv) Consumers as clown fishes and sea urchins (a) (i) and (iv) (b) (ii) and (iii) (c) (i) and (iii) (d) (ii) and (iv)
A self-sustaining aquarium needs aquatic plants (producers), small herbivorous fishes (consumers), and decomposers. Terrestrial plants/animals and large consumers like sea urchins are not required in such an aquatic ecosystem.
Q9: Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below (1 Mark) Assertion (A): Use of jute bags for shopping reduces pollution. Reason (R): Jute is biodegradable and its bag may be reused as and when needed.
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Ans: (a)
Using jute bags helps reduce pollution because jute is biodegradable and such bags can be reused, reducing non-biodegradable plastic waste.
Q10: In the food chains given below, select the most efficient food chain in terms of energy. (1 Mark)
(a)Grass → Grasshopper → Frog → Snake (b) Plants → Deer → Lion (c) Plants → Man (d) Phytoplankton → Zooplankton → Small Fish → Big Fish
The shorter the food chain, the less energy is lost between trophic levels. Since this chain has only two trophic levels, it is the most efficient in terms of energy transfer.
Q11: What are decomposers? Give two examples. State how they maintain a balance in an ecosystem. (3 Mark)
Ans: (c)Lakes are natural ecosystems, while gardens are human-made (artificial) ecosystems maintained by people.
Q13: Human activities that are affecting the environment are: (1 Mark)
(a) Minimising the use of chlorofluorocarbons (b) Excessive use of disposable cups and plates (c) Maximising the use of reusable utensils for eating food and drinking fluids (d) Segregating the wastes into biodegradable and non-biodegradable before disposal
Using disposable cups and plates increases non-biodegradable waste, leading to environmental pollution and harming ecosystems.
Q14: (a) Why are the organisms of first trophic level important in any food chain? (b) Justify the following statement: ‘The flow of energy in an ecosystem is unidirectional.’ (2 Marks)
Ans: (a) The organisms of the first trophic level, known as producers, are very important because they absorb solar energy and convert it into chemical energy through photosynthesis. This stored energy in food becomes the source of energy for all other organisms in the food chain such as herbivores, carnivores, and decomposers. Without producers, no other organisms could survive, as they depend directly or indirectly on them for food.
(b) The flow of energy in an ecosystem is unidirectional, meaning it moves in one direction only — from the Sun → producers → consumers → decomposers. The energy captured by producers does not return to the Sun, nor can it go backward in the chain. At each trophic level, some energy is lost as heat, so the energy available to the next level keeps decreasing, making the flow one-way and non-cyclic.
Q15: A gas ‘X’ is found in the upper layer of atmosphere. It is a deadly poison, but still essential for all life forms on earth. (a) Identify the gas and state the main factor for its depletion in the atmosphere. (b) How is this gas formed in the upper atmosphere? (2 Marks)
(a) The gas ‘X’ is ozone (O₃). It is getting depleted mainly due to synthetic chemicals like chlorofluorocarbons (CFCs) used in refrigerants and fire extinguishers.
(b) In the upper atmosphere, ozone is formed when ultraviolet (UV) radiations split some molecules of oxygen (O₂) into free oxygen atoms (O), which then combine with molecular oxygen to form ozone (O₃).
Q16: Study the food web given below:
(a) Identify the food chain(s) in which the eagle receives the highest energy from the producers. (b) Identify the organism in which a non-biodegradable pesticide will be found in maximum concentration. Name the term used for this phenomenon. (2 Marks)
(a) The food chain in which the eagle receives the highest energy from producers is: Grass → Mouse → Eagle. This is because the chain is shortest, involving fewer trophic levels, so there is less energy loss between steps, and the eagle receives more energy from the producers.
(b) The organism in which a non-biodegradable pesticide will be found in maximum concentration is the Eagle. This is due to biological magnification, where harmful chemicals accumulate progressively at each trophic level, reaching the highest concentration in top consumers.
Q17: Consider the following food chain: Grass → Grasshopper → Frog → Snake → Eagle. If the amount of energy available at third trophic level is 50 kJ, the available energy at the producer level was: (1 Mark)
According to the 10% law, only 10% of energy is transferred from one trophic level to the next. If the third trophic level (frog) has 50 kJ, then: Producer → 1st → 2nd → 3rd Energy at producer level = 50 × 10 × 10 × 10 = 50,000 kJ.
Q18: The incorrect statement about ozone is: (1 Mark)
(a) It is a deadly poisonous gas. (b) It shields the surface of the Earth from UV radiation from sun. (c) It is used as a refrigerant and in fire extinguishers. (d) It is formed by combining an oxygen molecule with a free oxygen atom.
Ozone itself is not used as a refrigerant or in fire extinguishers; CFCs (chlorofluorocarbons) are. Ozone protects the Earth from harmful UV radiation.
Q19: Two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below : Assertion (A): Food web is a network of several food chains operating in an ecosystem. Reason (R): Food web decreases the stability of an ecosystem. (1 Mark)
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Ans: Natural ponds and lakes are self-sustaining ecosystems that contain producers, consumers, and decomposers. These components interact and maintain a natural balance by recycling nutrients and cleaning the system on their own.
An aquarium or swimming pool, however, is a human-made system that lacks sufficient decomposers and natural processes of purification. Hence, it needs to be cleaned regularly to remove waste and maintain a healthy environment.
Q21: Green plants occupy the first trophic level in every food chain because they: (1 Mark)
(a) exist over a large area. (b) have very less concentration of harmful chemicals. (c) have to feed large number of herbivores. (d) can synthesize food by photosynthesis.
Green plants are producers; they make their own food from inorganic substances using sunlight through photosynthesis, so they always occupy the first trophic level in a food chain.
Previous Year Questions 2024
Q1: For Q. Nos., two statements are given – One labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below: (1 Mark) (2024) Assertion (A): Accumulation of harmful chemicals is maximum in the organisms at the highest trophic level of a food chain. Reason (R): Harmful chemicals are sprayed on the crops to protect them from diseases and pests. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.
Ans: (b) Harmful chemicals do accumulate in organisms at the highest trophic level due to a process called biomagnification, and these chemicals are often sprayed on crops to protect them. However, the reason does not directly explain why the accumulation is highest at the top of the food chain, so it’s not a correct explanation of the assertion.
Q2: (A) Plants -> Deer -> Lion. (1 to 3 Marks) (2024) In the given food chain, what will be the impact of removing all the organisms of the second trophic level on the first and third trophic levels? Will the impact be the same for the organisms of the third trophic level in the above food chain if they were present in a food web? Justify. OR (B) A gas ‘X’ which is a deadly poison is found at the higher levels of the atmosphere and performs an essential function. Name the gas and write the function performed by this gas in the atmosphere. Which chemical is linked to the decrease in the level of this gas? What measures have been taken by an international organization to check the depletion of the layer containing this gas?
Number of plants/organisms of first trophic level will increase.
Number of lions/ organisms of third trophic level will decrease.
No
As the organisms of that level will find alternative foods and will not starve to death / food web is more stable where other animals as prey may be available.
OR
Gas ‘X’ is Ozone.
Ozone shields the surface of the earth from ultra-violet (UV) radiations from the sun.
CFCs (Chlorofluorocarbons)
Succeeded in forging an agreement to freeze CFC production at 1986 levels / Manufacturing of CFC free refrigerators
Q3: Name the term used for the materials which cannot be broken down by biological processes. Give two ways by which they harm various components of an ecosystem. (1 to 3 Marks) (2024)
Two ways: (i) They are inert and persist in the environment for long time and cause pollution. (ii) Cause Biological magnification (iii) Affect the fertility of soil
Q4: Use of pesticides to protect our crops affects organisms at various trophic levels especially human beings. Name the phenomenon involved and explain how does it happen. (1 to 3 Marks) (2024)
Pesticides are washed down into the soil and water bodies.
From the soil pesticides are absorbed by crop plants along with water and minerals and enter the food chain.
These chemicals are non-biodegradable and get accumulated progressively at each trophic level.
As human beings occupy the top level in any food chain, the maximum concentration of these chemicals gets accumulated in our bodies.
Q5: Consider the following statements about ozone: (1 to 3 Marks) (2024) (A) Ozone is poisonous gas. (B) Ozone shields the earth’s surface from the infrared radiation from the sun. (C) Ozone is a product of UV radiations acting on oxygen molecule. (D) At the lower level of the earth’s atmosphere, ozone performs most essential function. The correct statements are (a) (A) and (B) (b) (A) and (C) (c) (B) and (C) (d) (B) and (D)
Ans: (b) Ozone can be harmful at ground level, making it a poisonous gas. Statement (C) is also correct since ozone is formed when ultraviolet (UV) radiation interacts with oxygen molecules. However, statement (B) is incorrect because ozone protects the Earth from UV radiation, not infrared radiation, and statement (D) is misleading because ozone’s essential functions are primarily in the upper atmosphere, not at lower levels.
Q6: Assertion (A) and Reason (R), answer these questions selecting the appropriate option given below: (1 Mark) (2024) Assertion (A): The waste we generate daily may be biodegradable or non-biodegradable. Reason (R): The waste generated, if not disposed off properly may cause serious environmental problems. (a) Both (A) and (R) are true and (R) is the correct explanation of (A). (b) Both (A) and (R) are true and (R) is not correct explanation of (A). (c) (A) is true, but (R) is false. (d) (A) is false, but (R) is true.
The answer is (b) because both statements are true: waste can be either biodegradable (which can break down naturally) or non-biodegradable (which cannot), and improper disposal of waste can indeed lead to serious environmental issues. However, the reason does not explain why waste is categorized as biodegradable or non-biodegradable, so it’s not a direct explanation of the assertion.
Q7: What are decomposers? List two consequences of their absence in an ecosystem. (1 to 3 Marks) (2024)
Ans: Decomposers are the microorganisms that break-down the complex organic substances into simple inorganic substances. Consequences: (i) No replenishment of soil (ii) Foul smell (iii) Breeding of flies (iv) Accumulation of dead plants and animals in the environment. (v) No recycling of nutrients
Q8: We water the soil but it reaches the topmost leaves of the plants. Explain in brief the process involved. (1 to 3 Marks) (2024)
Ans: When water is lost through stomata in the leaves by transpiration, it creates a suction force/transpiration pull, due to which water is pulled up through xylem of the roots to the leaves.
Q9: Some wastes are given below: (1 to 3 Marks) (2024) (i) Garden waste (ii) Ball point pen refills (iii) Empty medicine bottles made of glass (iv) Peels of fruits and vegetables (v) Old cotton shirt The non-biodegradable wastes among these are: (a) (i) and (ii) (b) (ii) and (iii) (c) (i), (iv) and (v) (d) (i), (iii) and (iv)
Ans: (b) Non-biodegradable wastes are those that do not break down naturally in the environment. In the given options, ballpoint pen refills (ii) and empty medicine bottles made of glass (iii) are non-biodegradable because they can persist in the environment for a long time. The other items like garden waste, fruit and vegetable peels, and old cotton shirts decompose naturally and are considered biodegradable.
Q10: Two statements are given one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (1 Mark) (2024) Assertion (A): Oxygen is essential for all aerobic forms of life. Reason (R): Free oxygen atoms combine with molecular oxygen to form ozone. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.
Ans: (b) Assertion (A) correctly states that oxygen is essential for aerobic life forms, while Reason (R) explains a process involving oxygen and ozone formation, which is also true. However, Reason (R) does not explain why oxygen is essential for life, so the correct answer is (b): both are true, but Reason (R) does not explain Assertion (A).
Q11: Which one of the following is not a natural ecosystem? (1 Mark) (2024) (a) Pond ecosystem (b) Grassland ecosystem (c) Forest ecosystem (d) Cropland ecosystem
Ans: (d) A natural ecosystem is one that occurs naturally without human interference. Among the options given, a cropland ecosystem (d) is not natural because it is created and managed by humans for farming purposes. In contrast, pond, grassland, and forest ecosystems develop naturally in the environment.
Q12: Differentiate between food chain and food web. In a food chain consisting of deer, grass and tiger, if the population of deer decreases, what will happen to the population of organisms belonging to the first and third trophic levels? (3 Marks) (2024)
Population of grass/ first trophic level will increase.
Population of tiger/ third trophic level will decrease.
Q13: Identify the food chain in which the organisms of the second trophic level are missing: (1 Mark) (2024) (a) Grass, goat, lion (b) Zooplankton, Phytoplankton, small fish, large fish (c) Tiger, grass, snake, frog (d) Grasshopper, grass, snake, frog, eagle
In a food chain, the second trophic level usually consists of primary consumers that eat producers (like plants). In option (c), tiger, grass, snake, frog, there are no primary consumers (like a herbivore) between the grass (producer) and the snake. Instead, the snake directly feeds on frogs, which means the second trophic level is missing.
Q14: In which of the following organisms, multiple fission is a means of asexual reproduction? (1 Mark) (2024) (a) Yeast (b) Leishmania (c) Paramoecium (d) Plasmodium
Ans: (d) Multiple fission is a type of asexual reproduction where an organism divides into many parts at once. In this case, Plasmodium (d), which causes malaria, reproduces through multiple fission in its life cycle, producing many daughter cells simultaneously. The other options like yeast, Leishmania, and Paramecium reproduce differently.
Q15: A food chain will be more advantageous in terms of energy if it has: (a) 2 trophic levels (b) 3 trophic levels (c) 4 trophic levels (d) 5 trophic levels (1 Mark) (CBSE 2024)
Ans: (a) In a food chain, energy is lost at each trophic level due to metabolic processes, typically around 90% of energy is lost as heat, while only about 10% is transferred to the next level. Therefore, a food chain with fewer trophic levels will retain more energy available to the organisms at the higher level. With 2 trophic levels (e.g., producers and primary consumers), there is minimal energy loss compared to longer chains, making it more advantageous in terms of energy efficiency. Thus, the correct answer is (a) 2 trophic levels.
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Previous Year Questions 2023
Q1: Use of several pesticides which results in excessive accumulation of pesticides in rivers or ponds, is a matter of deep concern. Justify this statement. (3 Marks) (2023)
Ans: The use of several pesticides results in accumulation of pesticides in rivers and ponds. These chemicals are either washed down into the soil or into the water bodies. From the soil, these are absorbed by the plants along with water and minerals, and from the water bodies these are taken up by aquatic plants and animals and enters the food chain. As these chemicals are not degradable, these get accumulated progressively at each trophic level. As human beings occupy the top level in any food chain, the maximum concentration of these chemicals get accumulated in our bodies i.e., biological magnification. Our food grains such as wheat and rice, vegetables and fruits, and even meat, contain varying amounts of pesticide residues cannot always be removed by washing or other means and causes health hazards.
Q2: How is ozone formed in the higher levels of the atmosphere? “Damage to the ozone layer is a cause of concern.” Justify this statement. (3 Marks) (2023)
Ans: Ozone (O3) is a molecule formed by three atoms of oxygen. It is formed in the stratosphere layer of atmosphere when high energy UV rays act on O2 molecule splitting it into free oxygen (O) atoms. These atoms then combine with molecular oxygen (O2) to form ozone (O3).
Q3: “Although gardens are created by man but they are considered to be an ecosystem.” Justify this statement. (2 Marks) (2023)
Ans: In a garden, various plants like grasses, trees, flower bearing plants such as jasmine, sunflower, rose, and animals like insects, frogs and birds are found. All these living organisms interact with each other and their growth, reproduction and other activities are affected by the abiotic components such as light, water, wind, soil, minerals, etc. of ecosystem. Thus, a garden is considered to be an ecosystem.
Q4: What is the difference between biodegradable and non-biodegradable substances? List two methods of safe disposal of biodegradable domestic waste. (2 Marks) (2023)
Ans: Differences between biodegradable and non-biodegradable substances are as follows : Domestic waste can be safely disposed off by composting. In composting, biodegradable domestic wastes, such as left-over food, fruit, vegetable peels, etc., can be buried in pit, dug into ground. They are converted into compost and used as manure. In landfill a huge pit is made in an open low lying area and wastes are dumped into the pits. Once the pits are full, they are covered with soil and left for decomposition.
Q5: How do harmful chemicals get accumulated progressively at each trophic level in a food chain? (3 Marks) (CBSE 2023)
Ans: The process by which harmful substances enter the food chain and becomes concentrated at each trophic level is known as biomagnification. Example:Biomagnification of DDT in an Aquatic Food Chain
Q6: (A) Why does a kitchen garden called an artificial ecosystem while a forest is considered to be a natural ecosystem? (B) While designing an artificial ecosystem at home, write any two things to be kept in mind to convert it into a selfsustaining system. Give reason to justify your answer. (3 Marks) (CBSE 2023)
Ans: (A) Kitchen gardens are referred to as artificial ecosystems because they are man-made and contain adjusted abiotic and biotic components. It is an ecosystem where plants are produced, including fruits and vegetables, whereas a forest is a place where various creatures live in harmony and rely on one another for food and other necessities. As a result, a forest creates an ecosystem that can support itself. (B) Two things to keep in mind while designing an artificial ecosystem: (1) Balance between biotic and abiotic factors. (2) Recycling of nutrients and wastes. Reason: Abiotic factors include elements like temperature, light, water, and nutrients, whereas biotic factors are things like plants, animals, fungus, and microorganisms. Due of their interdependence, any changes to one of these components can have an impact on the ecosystem as a whole. To support the development and survival of all organisms within the ecosystem, it is crucial to maintain a balance between various components in the ecosystem. Recycling is essential for maintaining the health and sustainability of the ecosystem.
Q7: (A) Construct a food chain of four trophic levels comprising the following:
Hawk, snake, plants, rat
(B) 20,000 J of energy was transferred by the producers to the organism of second trophic level. Calculate the amount of energy that will be transferred by organisms of the third trophic level to the organisms of the fourth trophic level. (3 Marks) (2023)
Ans: (A) The flow of nutrients and energy from one organism to another at different trophic levels forms a food chain. Plants → Rat → Snake → Hawk (B) Only 10% of energy will be transferred to the next trophic level, according to the 10% law of energy transfer, and 90% will be wasted as heat and incomplete digestion. According to this law, (1) Energy is transferred from producers to the second trophic level = 20,000 J. (2) Energy moved from the second to the third trophic level: 10% of 20,000 J = 2,000 J. (3) Energy moved from the third to the fourth trophic level is equal to: 10% of 2,000 J = 200 J
Previous Year Questions 2022
Q1: (i) Why are crop fields considered as artificial ecosystems? (ii) Write a common food chain of four steps operating in a terrestrial ecosystem. (2022 C)
Ans: (i) Crop fields are the artificial ecosystems because in crop fields, both biotic (living) and abiotic (non-living) components are manipulated by human beings. Humans can change edaphic factors by adding fertilisers, water, etc. Biotic components may be changed using biocides or adding useful organisms like earthworms etc. (ii) A food chain consists of various organism at various trophic levels. In terrestrial ecosystem, a common food chain is Grass Grasshopper→ Frog→ Snake
Q2: (i) List two human-made ecosystems. (ii) “We do not clean a pond in the same manner as we do in an aquarium.” Give reason to justify this statement. (Term II, 2021-22)
Ans: (i) The two human made ecosystems are aquarium and garden. (ii) We do not clean pond as we do in an aquarium because the waste generated in a pond is acted upon by the decomposers which convert them into simple soluble substances, whereas, in aquarium, the waste gets mixed with water and left untreated due to absence of decomposers.
Q3: In the following food chain, only 2J of energy was available to the peacocks. How much energy would have been present in Grass? Justify your answer. Grass → Grasshopper → Frog → Snake Peacock (Term II, 2021-22)
Ans: In the given food chain, 20,000 J of energy must have been present in grass. This is because, as per the 10% law of energy transfer, only 10% of energy is transferred to the next trophic level.
Q4: What are human-made ecosystems? Give an example. Can a human-made ecosystem become a self-sustaining ecosystem? Give reason to justify your answer. (2022)
Ans: Artificial ecosystems area unit human-made structures wherever organic phenomena and abiotic elements area unit created to act with one another for survival. it’s not self-sufficient and might decrease while not human facilitated. samples of artificial ecosystems embrace aquariums, agriculture fields, zoos, etc.
A system, that is formed and maintained by mortals, is named synthetic | a synthetic} or man-made system. Some samples of artificial ecosystems area unit aquariums, gardens, agriculture, apiary, poultry, piggery, etc.
Man-made ecosystems don’t seem to be self-sufficient as a result they rely on naturally created ecosystems just like the water bodies. just like the crop fields that could be an artificial system that depends on water bodies like rivers, and groundwater for water and life. Same approach gardens conjointly rely on nature for their property.
All organisms like plants, animals, microorganisms, and mortals further because the physical surroundings act with one another and maintain a balance in nature. All interacting organisms in a locality in conjunction with the non-living constitute the setting kind associate degree system.
All the organic phenomenon elements comprising living organisms and abiotic elements comprising physical factors like temperature, rainfall, soil, etc create an associate degree system.
An artificial system could be a variety of systems created by man by artificial means and not present sort of a forest, ponds, lakes, etc. samples of artificial ecosystems area unit crop fields and gardens.
Man-made ecosystems don’t seem to be self-sufficient as a result they rely on naturally created ecosystems just like the water bodies. just like the crop fields that could be an artificial system that depends on water bodies like rivers, and groundwater for water and life. Same approach gardens conjointly rely on nature for their property.
Hence, artificial ecosystems don’t seem to be self-sufficient.
Q5: (a) Name the group of organisms which form in the first trophic level of all food chains. Why are they called so? (b) Why are the human beings most adversely affected by biomagnification? (c) State one ill-effect of the absence of decomposers from a natural ecosystem. (2022)
Ans: (a) Producers form the first trophic level of all food chains. They are called producers because they are autotrophic organisms which alone are able to manufacture organic food from inorganic raw materials by the process of photosynthesis. They capture sun’s energy and convert it into chemical energy. The chemical energy is used in combining raw materials into organic food. This food is used up by themselves and rest enters the food chains as food for consumers. (b)Human beings are most adversely affected by biomagnification because they occupy the highest trophic level in any food chain. As in biomagnification, successive concentration of non-biodegradable substances increases in the trophic level of food chains, so, it leads to most toxicity at highest trophic level. Hence, maximum concentration of chemicals get accumulated most in their body. (c) Absence of decomposers will lead to the accumulation of dead remains and waste products of organisms in our natural ecosystem. The decomposers breakdown complex organic substances into simple inorganic substances, so, that it can go into the soil and can be used up by plants.
Q6: What is ozone? How is it formed in the upper layers of the earth’s atmosphere? How does ozone affect our ecosystem? (2022)
Ans: Ozone (O3) is a molecule formed by three atoms of oxygen. It is formed in the stratosphere layer of atmosphere when high energy U V rays act on O2 molecule splitting it into free oxygen (O) atoms. These atoms then combine with molecular oxygen (O2) to form ozone (O3).
Ozone shields the surface of the earth from U V radiations from the sun. The depletion of ozone layer will lead to global warming and some serious health issues such as damage of skin cells that leads to skin cancer, snow blindness or inflammation of cornea, increased fatality of young animals, mutations and reduced immunity.
Q7: (a) We do not clean ponds or lakes, but an aquarium needs to be cleaned regularly. Why? (b) Why is ozone layer getting depleted at the higher levels of the atmosphere? Mention one harmful effect caused by its depletion. (2022)
Ans: (a) We do not clean pond as we do in an aquarium because the waste generated in a pond is acted upon by the decomposers which convert them into simple soluble substances, whereas, in aquarium, the waste gets mixed with water and left untreated due to absence of decomposers. (b) The ozone layer is getting depleted at the higher levels of the atmosphere due to use of chlorofluorocarbons (CFCs) which are used in refrigerator. Other ozone depleting substances include carbon tetrachloride, hydrofluorocarbons used in fire extinguisher, air i conditioners, etc. Due to the ozone layer depletion, humans will be directly exposed to the ultraviolet radiations of sun. This will result in serious health issues like skin cancer, sunburns, quick ageing, mutations and weak immune system.
Q8: Kulhads (disposable cups made of day) and disposable paper cups both a re used as an alternative For disposable plastic cups. Which one of these two can be considered as a better alternative to plastic cups and why? (2022)
This was brought in use to replace the plastic cup in trains.
Plastic is a nonbiodegradable harmful substance that does not decompose in nature and affects the ecosystem or environment negatively.
Kulhads are made of biodegradable soil, therefore, this was used to replace the plastic and protect the environment and health of humans.
Discontinuation of kulhads:
Since kulhads are made using clay, it is a practice of harming the environment too.
The clay is fertile soil and kulhads were discontinued to avoid its reduction.
Making of kulhad on a large scale to serve tea passengers in the train was leading to the loss of this top fertile soil.
Q9: (a) What is meant by garbage? List two classes into which garbage is classified. (b) What do we actually mean when we say that “enzymes are specific in their action”? (2022)
Ans: (a) Garbage is the waste material (rubbish) especially of domestic refuse. The two classes into which garbage is classified are (i)Biodegradable (ii)Non-biodegradable. (b) Enzymes are specific in their action. For example, enzyme maltase acts on sugar maltose but not on lactose or sucrose. Different enzymes may act on the same substrate but give rise to different products. Similarly an enzyme may act on different substrates producing different end products.
Also read: Garbage, Waste Management & Depletion of the Ozone Layer
Previous Year Questions 2021
Q1: What are consumers? Name the four categories under which the consumers are further classified. (2021 C)
Ans: Consumers are the organisms which are unable to synthesise their own food. Therefore, they utilise materials and energy stored by the producers or eat other organisms. They are known as the heterotrophs. The consumers are of following categories: (i) Primary or first-order consumers: These include the animals which eat plants or plant products. They are called herbivores or primary (first order) consumers. E.g., Cattle, deer, goat, rabbit, hare, rats, mice, grasshoppers etc (ii) Secondary or second order consumers: These include the animals which depend on primary consumers for their food. They are called primary carnivores or secondary (second order) consumers. Secondary consumers can be carnivores or omnivores. E.g., Cats, dogs, foxes, small fish, etc. (iii) Tertiary or third order consumers: These are large carnivores (or top carnivores) which feed on primary and secondary consumers. These are termed as secondary carnivores or tertiary (third order) consumers. Common examples include shark and crocodile, wolves, lion, etc. (iv) Quaternary or fourth order consumers: These are even larger carnivores which feed on secondary carnivores (tertiary consumers). E.g., Tigers, lions and eagles/hawks etc.
Previous Year Questions 2020
Q1: How is ozone layer formed? State its importance to all life forms on earth. Why the amount of ozone in the atmosphere dropped sharply in the 1980s? (2020)
Ans: When high energy ultraviolet radiations react with oxygen present in stratosphere (the higher level of atmosphere) it splits into its constituent atoms. Since these atoms produced are very reactive, they react with molecular oxygen (O2) to form ozone (O3). Ozone shields the surface of the earth from UV radiations from the sun. The depletion of ozone layer will lead to global warming and some serious health issues such as damage of skin cells that leads to skin cancer, snow blindness or inflammation of cornea, increased fatality of young animals, mutations and reduced immunity. In 1980s, the production of CFCs increased which releases active chlorine in the atmosphere. The active chlorine then reacts with ozone molecules present there to convert them to oxygen. This results in thinning of ozone layer. CFCs are used as refrigerants and in fire extinguishers. That is why, amount of ozone in the atmosphere dropped sharply.
Q2: (a) Write two harmful effects of using plastic bags on the environment. Suggest alternatives to the usage of plastic bags. (b) List any two practices that can be followed to dispose off the waste produced in our homes. (2020)
Ans: (a) Two harmful effects of using plastic bags on the environment: (i) Plastic bags are non-biodegradable substances which are not acted upon by microbes. So, they cannot be decomposed and therefore persist in the environment for a long time causing harm to the soil fertility and quality. (ii) Plastic bags choke drains which result in waterlogging, that allows breeding of mosquitoes and hence leads to various diseases. Jute bags and cloth bags are the alternatives to the polyethene bags. (b) Practices that can be followed to dispose off the waste produced in our homes: (i) Separation of biodegradable and non-biodegradable wastes. (ii) The biodegradable waste can be converted to manure. (iii) Non-biodegradable waste should be disposed off at suitable places from where municipal authorities can pick them up and dispose properly and scientifically. (iv) Use discarded bottles and jars to store food items.
Q3: (A) Construct a terrestrial food chain comprising four trophic levels. (B) What will happen if we kill all the organisms in one trophic level? (C) Calculate the amount of energy available to the organisms at the fourth trophic level if the energy available to the organisms at the second trophic level is 2000 J. (CBSE 2020)
Ans: (A) A terrestrial food chain comprising of four trophic levels: Grass → Grasshopper → Frog → Snake (B) If we kill all the organisms in one trophic level then the transfer of food energy to next level will stop. Organisms of previous trophic level will also increase. For Example: If all herbivores in an ecosystem are killed, there will be no food available for the carnivores of that area. Consequently, they will also die or will shift to other areas. Populations of producers will also increase in absence of herbivores causing an imbalance in the ecosystem. (C) Consider the same food chain that we have made i.e., (A), Grass → Grasshopper → Frog → Snake In this food chain, organism at the second trophic level is grasshopper and the energy available at this trophic level is 2000 J. According to 10% law, 10% of energy will be available to frog (Third trophic level) which is 200 J. The energy available to the snakes will be available as 10% of 200 J. Thus, the energy available to the snake is 20 J. The 10% law states that during transfer of energy from one trophic level to the next trophic level, only about 10% of energy is available to the higher trophic level. To summarise:
Ans: (a) The number of trophic levels in a food chain are limited because at each trophic level only 10% of energy is utilised for the maintenance of organism which occur at that trophic level and the remaining large portion is lost as heat. As a result, organisms at each trophic level pass on lesser energy to the next trophic level, than they receive. The longer the food chain, the lesser is the energy available to the final member of food chain. (b) Biological magnification is characterised by the increase in the non-biodegradable substances (DDT, Hg, etc.) in successive trophic levels of a food chain. The level of such toxic substances will be different in different trophic levels of a food chain because these substances are accumulated more in higher trophic levels.
Q2: Define an ecosystem. Draw a block diagram to show the flow of energy in an ecosystem. (Delhi 2019)
Ans: An ecosystem is defined as a structural and functional unit of the biosphere. It comprises of living organisms and their non-living environment that interact by means of food chains and biogeochemical cycles resulting in energy-flow, biotic diversity and material cycling to form stable self-supporting system. Green plants capture about 1% of the solar energy incident on the earth to carry out the process of photosynthesis. A part of this trapped energy is used by plants in performing their metabolic activities and some energy is released as heat into the atmosphere. The remaining energy is chemical energy stored in the plants as photosynthetic products. When these green plants are eaten up by herbivores, the chemical energy stored in the plants is transferred to these animals. These animals (herbivores) utilise some of this energy for metabolic activities and some energy is released as heat while the remaining energy is stored in their body. This process of energy transfer is repeated till top carnivores. In an ecosystem, transfer of energy follows 10 per cent law, i.e., only 10 per cent of the energy is transferred to each trophic level from the lower trophic level. The given block diagram shows unidirectional flow of energy at different trophic levels in a freshwater ecosystem:
Q3: (a) How can we help in reducing the problem of waste disposal? Suggest any three methods. (Delhi 2019) (b) Distinguish between biodegradable and nonbiodegradable wastes. (DoE, A1 2011)
Recycling: solid, wastes like paper, plastics, metals can be sent to processing factories where they are remoulded or reprocessed to new materials.
Production of compost: Biodegradable wastes like fruit and vegetable peels, plant products, left over food, grass clippings, human and animal waste can be converted into compost by burying this waste into grund and can be used as manure.
Incineration: Burning dawn many household waste, chemical waste and biological waste into ash is known as incineration. A large amount of waste can be easily converted into ash which can be disposed off in landfill.
(b) Differences between:
Previous Year Questions 2017
Q1: (a) What is an ecosystem? List its two main components. (b) We do not clean ponds or lakes, but an aquarium needs to be cleaned regularly. Explain. (CBSE 2013,2017)
Ans: (a) A self-sustaining functional unit consisting of living and non-living components is called an ecosystem. Components: Biotic components like plants and animals. Non-biotic components like soil, wind, light etc. (b) A pond is a complete, natural and self-sustaining ecosystem whereas an aquarium is an artificial and incomplete ecosystem, without decomposers therefore it needs regular cleaning for proper running.
Q2: You have been selected to talk on “ozone layer and its protection” in the school assembly on ‘Environment Day.’ (Delhi 2017) (a) Why should ozone layer be protected to save the environment? (b) List any two ways that you would stress in your talk to bring in awareness amongst your fellow friends that would also help in protection of ozone layer as well as the environment.
Ans: (a) Ozone layer at the higher levels of the atmosphere, acts as a shield to protect earth from the harmful effects of the ultraviolet (UV) radiations; hence, it should be protected. (b)
Urging the people to not to buy aerosol products with CFC that are available in the market.
Conducting poster making competition or street plays presenting the importance of ozone layer on earth.
Q3: Your mother always through that fruit juices are very healthy for everyone. One day she read in the newspaper that some brands of fruit juices in the market have been found to contain certain level of pesticides in them. She got worried as pesticides are injurious to our health. (Foreign 2017) (а) How would you explain to your mother about fruit juices getting contaminated with pesticides? (b) It is said that when these harmful pesticides enter our body as well as in the bodies of other organisms they get accumulated and beyond a limit cause harm and damage to our organs. Name the phenomenon and write about it.
Pesticides are the chemicals used to protect our crops from diseases and pests.
These chemicals are washed down either into the soil or into water bodies.
From the soil, they are absorbed by the terrestial plants along with water and minerals.
From the water bodies, they are absorbed by the aquatic plants.
When the fruits of these plants are used to prepare fruit juices, they are contaminated with the pesticides.
(b)
The phenomenon is called biomagnification. It is the phenomenon in which certain harmful chemicals enter the food chain and get accumulated and increase in concentration at successive trophic levels.
It is because they are not degradable.
The maximum concentration of these chemicals is found in the top level consumers.
Q4: In the following food chain, 100 J of energy is available to the lion. How much energy was available to the producer? (AI 2017)
Ans: The biodegradable and non-biodegradable wastes must be discarded in two different dustbins because biodegradable wastes gets decomposed by the microorganisms whereas non-biodegradable wastes can be recycled and reused.
Q6: Name any two man-made ecosystems. (Foreign 2017)
Ans: Decomposers are microorganisms that derive their nutrition from dead remains and waste products of organisms. They play a vital role in our environment by breaking down the complex organic substance into simple inorganic substance which is made available for plants and other organisms. Hence they act as scavengers and not only keep the environment clean but also replenish the minerals.
Q3: We often use the word environment. What does it mean? (Foreign 2016)
Ans: It is the sum total of all external conditions and influences that affect the life and development of an organism, i.e. the environment includes all the physical or abiotic and biological or biotic factors.
Q4: Why are green plants called producers? (Delhi 2016)
Ans: Energy available at each successive trophic level of food chain is ten per cent of that at the previous level. This is called ten per cent law. Thus, 90 per cent energy is lost to the surroundings at each trophic level. However, plants absorb only one per cent of radiant energy of the Sun during photosynthesis. This is explained as under :
Q2: What is ozone? How and where is it formed in the atmosphere? Explain how does it affect ecosystem. (Foreign 2015)
Ans: Ozone is an isotope of oxygen, i.e. it is a molecule formed by 3 atoms of oxygen. Ozone exists in the ozone layer of stratosphere. At higher level o f atmosphere, O2 molecule breaks down to 2 oxygen atom. The oxygen atom then combines with the oxygen molecule to form ozone. Ozone layer in the atmosphere prevents UV rays from reaching earth. Exposure to excess UV rays causes skin cancer, cataract and damages eye and immune system. It also decreases crop yield and reduces population of phytoplankton, zooplankton and certain fish larvae which are an important constituent of aquatic food chain. It also disturbs rainfall, causing ecological disturbance and reduces global food production. Thus, it affect the ecosystem.
Q3: “Energy flow in food chains is always unidirectional.” Justify this statement. Explain how the pesticides enter a food chain and subsequently get into our body. (Foreign 2015)
Ans: The energy flow through different steps in the food chain is unidirectional. The energy captured by autotrophs does not revert back to the solar input and it passes to the herbivores, i.e. it moves progressively through various trophic levels. Thus energy flow from sun through producers to omnivores is in single direction only. Pesticides are sprayed to kill pests on food plants. The food plants are eaten by herbivores and alongwith the food, pesticides are also eaten by the herbivores. Herbivores are eaten by carnivores and alongwith the herbivore animal, pesticide also enters the body of the carnivore. Man eat both plants and animals and pesticide alongwith food enters the body of human. Concentration of pesticides increases as we move upward in the food chain and the process is called bio-magnification.
Q4: What is an ecosystem? List its two main components. We do not clean natural ponds or lakes but an aquarium needs to be cleaned regularly. Why is it so? Explain. (AI 2015)
Ans: Ecosystem: It is the structural and functional unit of biosphere, comprising of all the interacting organisms in an area together with the non-living constituents of the environment. Thus, an ecosystem is a self-sustaining system where energy and matter are exchanged between living and non-living components. Main components of ecosystem: (i) Biotic Component: It means the living organisms of the environment-plants, animals, human beings and microorganisms like bacteria and fungi, which are distinguished on the basis of their nutritional relationship. (ii) Abiotic Component: It means the non-living part of the environment-air, water, soil and minerals. The climatic or physical factors such as sunlight, temperature, rainfall, humidity, pressure and wind are a part of the abiotic environment. An aquarium is an artificial and incomplete ecosystem compared to ponds or lakes which are natural, self-sustaining and complete ecosystems where there is a perfect recycling of materials. An aquarium therefore needs regular cleaning.
Q5: Write the full name of the group of compounds mainly responsible for the depletion of ozone layer. (Foreign 2015)
Ans: Herbivores are always at the 2nd trophic level.
Q7: The following organisms form a food chain. Which of these will have the highest concentration of nonbiodegradable chemicals? Name the phenomenon associated with it. Insects, Hawk, Grass, Snake, Frog. (Foreign 2015)
Ans: Only green plants can make their own food from sunlight. Green plants, therefore, always occupy the 1st trophic level in a food chain.
Q9: What will be the amount of energy available to the organism of the 2nd trophic level of a food chain, if the energy available at the first trophic level is 10,000 joules? (AI 2015)
Ans: 100 Joules of energy will be available to the organism of the 2nd trophic level.
Q10: (a) What is biodiversity? What will happen if biodiversity of an area is not preserved? Mention one effect of it. (AI 2015) (b) With the help of an example explain that a garden is an ecosystem. (c) Why only 10% energy is transferred to the next trophic level?
Ans: (a) Biodiversity is the existence of a wide variety of species of plants, animals and microorganisms in a natural habitat within a particular environment or existence of genetic variation within a species. Biodiversity of an area is the number of species or range of different life forms found there. Forests are ‘biodiversity hotspots’. Every living being is dependent on another living being. It is a chain. If biodiversity is not maintained, the links of the chain go missing. If one organism goes missing, this will affect all the living beings who are dependent on it. (b) A garden comprises of different kind of flora and fauna such as grasses, flowering and nonflowering plants, trees, frogs, insects, birds, etc. All these living organisms depend and interact with each other and their growth, reproduction and other vital biological activities depend upon the abiotic component comprising of physical factors like temperature, rainfall, wind, soil and minerals. Therefore, we can say that a garden is an ecosystem. (c) Only 10% energy is transferred to the next trophic level because other 90 per cent is used for things like respiration, digestion, running away from predators.
Q11: Differentiate between biodegradable and non-biodegradable substances with the help of one example each. List two changes in habits that people must adopt to dispose non-biodegradable waste, for saving the environment. (CBSE 2015)
Ans: Biodegradable substances: These can be broken down into simpler substances by nature/ decomposers/bacteria/saprophytes/ saprobionts. Example: Human Excreta/Vegetable peels, etc. Non-biodegradable substances: These can’t be broken down into simpler substances by nature/decomposers. Example: Plastic/glass (or any other) (Any one) Habits: (1) Use of separate dustbins for iodegradable and non-biodegradable waste. (2) Recycling of waste. (3) Use of cotton/jute bags for carrying vegetables etc
Q1:Two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below : (1 Mark)
Assertion (A): No two magnetic field lines are found to cross each other. Reason (R): The compass needle cannot point towards two directions at the point of intersection of two magnetic field lines. (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Ans: (A) Both A and R are true, and R is the correct explanation of A.
Assertion (A): Magnetic field lines represent the direction and strength of the magnetic field. They never cross each other because at any point in space, the magnetic field has a unique direction. If two field lines crossed, it would imply two different directions for the magnetic field at that point, which is physically impossible. Thus, A is true.
Reason (R): A compass needle aligns itself along the magnetic field direction at a given point. If field lines intersected, the needle would have to point in two directions simultaneously, which is not possible. This explains why magnetic field lines do not cross. Thus, R is true and correctly explains A.
Conclusion: Option (A) is correct.
Q2:Two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below : (1 Mark) Assertion (A): The pattern of the magnetic field of a solenoid carrying a current is similar to that of a bar magnet. Reason (R): The pattern of the magnetic field around a current-carrying conductor is independent of the shape of the conductor. (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Assertion (A): A current-carrying solenoid produces a magnetic field similar to that of a bar magnet, with one end acting as the north pole and the other as the south pole. The field lines emerge from one end (north) and enter the other (south), resembling a bar magnet’s field. Thus, A is true.
Reason (R): The magnetic field pattern around a current-carrying conductor depends on its shape. For example, a straight conductor produces concentric circular field lines, while a solenoid produces a bar magnet-like field. Thus, R is false as the field pattern is shape-dependent.
Conclusion: Option (C) is correct, as A is true, but R is false and does not explain A.
Q3:Which one of the following statements is not true about a bar magnet? (1 Mark) (a) It sets itself in north-south direction when suspended freely. (b) It has attractive power for iron filings. (c) It produces magnetic field lines. (d) The direction of magnetic field lines inside a bar magnet is from its north pole to its south pole.
Ans: (d) The direction of magnetic field lines inside a bar magnet is from its north pole to its south pole.
Option (A): True. A freely suspended bar magnet aligns itself in the north-south direction due to Earth’s magnetic field, with its north pole pointing toward the geographic north.
Option (B): True. A bar magnet attracts ferromagnetic materials like iron filings due to its magnetic field.
Option (C): True. A bar magnet produces magnetic field lines that emerge from the north pole and enter the south pole, forming closed loops.
Option (D): False. By convention, magnetic field lines outside a bar magnet go from the north pole to the south pole, but inside the magnet, they travel from the south pole to the north pole to form continuous loops.
Conclusion: Option (D) is not true.
Q4:The strength of the magnetic field produced inside a long straight current-carrying solenoid does not depend upon: (1 Mark) (A) Number of turns in the solenoid (B) Direction of current flowing through the solenoid (C) Material of the core filled inside the solenoid (D) Magnitude of current in the solenoid
Ans: (B) Direction of current flowing through the solenoid
For a long current-carrying solenoid, the magnetic field inside is Hence, B depends on:
Number of turns per unit length n (more turns → stronger field),
Core material via μr (e.g., iron increases B),
Magnitude of current I (largerI → larger B).
Changing the direction of current only reverses the polarity (which end is north/south) but does not change the magnitude of BB. Therefore, the strength of the magnetic field does not depend on the direction of current. (Option B).
Q5: (i) The given figure shows the current passing through the straight conductor XY.
(ii) Name and state the rule used in determining the direction of the magnetic field lines in the situation given above. (iii) State Fleming’s left-hand rule. Using this rule, determine the direction of force applied on an electron entering a uniform magnetic field as shown in the figure. (5 Marks)
i) Magnetic Field Lines Due to Current in Conductor XY
When current flows from X to Y in the straight conductor XY, the magnetic field around the wire consists of concentric circles centred on the wire. By the Right-hand thumb rule (thumb along current X→Y; curled fingers give field direction), the field is clockwise around the wire as seen from the X→Y end.
ii) Rule Used to Determine Magnetic Field Direction
Rule Name: Right-Hand Thumb Rule (Maxwell’s Corkscrew Rule) Statement: If you hold a straight conductor in your right hand so that your thumb points in the direction of current, the direction in which your fingers wrap around the conductor gives the direction of the magnetic field lines.
iii) Fleming’s Left Hand Rule and Direction of Force on Electron
Fleming’s Left Hand Rule Statement: Stretch the thumb, forefinger, and middle finger of your left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the middle finger in the direction of current (conventional, positive to negative), then the thumb gives the direction of force (motion) on the conductor.
Magnetic field: to the right.
Electron velocity: upwards ⇒ conventional current is downwards.
Set forefinger → right (B), middle finger → down (I).
Thumb then points out of the plane of the paper (towards the observer).
Therefore, the force on the electron is out of the page, perpendicular to both (For a positive charge it would be into the page.)
Q6: (a) Name and state the rule which determines the force on a current-carrying conductor placed in a uniform magnetic field. (b) Consider three diagrams in which the entry of a positive charge (+Q) in a magnetic field is shown. Identify the case in which the force experienced by the charge is (i) maximum, and (ii) minimum. (3 Marks)
Ans: (a) Rule: Fleming’s left-hand rule. (b) (i) Maximum force: When velocity is perpendicular to the magnetic field. (ii) Minimum force: When velocity is parallel to the magnetic field.
(a) Rule:
Name: Fleming’s left-hand rule.
Statement: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular. If the forefinger points in the direction of the magnetic field (B) and the middle finger in the direction of the current (I), the thumb points in the direction of the force (F).
(b) Force on a positive charge:
The force on a moving charge is given by F = qvB sinθ, where q is the charge, v is velocity, B is the magnetic field strength, and θ is the angle between v and B.
(i) Maximum force: F is maximum when sinθ = 1, i.e., θ = 90° (velocity perpendicular to the magnetic field). In the diagram where +Q’s velocity is perpendicular to B, the force is maximum.
(ii) Minimum force: F is minimum (zero) when sinθ = 0, i.e., θ = 0° or 180° (velocity parallel or antiparallel to the magnetic field). In the diagram where +Q’s velocity is parallel to B, the force is zero.
Conclusion: Maximum force at θ = 90°, minimum at θ = 0° or 180°.
Q7:What are magnetic field lines? List two important properties of magnetic field lines. (2 Marks)
Magnetic field lines: These are visual representations of the magnetic field, showing the direction a north pole would move and indicating field strength by their closeness.
Properties:
Closed loops: Field lines emerge from the north pole, travel outside to the south pole, and continue inside from south to north, forming continuous loops.
Non-intersecting: No two field lines cross, as the magnetic field has a unique direction at each point. Intersection would imply multiple directions, which is impossible.
Q8:In order to obtain magnetic field lines around a bar magnet, a student performed an experiment using a magnetic compass and a bar magnet. The magnet was placed on a sheet of white paper fixed on a drawing board. Using a magnetic needle, he obtained on the paper a pattern of magnetic field lines around the bar magnet. (a) By convention, the field lines emerge from the north pole and merge at the south pole. Why? Give reason. (b) State the relationship between the strength of the magnetic field and the degree of closeness of the field lines. (c) (A) (i) No two field lines can ever intersect each other. Give reason. (ii) The magnetic field in a given region is uniform. Draw a diagram to represent it. (4 Marks)
(a) Why field lines emerge from north and merge at south:
By convention, magnetic field lines show the direction a free north pole would move. The north pole of a compass is attracted to the south pole of a magnet and repelled by its north pole. Thus, field lines are drawn emerging from the north pole and merging at the south pole, forming closed loops (inside the magnet, they go from south to north).
(b) Relationship:
The strength of the magnetic field is proportional to the density of field lines. Where field lines are closer together (e.g., near the poles), the field is stronger; where they are farther apart, the field is weaker.
(c) (A):
(i) No intersection: Magnetic field lines do not intersect because the magnetic field at any point has a unique direction. If lines crossed, it would imply two different directions at the intersection point, which is physically impossible, as a compass needle can only point in one direction.
(ii) Uniform magnetic field:
Diagram: Draw parallel, equally spaced straight lines with arrows in the same direction (e.g., left to right). This represents a uniform magnetic field, as seen inside a solenoid or between flat magnet poles.
Conclusion: As described above.
OR
(c) (B) Draw the pattern of the magnetic field lines through and around a current carrying solenoid. What does the pattern of field lines inside the solenoid represent ?
The magnetic field pattern of a current-carrying solenoid resembles that of a bar magnet. One end of the current-carrying solenoid behaves as a magnetic north , while the other behaves as the magnetic south just like in the bar magnet.
Q9: (a) What are magnetic field lines? How is the direction of the magnetic field at a point determined? Draw the pattern of magnetic field lines of the magnetic field produced by a current-carrying circular loop. Mark on it the direction of (i) current and (ii) magnetic field lines. Name the two factors on which the magnitude of the magnetic field due to a current-carrying coil depends.
Magnetic field lines: These are imaginary lines that indicate the direction a north pole would move and show field strength by their density.
Direction determination: Place a small compass needle at the point; the needle’s north pole points in the direction of the magnetic field at that point.
Factors affecting field magnitude: B = (μ₀NI)/(2r).
Current (I): Higher current increases the field strength.
Number of turns (N): More turns increase the field strength.
Q10: (a) Study the following electric circuit diagram and answer the questions that follow : (i) What does the circuit diagram show?
(ii) What will happen if the direction of current is reversed? Justify your answer giving a circuit diagram. (b) Name and state the rule to determine the direction of magnetic field associated with a straight current carrying conductor. (5 Marks)
(a) (i) The circuit diagram shows a current-carrying conductor placed near a compass needle. The conductor appears to form a trapezoidal loop with a key (K) to control the current flow. The compass needle, positioned near the conductor, is used to detect the magnetic field generated by the current. This setup is typically used to demonstrate the relationship between electric current and the magnetic field it produces, such as in an experiment to verify the magnetic effect of current.
(a) (ii) Reversing current:
Effect: Reversing the current direction reverses the force direction on the conductor, per Fleming’s left-hand rule. For example, if the original force is upward, it becomes downward.
Justification: The force direction depends on the current direction (middle finger in Fleming’s rule). Reversing the current (e.g., by swapping battery terminals) changes the middle finger’s direction, reversing the thumb (force).
(b) Rule:
Name: Right-hand thumb rule.
Statement: Hold the conductor with the right hand, thumb pointing in the current direction. The curled fingers indicate the magnetic field direction (concentric circles around the conductor).
OR
(a) Draw the pattern of magnetic field lines of (i) a bar magnet (ii) a current carrying solenoid. List two distinguishing features between the two magnetic fields. (b) Study the following three diagrams in which the entry of an electron in a magnetic field is shown. Identify the case in which the magnetic force experienced by the electron is (i) maximum, and (ii) minimum. Give reason for your answers in each case. (5 Marks)
Two distinguishing features between the two fields:
The magnetic field of the solenoid can be varied as per our requirements just by changing the current or core of the solenoid whereas the magnetic field of the bar magnet is fixed.
The magnetic field outside the solenoid is negligible as compared to the bar magnet.
(b) Let’s analyze each case using the magnetic force on a moving charge:
F=q v B sin θ
Where:
q = charge of particle (electron has negative charge) v = velocity of electron B = magnetic field strength θ = angle between velocity of electron and magnetic field
Case (A): Electron velocity is perpendicular to magnetic field (θ = 90∘)
sin90∘=1
Force is maximum.
Case (B): Electron velocity is at an angle (oblique) to magnetic field (0° < θ < 90°).
sinθ is less than 1 but not zero.
Force is less than maximum but not zero.
Case (C): Electron velocity is parallel (or anti-parallel) to magnetic field (θ = 0∘ or 180∘)
Sin 0∘= 0 or sin 180∘= 0.
Force is minimum (zero).
Q11: Answer the following questions for a case in which a current carrying conductor is placed in a uniform magnetic field : (a) List three factors on which the magnitude of the force acting on a current-carrying conductor in a uniform magnetic field depends. (b) When is the magnitude of the force on the conductor maximum? (c) Name the rule which helps in determining the direction of force on the conductor and give its one application. (3 Marks)
(a) Factors: The force on a current-carrying conductor is given by F = BIL sinθ.
Current (I): Higher current increases the force.
Magnetic field strength (B): Stronger field increases the force.
Length (L): Longer conductor in the field increases the force. [Note: The angle θ also affects the force, but the question asks for three factors.]
(b) Maximum force: F = BIL sinθ is maximum when sinθ = 1, i.e., θ = 90° (conductor perpendicular to the magnetic field).
(c) Rule and application:
Name: Fleming’s left-hand rule.
Statement: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular. Forefinger points to the magnetic field (B), middle finger to the current (I), and thumb to the force (F).
Application: In an electric motor, the rule determines the direction of force on the current-carrying coil, causing it to rotate in the magnetic field, converting electrical energy to mechanical energy.
Q12: Why can’t two magnetic field lines cross each other? Draw magnetic field lines showing the direction of the magnetic field due to a current-carrying long straight solenoid. State the conclusion which can be drawn from the pattern of magnetic field lines inside the solenoid. Name any two factors on which the magnitude of the magnetic field due to this solenoid depends. (5 Marks)
If two magnetic field lines will cross each other, then there will be two directions of magnetic field at the point of crossing, which is not possible in a magnetic field. The field around a solenoid is like that of a bar magnet.
Diagram:
Conclusion: The field lines inside the solenoid are straight, parallel, and equally spaced, indicating a uniform magnetic field, similar to a bar magnet’s interior field.
Factors: The magnetic field inside a solenoid is B = μ₀nI.
Current (I): Higher current increases the field strength.
Number of turns per unit length (n): More turns per unit length increase the field strength.
Previous Year Questions 2024
Q1: (i) Two magnetic field lines do not intersect each other. Why? (ii) How is a uniform magnetic field in a given region represented? Draw a diagram in support of your answer. (1 to 3 Marks) (2024)
Ans:(i) If they intersect then at the point of intersection, there would be two directions of magnetic field or compass needle would point towards two directions, which is not possible. (ii) Uniform magnetic field is represented by equidistant parallel straight lines
Q2: Strength of magnetic field produced by a current carrying solenoid DOES NOT depend upon: (1 Mark) (2024) (a)number of turns in the solenoid (b)direction of the current flowing through it (c)radius of solenoid (d)material of core of the solenoid
Ans: (b) The strength of the magnetic field in a solenoid depends on factors like the number of turns of wire and the radius of the solenoid. However, the direction of the current (whether it flows one way or the opposite) does not affect the strength of the magnetic field itself; it only changes the direction of the field. Therefore, the correct answer is (b).
Q3: Assertion – Reason based questions: (1 Mark) (2024) These questions consist of two statements — Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below: (a) Both (A) and (R) are true and (R) is the correct explanation of (A). (b)Both (A) and (R) are true, but (R) is not the correct explanation of (A). (c) (A) is true, but (R) is false. (d) (A) is false, but (R) is true. Assertion (A): The deflection of a compass needle placed near a current carrying wire decreases when the magnitude of an electric current in the wire is increased. Reason (R): Strength of the magnetic field at a point due to a current carrying conductor increases on increasing the current in the conductor.
Ans: (d) The assertion states that the deflection of a compass needle decreases as the current increases, which is false. The reason explains that the magnetic field strength increases with more current, which is true. Therefore, the correct answer is (d), as the assertion is false while the reason is true.
Q4: Draw a diagram to show the pattern of magnetic field lines on a horizontal sheet of paper due to a straight conductor passing through its centre and carrying current vertically upwards. Mark on it (i) the direction of current in the conductor and (ii) the corresponding magnetic field lines. State right hand thumb rule and check whether the directions marked by you are in accordance with this rule or not. (1 to 3 Marks) CBSE 2024)
Right-Hand Thumb Rule: When a current-carrying straight conductor is being held in right-hand such that the thumb points towards the direction of current, then fingers will wrap around the conductor in the direction of the magnetic field lines.
Q5: Name the device used to magnetise a piece of magnetic material. Draw a labelled diagram to show the arrangement used for the magnetisation of a cylinder made of soft iron. (1 to 3 Marks) (CBSE 2024)
Ans: (a) (b) Permanent magnet / Current carrying solenoid/ Electromagnet is used to magnetise a piece of magnetic material.
Q6: A rectangular loop ABCD carrying a current I is situated near a straight conductor XY, such that the conductor is parallel to the side AB of the loop and is in the plane of the loop. If a steady current I is established in the conductor as shown, the conductor XY will (1 Mark) (2024) (a) remain stationary. (b) move towards the side AB of the loop. (c) move away from the side AB of the loop. (d) rotate about its axis.
Ans: (b) A rectangular loop ABCD carrying a current I is situated near a straight conductor XY, such that the conductor is parallel to the side AB of the loop and is in the plane of the loop. If a steady current I is established in the conductor as shown, the conductor XY will move towards the side AB of the loop.
Q7: Two statements are given one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (1 Mark) (2024) (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true. Assertion (A): Magnetic field lines never intersect each other. Reason (R): If they intersect, then at the point of intersection, the compass needle would point towards two directions, which is not possible.
The assertion states that magnetic field lines never intersect, which is true. The reason explains that if they did intersect, a compass needle would point in two different directions at that point, which is also true. Since the reason correctly explains why the assertion is true, the correct answer is (a).
Q8: A student fixes a sheet of white paper on a drawing board. He places a bar magnet in the centre of it. He sprinkles some iron filings uniformly around the bar magnet. Then he taps the drawing board gently and observes that the iron filings arrange themselves in a particular pattern. (4 to 5 Marks) (2024) (a) Why do iron filings arrange in a particular pattern? (b) What does the crowding of iron filings at the ends of the magnet indicate? (c) What do the lines, along which the iron filings align, represent? (d) If the student places a cardboard horizontally in a current carrying solenoid and repeats the above activity, in what pattern would the iron filings arrange? State the conclusion drawn about the magnetic field based on the observed pattern of the lines.
Ans: (a) The iron filings arrange themselves in a particular pattern because they align with the magnetic field lines created by the bar magnet. The filings act like tiny magnets and are attracted to the lines of force, forming a pattern that shows the shape of the magnetic field around the magnet.
(b) The crowding of iron filings at the ends of the magnet indicates the location of the magnetic poles. The magnetic field is stronger at the poles of the magnet, which is why the filings are more concentrated in these regions. This shows the magnetic field lines emerging from the north pole and entering the south pole.
(c)The lines along which the iron filings align represent the magnetic field lines. These are the paths along which the magnetic force is exerted. The pattern formed by the iron filings shows the direction and shape of the magnetic field around the magnet.
(d) If the student places a cardboard horizontally in a current-carrying solenoid and repeats the activity, the iron filings would arrange themselves in concentric circles around the solenoid, with the pattern showing loops around the solenoid. The magnetic field of a solenoid is similar to that of a bar magnet, with distinct field lines forming a pattern resembling that of a bar magnet’s poles.
Conclusion: The magnetic field inside a solenoid is uniform and parallel to its axis, while outside the solenoid, the magnetic field resembles that of a bar magnet, with a clear direction from one end (north) to the other (south)
Q9: The current-carrying device which produces a magnetic field similar to that of a bar magnet is: (1 Mark) (2024) (a) A straight conductor (b) A circular loop (c) A solenoid (d) A circular coil
Ans: (c) A solenoid is a coil of wire that, when carrying an electric current, creates a magnetic field similar to that of a bar magnet, with distinct north and south poles. This means that a solenoid can generate a uniform magnetic field inside it, making it behave like a bar magnet. Therefore, the correct answer is (c).
Q10:
A uniform magnetic field exists in the plane of the paper as shown in the diagram. In this field, an electron (e–) and a positron (p+) enter as shown. The electron and positron experience forces: (1 Mark) (2024) (a) both pointing into the plane of the paper. (b) both pointing out of the plane of the paper. (c) pointing into the plane of the paper and out of the plane of the paper respectively. (d) pointing out of the plane of the paper and into the plane of the paper respectively.
pointing out of the plane of the paper and into the plane of the paper respectively.
The forces on the electron and positron will be in opposite directions according to the right-hand rule, as they have opposite charges. This means the force on the electron will be in one direction and the force on the positron will be in the opposite direction.
Q11: (a) What happens when a bundle of wires of soft iron is placed inside the coil of a solenoid carrying a steady current? Name the device obtained. Why is it called so? (b) Draw the magnetic field lines inside a current carrying solenoid. What does this pattern of magnetic field lines indicate? (1 to 3 Marks) (2024)
Ans: (a) The pattern of the magnetic field inside a solenoid is uniform and parallel, resembling straight lines. This means that the magnetic field strength is consistent throughout the solenoid, making it effective for applications requiring a stable magnetic field. Therefore, the correct answer is (a).
Previous Year Questions 2023
Q1: Assertion (A) : The magnetic field lines around a current carrying straight wire do not intersect each other. Reason (R) : The magnitude of the magnetic field produced at a given point increases as the current through the wire increases. (1 Mark) (2023) (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A) (c) Assertion (A) is true, but Reason (R) is False. (d) Assertion (A) is false, but Reason (R) is true. Ans: (b)
Sol: Assertion is true and reason is also true, but reason is not the correct explanation of assertion. Magnetic field lines around a current carrying straight wire do not intersect each other because at the point of intersection there will be two directions which is not possible. Also, the strength of magnetic held increased by increasing the magnitude of the current in the wire.
Q2: Draw the pattern of the magnetic field produced around a vertical current carrying straight conductor passing through a horizontal cardboard. Mark the direction of current and the magnetic field lines. Name and state the rule which is used to determine the direction of magnetic field associated with a current carrying conductor. (3 Marks) (2023)
Ans: The pattern of magnetic field produced around a vertical current carrying straight conductor passing through a horizontal cardboard is shown in figure. The magnitude of magnetic field produced is (i) ∝ I (ii) ∝ 1/r Right hand thumb rule is used to determine the direction of magnetic field associated with a current carrying conductor. It states that you are holding a current carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your finger will wrap around the conductor in the direction of the field lines of the magnetic field.
Q3: An alpha particle enters a uniform magnetic field as shown. The direction of force experienced by the alpha particle is ____ (1 Mark) (2023)
(a) Towards right (b) towards left (c) Into the page (d) Out of the page Ans: (d)
Sol: According to the Fleming’s left hand rule, the direction of force is out of the page.
Q4: A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in the figure. The direction of the magnetic field will be north to south at a point N (1 Mark) (2023)
(a) Directly above the wire (b) Directly below the wire (c) Located in the plane of the paper on the north side of the wire (d) Located in the plane of the paper on the south side of the wire
Ans: (a) The direction of the magnetic field at point N, located above a horizontal wire carrying a constant current from east to west, is from north to south. This can be determined using the right-hand thumb rule, which states that if you hold the wire with your right hand and point your thumb in the direction of the current, your fingers will curl in the direction of the magnetic field lines.
Q5: Assertion : A current carrying straight conductor experiences a force when placed perpendicular to the direction of magnetic field. Reason : The net charge on a current carrying conductor is always zero, (1 Mark) (2023) (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Assertion (A) is true because a current-carrying conductor placed perpendicular to a magnetic field experiences a force (Fleming’s Left-Hand Rule).
Reason (R) is true because a current-carrying conductor remains electrically neutral (equal protons and electrons).
However, R does not explain A. The force arises due to the motion of charges (current), not the net charge being zero. Thus, (b) is correct.
Q6: (i) Why is an alternating current (A.C.) considered to be advantageous over direct current (D.C.) for the long distance transmission of electric power? (ii) How is the type of current used in household supply different from the one given by a battery of dry cells? (iii) How does an electric fuse prevent the electric circuit and the appliances from the possible damage due to short circuiting or overloading. (3 Marks) (2023)
Ans: (i) Alternating current (A.C.) is preferred for long-distance power transmission because it experiences significantly lower power loss compared to direct current (D.C.). This efficiency makes A.C. more suitable for delivering electricity over vast distances. (ii) The current used in household supply is alternating in nature while the current given by battery is direct in nature. (iii) Electric fuse protects circuits and appliances by stopping the flow of any unduly high electric current. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.
Q7: (A) State the rule used to find the force acting on a current carrying conductor placed in a magnetic field. (B) Given below are three diagrams showing entry of an electron in a magnetic field. Identify the case in which the force will be (1) maximum and (2) minimum respectively. Give reason for your answer. (1 to 3 Marks) (CBSE 2023)
Ans: (A) Fleming’s left-hand rule is used to determine the direction of force experienced by a current carrying conductor placed in a uniform magnetic field. Acc to Fleming’s Left Hand Rule: When a current carrying conductor is placed in a magnetic field, it experiences a force, whose direction is given by Fleming’s left hand rule, which states that “Stretch the forefinger, the central finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger shows the direction of the field and the central finger that of the current, then the thumb will point towards the direction of motion of the conductor, i.e., force.”
(B) Force on electron is maximum in (i) case because the electron direction show that it is moving at right angle to the direction of a magnetic field. Force on electron is minimum in (iii) case as the electron shown is moving parallel to the direction of a magnetic field. The direction of maximum force acting on an electron in (i) case which is into the plane of paper in accordance with Fleming’s left hand rule.
Q8: (A) Draw the pattern of magnetic field lines of: (i) a current carrying solenoid (ii) a bar magnet (B) List two distinguishing features between the two fields. (1 to 3 Marks) (CBSE 2023)
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Previous Year Questions 2022
Q1: Give reason for the following (i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid. (ii) The current carrying solenoid when suspended freely rests along a particular direction. (2022)
Ans: (i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid because it behaves similar to that of a bar magnet and has a magnetic field line pattern similar to that of a bar magnet. Thus the ends of the straight solenoid behaves like poles of the magnet, where the converging end is the south pole and the diverging end is the north pole. (ii) The current carrying solenoid behaves similar to that of a bar magnet and when freely suspended aligns itself in the north-south direction.
Q2: A student fixes a sheet of white paper on a drawing board using some adhesive materials. She places a bar magnet in the centre of it and sprinkles some iron filings uniformly around the bar magnet using a salt-sprinkler. On tapping the board gently, she observes that the iron filings have arranged themselves in a particular pattern. (a) Draw a diagram to show this pattern of iron filings. (b) What does this pattern of iron filings demonstrate? (c) (i) How is the direction of magnetic field at a point determined using the field lines? Why do two magnetic field lines not cross each other? (2022)
Ans: (a) The pattern of iron filings is shown below.
(b) This pattern of iron filings demonstrate that the magnet exerts its influence in the region surrounding it. Therefore, the iron filings experience a force. The lines along which the iron filings align themselves represent magnetic field lines. (c) (i) The direction of magnetic field is determined by placing a small compass needle in the magnetic field. The North-pole of the compass indicates the direction of magnetic field at that point. Magnetic field lines never intersect each other because it is not possible to have two directions of magnetic field at the same point.
Q3: (i) What is a solenoid? (ii) Draw the pattern of magnetic field lines of the magnetic field produced by a solenoid through which a steady current flows. (2022)
Ans: The magnitude of force experienced by a current carrying conductor placed in a uniform magnetic field is (i)maximum when the conductor is placed perpendicular to the magnetic field, (ii) minimum when the conductor is placed parallel to the magnetic field.
Q5: (i) Name and state the rule to determine the direction of force experienced by a current carrying straight conductor placed in a uniform magnetic field which is perpendicular to it. (ii) An alpha particle while passing through a magnetic field gets projected towards north. In which direction will an electron project when it passes through the same magnetic field? (2022)
Ans: (i) Fleming’s left-hand rule is used to determine the direction of magnetic force experienced by a current carrying straight conductor placed perpendicularly in a uniform magnetic field. Fleming’s left-hand rule, states that when left hand’s thumb, forefinger and centre finger are held mutually perpendicular to one another and adjusted in such a way that the forefinger points in the direction of magnetic field, and the centre finger points in the direction of the current, then the direction in which thumb points, gives the direction of force acting on the conductor. (ii)As we know that, an alpha particle is positively charged. It is given that an alpha particle while passing through a magnetic field gets deflected (projected) towards north. Since an electron is negatively charged, it will deflect in Opposite direction i.e.. south.
Q6: A student was asked to perform an experiment to study the force on a current carrying conductor in a magnetic field. He took a small aluminium rod AB, a strong horseshoe magnet, some connecting wires, a battery and a switch and connected them as shown. He observed that on passing current, the rod gets displaced. On reversing the direction of current, the direction of displacement also gets reversed. On the basis of your understanding of this phenomenon, answer the following questions:
(a) Why does the rod get displaced on passing current through it? (b) State the rule that determines the direction of the force on the conductor AB. (c) If the U shaped magnet is held vertically and the aluminium rod is suspended horizontally with its end B towards due north, then on passing current through the rod B to A as shown, in which direction will the rod be displaced? (2022)
Ans: (a) On passing current, the rod gets displaced because of a magnetic force exerted on the rod when it is placed in the magnetic field. (b) Fleming’s left hand rule is used to determine the direction of magnetic force exerted on the conductor AB. (c) The rod will be displaced towards left according to Fleming’s left-hand rule.
Previous Year Questions 2021
Q1: Why do two magnetic field lines not intersect each other? (2021C)
Ans: The direction of magnetic field (B) at any point is obtained by drawing a tangent to the magnetic field line at that point. In case, two magnetic field lines intersect each other at the point P as shown in figure, magnetic field at P will have two directions, shown by two arrows, one drawn to each magnetic field line at P, which is not possible.
Q2: Name the instrument used to detect the presence of a current in a circuit. (2021C)
Ans: (a) In Figure ‘a’, poles P and Q of the magnet represents north pole and south pole respectively. In figure ‘b’, poles R and S of the magnet also represents north pole and south pole respectively. (b) Magnetic field lines are closed continuous curves directed from north pole to south pole outside the magnet but from south pole to north pole inside the magnet.
Q5: List two factors on which the strength of magnetic field at a point due to a current carrying straight conductor depends. State the rule that determines the direction of magnetic field produced in this case. (Term II, 2021-22C)
Ans: (i) Strength of magnetic field produced by a straight current-carrying wire at a given point is directly proportional to the current passing through it. inversely proportional to the distance of that point from the wire. (ii) Right-hand thumb rule: The straight thumb of right hand points in the direction of electric current. The direction of the curl of fingers represents the direction of magnetic field.
Previous Year Questions 2020
Q1: (a) What is an electromagnet? List any two uses. (b) Draw a labelled diagram to show how an electromagnet is made. (c) State the purpose of soft iron core used in making an electromagnet. (d) List two ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed. (2020)
Ans: (a) An electromagnet is a current-carrying solenoid coil which is used to magnetise steel rod inside it. Electromagnets are used in electric bells and buzzers, loudspeakers and headphones etc. (b) A strong magnetic field produced inside a solenoid can be used to magnetise a piece of magnetic material, like soft iron, when placed inside the coil. The magnet so formed is called an electromagnet. (c) The soft iron core placed in an electromagnet increases the strength of the magnetic field produced. Thus increasing the strength of electromagnet. (d) The strength of electromagnet can be increased by (i) Increasing the current passing through the coil. (ii) Increasing the number of turns in the coil.
Q2: Give reasons for the following: (A) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid. (B) The current carrying solenoid when suspended freely rests along a particular direction. (CBSE 2020)
Ans: (A) There is a divergence of magnetic field lines near the ends of a current carrying straight solenoid as the ends of the solenoid behave as poles. So, lines emerge and enter the ends, crowding the space and appearing divergent. (B)The current carrying solenoid when suspended freely rests along a particular direction because a current carrying solenoid behaves like a bar magnet with fixed polarities at its ends. The magnetic field lines are exactly identical to those of a bar magnet with one end of solenoid acting as a south-pole and its other end as north-pole
Q3: (A)Draw the pattern of magnetic field lines due to a magnetic field through and around a current carrying circular loop. (B) Name and state the rule to find out the direction of magnetic field inside and around the loop. (CBSE 2020)
(B)The rule used to find the direction of magnetic field lines is right hand thumb rule, which states that if we had the loop wire in our hand such that our thumb points in direction of current then our curled finger show the direction of field at that point.
Q4: What is the role of a fuse used in series with any electrical appliance? Why should a fuse with a defined rating not be replaced by one with a larger rating? (CBSE 2020, 19, 17, 10)
A fuse in a circuit prevents damage to the appliances and the circuit due to overloading and short-circuiting.
Overloading can occur when the live wire and the neutral wire come into direct contact.
In such a situation, the current in the circuit abruptly increases. This is called short-circuiting.
The use of an electric fuse prevents the electric circuit and the appliance from a possible damage by stopping the flow of unduly high electric current.
According to Joule’s Law of heating that takes place in the fuse, it melts to break the electric circuit.
So, a fuse is always connected in series with an appliance.
If it is connected in parallel, then it will not be able to break the circuit and the current keeps on flowing.
Overloading can also occur due to an accidental hike in the supply voltage.
Sometimes, overloading is caused by connecting too many appliances to a single socket.
The fuse with defined rating means the maximum current that can flow through the fuse without melting it.
It blows off when a current more than the rated value flows through it.
If a fuse is replaced by one with larger ratings, then large current will flow through the circuit without melting the fuse.
This large current may damage the appliances.
Q5: (A) What is an electromagnet? List any two uses. (B) Draw a labelled diagram to show how an electromagnet is made. (C) State the purpose of soft iron core used in making an electromagnet. (D) List two ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed. (CBSE 2020)
Ans: (A) An electromagnet is a temporary strong magnet. Its magnetism is only for the duration of current passing through it. The polarity and strength of an electromagnet can be changed. Uses of electromagnet: Electromagnets are used: (1) In electrical appliances like electric bell, electric fan etc. (2) In electric motors and generators. (B)
(C) Soft iron core is used in electromagnets to enhance their strength. The soft iron becomes magnetised when current flows through the coil, significantly increasing the overall magnetic field. (D) Ways of increasing the strength of an electromagnet: (1) If we increase the number of turns in the coil, the strength of electromagnet increases. (2) If the current in the coil is increased, the strength of electromagnet increases
Also watch: Introduction: Magnetic Field
Previous Year Questions 2019
Q1: Two circular coils P and Q are kept close to each other, of which coil P carries a current. What will you observe in the galvanometer connected across the coil Q, (a) if current in the coil P is changed? (b) if both the coils are moved in the same direction with the same speed? Give reason to justify your answer in each case. (CBSE 2011, 2019)
Ans: (a) When the current in coil P is increased, the galvanometer connected to coil Q shows a momentary deflection in one direction. Conversely, if the current in coil P is decreased, the galvanometer deflects in the opposite direction. This occurs because a change in current alters the magnetic field around coil P, which in turn induces a current in coil Q, resulting in the galvanometer’s deflection. (b) If both the coils P and Q are moved in the same direction with the same speed, the magnetic field of both the coils remain unchanged. Hence no induced current is set up in coil Q and there is no deflection in the galvanometer.
Q2: What is the function of a galvanometer in a circuit? (CBSE 2019)
Ans: A Galvanometer is an instrument used to detect the presence of current in a circuit. It indicates whether current is flowing and, if so, its direction.
Q3: One of the major causes of fire in office buildings is short-circuiting. List three reasons which may lead to short-circuiting. How can it be prevented? (CBSE 2019)
Ans: Three possible reasons of short-circuiting of an electrical circuit are as follows:
The insulation of electrical wirings is damaged.
The electrical appliance used in the circuit is defective.
An appliance of higher power rating is being run on an electrical line of lower power rating.
Short-circuiting can be prevented by the use of electrical fuse of appropriate capacity.
Q4: Draw magnetic field lines produced around a current carrying straight conductor passing through a cardboard. Name, state and apply the rule to mark the direction of these field lines. How will the strength of the magnetic field change when the point where magnetic field to be determined is moved away from the straight conductor? Give reason to justify your answer. (Allahabad 2019)
The Right-Hand Thumb Rule helps determine the direction of the magnetic field around a straight current-carrying conductor.
To apply this rule, hold the conductor in your right hand with your thumb pointing in the direction of the current.
Your curled fingers will then indicate the direction of the magnetic field lines surrounding the wire.
To assess the strength of the magnetic field, a compass needle can be used.
As the needle is moved further away from the conductor, the deflection decreases.
This indicates that the strength of the magnetic field diminishes with increasing distance from the wire.
The magnetic field lines form concentric circles around the conductor, which become larger as one moves away from it.
Previous Year Questions 2018
Q1: (a) What are magnetic field lines? How is the direction of magnetic field at a point in a magnetic field determined using field lines? (b) Two circular coils ‘X’ and ‘Y’ are placed close to each other. If the current in the coil ‘X’ is changed, will some current be induced in the coil ‘Y’? Give reason. (c) State ‘Fleming’s right hand rule. (CBSE 2018C) (5 marks)
Ans: (a) Magnetic field lines are imaginary lines along which the north magnetic pole would move in a magnetic field. The direction of a magnetic field at a point is determined by placing a small compass needle. The North pole of the compass indicates the direction of the magnetic field at that point
(b) Yes, if the current in coil X is changed, the magnetic field associated with it also changes around the coil ‘Y’ placed close to ‘X’. This change in magnetic field lines linked with ‘Y’, according to Faraday law of electric magnetic induction, induces a current in the coil Y. (c) Right-Hand Thumb Rule. This rule is used to find the direction of magnetic field due to a straight current carrying wire. It states that if we hold the current carrying conductor in the right hand in such a way that the thumb is stretched along the direction of current, then the curly finger around the conductor represent the direction of magnetic field produced by it. This is known as right-hand thumb rule. Direction of Field Lines due to current carrying straight conductor as shown in figure.
Q2: (a) State Fleming’s left hand rule. (b) Write the principle of working of an electric motor. (c) Explain the function of the following parts of an electric motor. (i) Armature (ii) Brushes (iii) Split ring (CBSE 2018)
Ans: (a) Fleming’s Left-Hand Rule: Stretch the thumb, forefinger and middle finger of the left hand such that they are mutually perpendicular to each other. If the forefinger pointed towards the direction of magnetic field and middle finger in the direction of current, then the thumb will indicate the direction of motion or force experienced by the conductor. It is to be applied only when the current and magnetic field, both are perpendicular to each other. (b) Principle of an electric motor: It works on the principle of magnetic effect of current. When a current carrying conductor is placed perpendicular to the magnetic field, it experiences a force. The direction of this force is given by Fleming’s left hand rule. (c)(i) Armature: It consists of a large number of turns of insulated current-carrying copper wires wound over a soft iron core and rotated about an axis perpendicular to a uniform magnetic field supplied by the two poles of permanent magnet. (ii) Brushes: Two conducting stationary carbon or flexible metallic brushes constantly touches the revolving split rings or commutator. These brushes act as a contact between commutator and terminal battery. (iii) Split ring: The split ring acts as a commutator in an electric motor. With the help of split ring, the direction of current through the coil is reversed after every half of its rotation and make the direction of currents in both the arms of rotating coil remains same. Therefore, the coil continues to rotate in the same direction.
Q1: The most important safety method used for protecting home appliances from short circuiting or overloading is: (a) earthing (b) use of fuse (c) use of stabilizers (d) use of electric meter (CBSE 2017, 13)
Ans: (b) A fuse is a safety device that protects electrical appliances and circuits from damage caused by overloading or short circuits. It contains a metal wire that melts when excessive current flows through it, breaking the circuit and stopping the flow of electricity, which prevents potential damage to appliances and reduces fire risk. Here’s why the other options are not as suitable for preventing short circuits or overloading: (a) Earthing: Protects against electric shocks but does not prevent overloading. (c) Use of stabilizers: Helps maintain a stable voltage but doesn’t protect against overloading or short circuits. (d) Use of electric meter: Measures electricity usage but doesn’t offer any protective function. Therefore, the most important safety method is the use of a fuse.
Previous Year Questions 2016
Q1: Under what condition is the force by a current-carrying conductor placed in a magnetic field maximum? (2016)
Ans: The force acting on a current-carrying conductor placed in a magnetic field is maximum when the direction of current is at right angles to the direction of the magnetic field.
Q2: What is an electric fuse ? Briefly describe its function. Or Explain the use of electrical fuse. What type of fuse material is used for fuse wire and why? (2016)
Ans: An electric fuse is a device that is used ahead of and in series of an electric circuit as a safety device to prevent the damage caused by short-circuiting or overloading of the circuit. It is a small, thin wire of a material whose melting point is low. Generally, wire of tin or tin-lead alloy or tin-copper alloy is used as a fuse wire. If due to some fault electric circuit gets short-circuited, then a strong current begins to flow. Due to such a strong flow of current, the fuse wire is heated up and gets melted. As a result, the electric circuit is broken and current flow stops. Thus, possible damage to the circuit and appliances is avoided.
Q3: (a) List four factors on which the magnitude of magnetic force acting on a moving charge in a magnetic field depends. (b) How will a fine beam of electrons streaming in west to east direction be affected by a magnetic field directed vertically upwards? Explain with the help of a diagram mentioning the rule applied. (2016)
Ans: (a) The magnitude of the magnetic force acting on a moving charge in a magnetic field depends on:
The magnitude of charge
Speed of moving charge
Strength of magnetic field
The angle between the direction of motion of charge and the direction of magnetic field.
(b) The electron streaming from west to east is equivalent to a current from east to west. The magnetic field B is vertically upwards and shown by cross marks (x). Hence, in accordance with Fleming’s left-hand rule, the electron will experience a force in north direction and deflected in that direction.
Q4: What is a solenoid? Draw magnetic field lines due to a current-carrying solenoid. Write three important features of the magnetic field obtained. (2016)
Ans: A solenoid is a coil of large number of circular turns of wire wrapped in the shape of a cylinder. On passing electric current, a magnetic field is developed. Magnetic field lines are drawn below. The field is along the axis of solenoid such that one end of solenoid behaves as north pole and the other south pole. Thus, field of a solenoid is similar to that of a bar magnet. Important features of magnetic field due to a current-carrying solenoid are:
Magnetic field lines inside the solenoid are nearly straight and parallel to its axis. It shows that magnetic field inside a solenoid is uniform.
Magnetic field of solenoid is identical to that due to a bar magnet with one end of solenoid behaving as a north pole and other end as a south pole.
A current-carrying solenoid exhibits the directive and attractive properties of a bar magnet.
Q5: Describe an activity with a neat diagram to demonstrate the presence of magnetic field around a current-carrying straight conductor. (2016)
Take a long straight thick copper wire and insert it through the centre of a plain cardboard. Cardboard is fixed so that it is perfectly horizontal and the thick wire is in vertical direction.
Prepare an electric circuit consisting of a battery, a variable resistance, an ammeter and a plug key.
Connect the copper wire in the circuit between the points X and Y. Sprinkle some iron filings uniformly on the cardboard. Put the plug in key K so that a current flows in the circuit.
Gently tap the cardboard using a finger. We observe that the iron filings align themselves forming a pattern of concentric circles around the copper wire.
These concentric circles represent the magnetic field lines. Direction of field lines is determined by the use of a compass needle. If current is flowing vertically downward then the magnetic field lines are along clockwise direction. If current is flowing vertically upward then the field lines are along anticlockwise direction.
Q6: (a) Describe an activity todraw a magnetic field line outside a bar magnet from one pole to another pole. (2016) (b) What does the degree of closeness of field lines represent?
Ans: (a) Take a bar magnet and place it on a sheet of white paper fixed on a drawing sheet. Mark the boundary of magnet. Take a small compass and place it near the north pole of the magnet. Mark the position of two ends of compass needle. Move the needle to a new position so that its south pole occupies the position previously occupied by its north pole. In this way proceed step by step till we reach the south pole of magnet. Join the points marked on the paper by a smooth curve. This curve represents a magnetic field line as shown in Fig.
(b) The degree of closeness of magnetic field lines signifies the strength of magnetic field.
Q7: What type of current is given by a cell? (CBSE 2016)
Ans: The magnetic field lines around a straight current-carrying conductor form concentric circles with the conductor at the centre. As you move away from the conductor, these circles expand. This pattern indicates that the strength of the magnetic field decreases with distance from the wire.
Ans: A magnetic field line around a magnet is the path along which north pole of a magnetic compass needle points. A magnetic field line gives the direction of magnetic field at a point.
Q3: (a) Describe an activity to show that an electric current-carrying wire behaves like a magnet. (b) Write the rule which determines the direction of magnetic field held developed around a current-carrying straight conductor. (CBSE 2015)
Ans: (a) Take a straight, thick copper wire X Y and connect it to an electric circuit consisting of a battery, key and resistor. Place a small compass near the copper wire. Now put the plugin key K so that a current begins to flow. The compass needle is deflected. It shows that the current-carrying copper wire is behaving like a magnet. (b) The direction of the magnetic field around a straight current-carrying conductor is given by the “right-hand thumb rule”. According to the rule, imagine holding the current-carrying straight conductor in your right hand such that the thumb points towards the direction of the current. Then the fingers of the right hand wrap around the conductor in the direction of the field lines of the magnetic field.
Q4: Out of the three wires live, neutral or earth, which one goes through ON/ OFF switch? (CBSE 2015)
Ans: The live wire is the one that goes through the ON/OFF switch. This wire carries the current to the appliance, allowing it to function when the switch is turned on. When the switch is off, the live wire is disconnected from the appliance, stopping the flow of electricity.
Q5: Why does a current-carrying conductor experience a force when it is placed in a magnetic field? (CBSE 2015)
Ans: A current-carrying conductor produces a magnetic field around it. This magnetic field interacts with the externally applied magnetic field and as a result the conductor experiences a force.
Q6: What is the frequency of A.C. being supplied in our houses? (CBSE 2015)
Ans: The frequency at which A.C. is supplied to residential houses is 50 Hz.
Q7: Describe an activity to explain how a moving magnet can be used to generate electric current in a coil. (CBSE 2015) Or A coil made of insulated copper wire is connected to a galvanometer. What will happen to the deflection of the galvanometer if a bar magnet is pushed into the coil and then pulled out of it? Give reason for your answer and name the phenomenon involved.
Ans: Take a coil AB of insulated copper wire having a number of turns. Connect the ends of coil to a sensitive galvanometer G. Now take a bar magnet NS and rapidly bring the magnet towards the end B of coil as shown in Fig. The galvanometer gives momentary deflection in one direction. Now take the magnet away from the coil, the galvanometer again gives momentary deflection but in the opposite direction. It clearly shows that motion of magnet induces, a current in the coil. Again keep the magnet fixed and gently move the coil AB either towards the magnet or away from the magnet. We get deflection in the galvanometer even now. Thus, an induced current is produced in a coil whenever there is relative motion between the coil and the magnet. This phenomenon is known as electromagnetic induction.
Q8: A metallic conductor is suspended perpendicular to the magnetic field of a horse-shoe magnet. The conductor gets displaced towards left when a current is passed through it. What will happen to the displacement of the conductor if the : (i) current through it is increased? (ii) horse-shoe magnet is replaced by another stronger horse-shoe magnet? (iii) direction of current through it is reversed ? (CBSE 2015)
Ans: (i) On increasing the current flowing through metallic conductor, the force experienced by it is proportionately increased because F ∝ I. (ii) On using a stronger horse-shoe magnet the magnetic force increases because F ∝ Magnetic Field. (iii) On reversing the direction of current the direction of force is reversed and conductor is displaced towards right instead of left direction.
Q9: For the current carrying solenoid as shown below, draw magnetic field lines and giving reason explain that out of the three points A, B and C at which point the field strength is maximum and at which point it is minimum. (CBSE 2015)
Ans: Outside the solenoid magnetic field is minimum. At the ends of solenoid, magnetic field strength is half that of inside it. So Minimum – at point B; Maximum – at point A
Q10: What is meant by solenoid? How does a current carrying solenoid behave? Give its main use. (CBSE 2015)
Ans: Solenoid: A coil of many circular turns of insulated copper wire wound on a cylindrical insulating body (i.e. cardboard etc.) such that its length is greater than its diameter is called solenoid. When current is flowing through the solenoid, the magnetic field line pattern resemble exactly with those of a bar magnet with the fixed polarity North and South pole at its ends and it acquires the directive and attractive properties similar to bar magnet. Hence the current carrying solenoid behaves a bar magnet. Use of current carrying solenoid: It is used to form a temporary magnet called electromagnet as well as permanent magnet.
Q11: What are magnetic field lines? Justify the following statements (a) Two magnetic field lines never intersect each other. (b) Magnetic field lines are closed curves. (CBSE 2015)
Ans: Magnetic field lines: It is defined as the path along which the unit North pole (imaginary) tends to move in a magnetic field if free to do so. (a) The magnetic lines of force do not intersect (or cross) one another. If they do so then at the point of intersection, two drawn tangents at that point indicate that there will be two different directions of the same magnetic field, i.e. the compass needle points in two different directions which is not possible. (b) Magnetic field lines are closed continuous curves. They emerge out from the north pole of a bar magnet and go into its south pole. Inside the magnet, they move, from south pole to north pole.
Q12: A current-carrying conductor is placed in a magnetic field. Now answer the following : (i) List the factors on which the magnitude of force experienced by conductor depends. (ii) When is the magnitude of this force maximum? (iii) If initially this force was acting from right to left, how will the direction of force change, if : (a) direction of magnetic field is reversed? (b) direction of current is reversed? (CBSE 2015)
Ans: (i) The magnitude of force experienced by the current-carrying conductor when placed in a magnetic field depends on current flowing, length of the conductor, the strength of magnetic field, orientation of conductor in the magnetic field. (ii) Magnitude of force is maximum when current-carrying conductor is placed at right angles to the direction of magnetic field. (iii) (a) Direction of force is reversed that is now the force acts from left to right. (b) Direction of force is reversed that is now the force acts from left to right.
Q13: In our daily life, we use two types of electric current whose current-time graphs are given in Fig. 13.29. (i) Name the type of current in two cases. (ii) Identify any one source for each type of current. (iii) What is the frequency of current in case (iv) in our country? (v) Out of the two which one is used in transmitting electric power over long distances and why? (CBSE 2015). Fig (a)Fig (b)
Ans: (i) Current shown in Fig. (a) is direct current (D.C.) but current shown in Fig. (b) is an alternating current (A.C.). (ii) A cell/battery produces D.C. but an A.C. generator produces A.C. (iii) In India frequency of A.C. is 50 Hz. (iv) A.C. is used in transmitting electric power over long distances. It is so because transmission loss of electric power can be minimised for A.C. by employing suitable transformers at generating stations and consuming centres.
Q14: A student fixes a sheet of white paper on a drawing board. He places a bar magnet in the centre of it. He sprinkles some iron filings uniformly around the bar magnet. Then he taps the board gently. Now answer the following questions : (i) What does the student observe? Draw a diagram to illustrate your answer. (ii) Why do the iron filings arrange in such a pattern? (iii) What does the crowding of the iron filings at the ends of the magnet indicate? (CBSE 2015) (5 marks)
Ans: (i) The student observes the magnetic field due to the given bar magnet. The pattern of the magnetic field is shown here. (ii) There is a magnetic field around the given bar magnet. The iron filings experience a force due to the magnetic field and thus align themselves along the magnetic field lines.
(iii) The crowding of the iron filings at the ends of the magnet indicates the position of two magnetic poles N and S of a given bar magnet.
Also watch: Introduction: Magnetic Field
Previous Year Questions 2013
Q1: If the key in the arrangement as shown below is taken out (the circuit is made open) and magnetic field lines are drawn over the horizontal plane ABCD, the lines are:
(a) concentric circles (b) elliptical in shape (c) straight lines parallel to each other (d) concentric circles near the point O but of elliptical shapes as we go away from it. (CBSE 2013)
Ans: (c) In the arrangement described, if the key is taken out and the circuit is open, no current flows through the wire. This means that no magnetic field is generated by the wire. However, if magnetic field lines are drawn over the horizontal plane ABCD in the absence of a current (or any other magnetic source), they would represent the Earth’s magnetic field, which appears as straight, parallel lines over a small horizontal area. Thus, the correct answer is (c) straight lines parallel to each other.
Previous Year Questions 2010
Q1: A magnetic compass needle is placed in the plane of paper near point A, as shown in the figure. In which plane should a straight current carrying conductor be placed so that it passes through A and there is no change in the deflection of the compass? Under what condition is the deflection maximum and why? (CBSE 2010)
Ans: The straight current carrying conductor should be placed in the paper plane, passing through A, to produce a magnetic field perpendicular to the paper plane. This ensures the compass needle remains undeflected due to the vertical magnetic field produced by the wire. The maximum deflection in the compass needle occurs when the conductor is perpendicular to the paper plane and the magnetic field is in the paper plane.
Q1: A wire of resistance R is cut into three equal parts. If these three parts are then joined in parallel, calculate the total resistance of the combination so formed. (2 Mark)
Ans: The total resistance of the combination is R/9.
When a wire of total resistance R is cut into three equal parts, each part has a resistance of R/3 because resistance is directly proportional to length. When these three parts are connected in parallel, the reciprocal of equivalent resistance isHence, the total equivalent resistance becomes
Therefore, the total resistance of the parallel combination is R/9, which shows that connecting smaller equal parts of a wire in parallel greatly reduces the overall resistance.
Q2:Define electric power. When do we say that the power consumed in an electric circuit is 1 watt? (2 Mark)
Electric Power: Power (P) is defined as the rate of doing work or the rate at which electrical energy is consumed or supplied. It is given by the formula: P = V × I, where V is the potential difference (volts) and I is the current (amperes). Alternatively, P = I²R or P = V²/R, where R is resistance (ohms). The SI unit of power is the watt (W). 1 Watt: Power is 1 watt when 1 joule of energy is consumed or transferred in 1 second. Mathematically, 1 W = 1 J/s. For example, if a circuit has a potential difference of 1 volt and a current of 1 ampere, the power is: P = V × I = 1 V × 1 A = 1 W.
Q3: (a) Write the relationship between resistivity and resistance of a cylindrical conductor of length l and area of cross-section A. Hence derive the SI unit of resistivity. (b) Why are alloys used in electrical heating devices? (3 marks)
Ans: (a) Relationship: R = ρl/A; SI unit of resistivity: ohm-meter (Ω·m). (b) Alloys are used due to high resistivity and high melting points.
(a) For a uniform cylindrical conductor of length l and cross-sectional area A, the resistance R is proportional to l and inversely proportional to A. Hence
SI unit derivation:
Resistance (R) is in ohms (Ω).
Area (A) is in square meters (m²).
Length (l) is in meters (m).
Unit of ρ = (Ω × m²)/m = Ω·m. Thus, the SI unit of resistivity is ohm-meter (Ω·m).
(b) Alloys are used in electrical heating devices because their resistivity is generally higher than that of constituent metals and, importantly (as stated in the chapter), alloys do not oxidise (burn) readily at high temperatures — making them suitable and durable for heating elements.
Q4: The following question is Source-based/Case-based question. Read the case carefully and answer the question that follow. In our homes, we receive the supply of electric power through a main supply also called mains, either supported through overhead electric poles or by underground cables. In our country the potential difference between the two wires (live wire and neutral wire) of this supply is 220 V.
(a) Write the colours of the insulation covers of the line wires through which supply comes to our homes. (1 Mark) (b) What should be the current rating of the electric circuit (220 V) so that an electric iron of 1 kW power rating can be operated? (1 Mark) (c) (i) What is the function of the earth wire? State the advantage of the earth wire in domestic electric appliances such as electric iron. (2 marks)
Ans: (a) Red (or brown) for live, black (or blue) for neutral, green (or green-yellow) for earth. (b) Current rating = 4.55 A (minimum 5 A fuse). (c) (i) Function: Provides a low-resistance path for leakage current to prevent electric shock; Advantage: Enhances user safety by grounding excess current.
(a) Colours: In domestic wiring, the standard colours are:
Live wire: Red or brown, carries current from the supply.
Neutral wire: Black or blue, completes the circuit back to the supply.
Earth wire: Green or green-yellow, provides a safety path for leakage current.
These colours help identify wires for safe installation and maintenance.
(b) Current rating:
Power of electric iron, P = 1 kW = 1000 W.
Voltage, V = 220 V.
Power formula: P = V × I.
The circuit needs a fuse with a current rating slightly higher than 4.54 A, typically 5 A, to safely operate the iron without frequent blowing.
(c) (i) Earth wire:
Function: The earth wire connects the metallic body of an appliance to the ground via a metal plate buried in the earth. If there’s a fault (e.g., live wire touching the metal body), it provides a low-resistance path for the leakage current to flow to the ground, preventing electric shock.
Advantage: In appliances like an electric iron, the earth wire ensures that any leakage current is safely diverted, keeping the appliance’s body at earth potential (0 V), thus protecting the user from severe electric shock.
Q5: Consider the following circuits :
In which circuit will the power dissipated in the circuit be (I) minimum (II) maximum ? Justify your answer. (2 Marks)
Ans: (a) Analyze the circuits to determine power dissipation. Use the formula for power dissipation: Minimum power dissipation is in circuit (ii) and maximum is in circuit (iii).
OR
Two lamps, rated 100 W; 220 V and 60 W; 220 V, are connected in parallel to an electric main supply of 220 V. Find the current drawn by the two lamps from the supply. (2 marks)
Ans: Total current = 0.727 A. Given: Lamp 1: 100 W, 220 V; Lamp 2: 60 W, 220 V; connected in parallel to 220 V supply.
Step 1: Current for each lamp
Power formula: P = V × I.
For Lamp 1:
For Lamp 2:
Step 2: Total current in parallel
In a parallel circuit, total current is the sum of individual currents:
Conclusion: The total current drawn from the supply is approximately 0.727 A.
Q6: (a) Define one ampere. (b) The resistance of a wire of 0.01 cm radius is 14 Ω. If the resistivity of the material of the wire is 44 × 10⁻⁸ Ω m, find the length of the wire. (Given π = 22/7) (3 Marks)
Ans: (a) One ampere is the current that flows through a conductor when one coulomb of charge passes through it in one second. (b) Length of the wire = 0.159 m (15.9 cm).
(a) One ampere:
One ampere is defined as the current in a conductor when one coulomb of electric charge flows through it per second. Mathematically, 1 A = 1 C/s. It is the SI unit of electric current.
(b) Length calculation:
Q7: (a) Consider two lamps A and B of rating 50 W; 220 V and 25 W; 220 V respectively. Find the ratio of the resistances of the two lamps (i.e., RA : RB). (2 Marks)
Given: Heat produced per second = 400 J/s (power, P = 400 W); Resistance, R = 4 Ω.
Power formula:
Conclusion: The potential difference across the resistor is 40 V.
Q9:An electric bulb is connected to a power supply of 220 V. If the current drawn by the bulb from the supply is 500 mA, the power of the bulb is: (1 Mark) (a) 11 W (b) 110 W (c) 220 W (d) 1100 W
Given: Voltage, V = 220 V; Current, I = 500 mA = 0.5 A.
Power formula: P = V × I. P = 220 × 0.5 = 110 W.
Conclusion: The power of the bulb is 110 W, option (B).
Q10:Four identical resistors of 12 Ω each are connected in series to form a square ABCD as shown in the figure. The resistance of the network between the two points 1 and 2 is : (a) 48 Ω (b) 36 Ω (c) 9 Ω (d) 6 Ω
Assuming points 1 and 2 are adjacent vertices (e.g., A and B in square ABCD with resistors on each side). The direct path AB is 12 Ω. The other path A-D-C-B is three resistors in series: 12 + 12 + 12 = 36 Ω. These two paths are in parallel:
Q11: In the following Question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (1 Mark)
Assertion (A): Nichrome is an alloy commonly used in electrical heating devices such as electric irons, toasters, etc. Reason (R): The resistivity of nichrome is high, and its resistance decreases with an increase in temperature. (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.
Assertion (A): Nichrome is widely used in heating devices like electric irons and toasters due to its high resistivity and high melting point, which allow it to generate significant heat and withstand high temperatures. Thus, A is true.
Reason (R): Nichrome has high resistivity, but its resistance increases slightly with temperature (as it’s a metal alloy with a positive temperature coefficient), not decreases. Thus, R is false.
Conclusion: Option (C) is correct.
Q12:An electric kettle is rated 750 W; 220 V. Can this kettle be used in a circuit which has a fuse of current rating 3 A? Give reason for your answer. (2 Marks)
Ans: No, the kettle cannot be used with a 3 A fuse. Given: Power, P = 750 W; Voltage, V = 220 V; Fuse rating = 3 A.
Current calculation: P = V × I =>
Comparison: The kettle draws 3.41 A, which exceeds the fuse rating of 3 A. A fuse blows if the current exceeds its rating, so the 3 A fuse will blow, disconnecting the circuit.
Conclusion: The kettle requires a fuse rating higher than 3.41 A (e.g., 5 A), so it cannot be used with a 3 A fuse.
Q13:Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in (i) series, and (ii) parallel. Draw the arrangements of the resistors and find the equivalent resistance of each arrangement. (3 Marks)
Q14:The resistivity of a wire made of an alloy is generally: (1 Mark) (a) Lower than that of its constituent metals. (b) Higher than that of its constituent metals. (c) Decreases with an increase in its area of cross-section. (d) Increases with an increase in its length.
Ans: (B) Higher than that of its constituent metals.
Alloys (e.g., nichrome) are made by mixing metals, which disrupts the regular lattice structure, increasing resistance to electron flow. Thus, the resistivity of alloys is generally higher than that of their constituent metals (e.g., nickel or chromium).
Other options:
(A) Incorrect: Alloys have higher resistivity.
(C) Incorrect: Resistivity is a material property, independent of area.
(D) Incorrect: Resistivity is independent of length; resistance increases with length.
Conclusion: Option (B) is correct.
Q15: Resistance of a wire of length 1 m is 35 Ω at 20°C. If the diameter of the wire is 0.2 mm, determine the resistivity of the material of the wire at that temperature. How will the resistivity of the wire change if the length and diameter of the wire both are doubled? Justify your answer. (Given π = 22/7) (3 Marks)
Use: An electric fuse is used in series to protect circuits from excessive current.
Why: To prevent damage due to overloading or short circuits.
Function: It melts and breaks the circuit if the current exceeds its rating, stopping current flow.
Use: A fuse is a safety device connected in series with an electrical circuit or appliance. It consists of a wire with a low melting point.
Why: It protects circuits and appliances from damage due to excessive current (e.g., from overloading or short circuits) and prevents fire hazards or electric shocks.
Function: When the current exceeds the fuse’s rating, the fuse wire heats up (due to I²R heating) and melts, breaking the circuit and stopping current flow, thus protecting the circuit and appliances.
Q17:A wire of length ‘l’ is gradually stretched so that its length increases to 3l. If its original resistance is R, then its new resistance will be: (1 Mark) (a) 3R (b) 6R (c) 9R (d) 27R
Ans: (C) 9R Given: Original length = l, resistance = R; New length = 3l. Resistance, R = ρl/A. When the wire is stretched to 3l, its volume remains constant (V = l × A = constant).
New length, l’ = 3l.
New area,
New resistance,
Conclusion: The new resistance is 9R, option (C).
Q18:An electric kettle is rated 230 V; 1000 W. Calculate the resistance of its heating element when in operation. (2 Marks)
Conclusion: The resistance of the heating element is 52.9 Ω.
Q19:A voltage source sends a current of 2 A to a resistor of 40 Ω connected across it for 5 minutes. Calculate the electrical energy supplied by the source. (2 Marks)
Electrical energy supplied:E = P × t = 160 × 300 = 48,000 J
Q20: Four resistors, each of resistance 2.0 Ω, are joined end to end to form a square ABCD. Using the appropriate formula, determine the equivalent resistance of the combination between its two ends A and B.
Given: Four resistors, each 2 Ω, form a square ABCD
Circuit analysis:
The square has resistors: AB = 2 Ω, BC = 2 Ω, CD = 2 Ω, DA = 2 Ω.
For resistance between A and B:- Path 1: Direct resistor AB = 2 Ω. – Path 2: Via B → C → D → A (series: BC + CD + DA = 2 + 2 + 2 = 6 Ω). – AB (2 Ω) is in parallel with BCD (6 Ω):
Q21:An electric bulb is rated 220 V; 11 W. The resistance of its filament when it glows with a power supply of 220 V is: (1 Mark) (a) 4400 Ω (b) 440 Ω (c) 400 Ω (d) 20 Ω
Q22:The minimum number of identical bulbs of rating 4 V; 6 W that can work safely with desired brightness when connected in series with a 240 V mains supply is: (1 Mark) (a) 20 (b) 40 (c) 60 (d) 80
Series connection: For n bulbs in series, total voltage = n × 4 V.
Conclusion: Minimum 60 bulbs, option (c).
Q23: The electrical resistivity of three materials A, B, and C at 20°C is given below: (i) Classify these materials as conductor, alloy, and insulator. (ii) Give one example of each of these materials and state one use of each material in the design of an electrical appliance say an electric stove or an electric iron. (3 Marks)
(ii) Example of Insulator: Rubber Uses: Coating of electrical wires
Example of conductor: Tungsten Uses: Tungsten is used as the filament of an electric bulb.
Example of alloy: Nichrome Uses: The heating element of an electric iron
Q24:Define the term “potential difference” between two points in an electric circuit carrying current. Name and define its S.I. unit. Also express it in terms of S.I. unit of work and charge. ( 3 Marks)
Potential Difference: The work done per unit charge to move a charge between two points.
S.I. Unit: Volt, defined as 1 joule of work per 1 coulomb of charge.
Expression: 1 V = 1 J/C.
Potential Difference: It is the energy required or released per unit charge when moving a charge between two points in a circuit. It drives current flow.
S.I. Unit: The volt (V), where 1 volt is the potential difference when 1 joule of work is done to move 1 coulomb of charge.
Expression: V = W/Q, where W is work (joules, J) and Q is charge (coulombs, C). Thus, 1 V = 1 J/C.
Q25: (a) State two applications of Joule’s heating in domestic electric circuits. (b) (i) Establish the relationship between the commercial unit of electric energy and the SI unit of electric energy. (2 Marks)
Ans: (a) Applications: (1) Electric iron for pressing clothes, (2) Electric toaster for heating bread. (b) 1 kWh = 3.6 × 10⁶ J.
(a) Applications:
1. Electric Iron: The electric iron uses Joule’s heating effect to heat up the iron plate, which is then used to press clothes.
2. Electric Heater: Electric heaters use Joule’s heating effect to convert electrical energy into heat energy, which is used to warm up the surroundings.
(b) Relationship:
Commercial unit: Kilowatt-hour (kWh), the energy consumed by a 1 kW device in 1 hour.
SI unit: Joule (J).
1 kWh = 1000 W × 3600 s = 1000 × 3600 J = 3.6 × 10⁶ J.
Conclusion: 1 kWh = 3.6 million joules.
Q26: (a) “The third wire of earth connection is very important in domestic electric appliances.” Justify this statement. (b) List two precautions to be taken to avoid the overloading of domestic electric circuits. (3 Marks)
Ans: (a) The earth wire prevents electric shock by grounding leakage current. (b) Precautions: (1) Avoid connecting too many appliances to one socket, (2) Use fuses with appropriate ratings.
(a) Earth wire importance:
The earth wire connects the metallic body of appliances to the ground. If a fault occurs (e.g., live wire touches the metal body), the earth wire provides a low-resistance path for the current to flow to the ground, preventing electric shock to the user and protecting the appliance from damage.
(b) Precautions:
Avoid multiple appliances per socket: Connecting too many high-power devices increases current, risking overheating and fire.
Use appropriate fuses: A fuse with the correct rating (e.g., 5 A for low-power circuits) blows if current exceeds safe limits, preventing circuit damage or fire.
Q27:(A) Use Ohm’s law to determine the potential difference across the 6 ohm resistor in the following circuit when key is closed : (2 Marks) OR (B) Prove that if the current through a resistor is increased by 100%, then the increase in power dissipated through the resistor will be 300%. (2 Marks)
Finding Total Resistance: The circuit has a 2Ω and a 4Ω resistor in series with the 6Ω resistor. The total resistance can be calculated as follows:Total Resistance,
Finding Total Current: Using Ohm’s law, the total current (I) in the circuit can be calculated using the total voltage (V=6V) and total resistance:
Finding Voltage Across 6Ω Resistor: Now, we can find the voltage across the 6Ω resistor using Ohm’s law:
Therefore, the potential difference across the 6Ω resistor is 3 V.
(B) Power increases by 300%.
Given: Current increases by 100%, i.e., new current I’ = 2I (doubled).
Conclusion: Doubling the current increases power by 300%.
Q28:What is short circuiting? State its possible causes. What is likely to happen if a domestic circuit gets short circuited? Give reason for the justification of your answer. (3 Marks)
Effect: Excessive current flow, causing overheating, fire, or appliance damage.
Reason: Low resistance in the circuit increases current significantly.
Short circuiting: Occurs when the live and neutral wires touch directly, creating a low-resistance path, bypassing the appliance (load).
Causes:
Worn insulation: Damaged insulation on wires allows them to touch.
Faulty wiring: Incorrect connections in appliances or circuits.
Effect:
A short circuit drastically reduces resistance, causing a large current (I = V/R) to flow. This generates excessive heat (H = I²Rt), which can melt wires, cause fires, or damage appliances.
A fuse or circuit breaker trips to prevent damage by breaking the circuit.
Reason: The low resistance in a short circuit allows a very high current, overwhelming the circuit’s capacity, leading to overheating or fire hazards.
Q29:Determine the maximum and minimum resistance which can be obtained by joining five resistors of 1/5 Ω each. (2 Marks)
(B) Calculate potential difference across a 4 ohm resistor that produces 100 W of heat every second. (2 Marks) Ans: Given, Power = 100 W, Resistance = 4Ω. Using Formula,We can rearrange formula to solve for V:Substitute the values: ⇒ The potential difference across the 4 Ω resistor is 20 volts.
Q30: (a) Explain the statement “Potential difference between two points is 1 volt”. (b) What do the symbols given below represent in an electric circuit? Write one function of each. (3 Marks)
(a) 1 volt: Potential difference is 1 volt when 1 joule of work is required to move 1 coulomb of charge between two points. Mathematically, 1 V = 1 J/C.
(b) Symbols:
(i) The first symbol represents an Ammeter. Function: It is used to measure the electric current flowing through a circuit. (ii) The second symbol represents a Variable Resistor (Rheostat). Function: It is used to vary or control the current in a circuit by changing the resistance..
Q31:Study the circuit shown in which two resistors X and Y of resistances 3 Ω and 6 Ω respectively are joined in series with a battery of 2 V. (I) Draw a circuit diagram showing the above two resistors X and Y joined in parallel with the same battery and same ammeter and voltmeter. (1 Mark) (II) In which combination of resistors will the (i) potential difference across X and Y, and (ii) current through X and Y, be the same? (1 Mark) (III) Find the current drawn from the battery by the series combination of the two resistors (X and Y). (2 Marks)
Ans: (A) When two 6 Ω resistances are connected in parallel and the third resistance of 6 Ω is connected in series combinations to this, then equivalent resistance will be 9 Ω
The circuit contains a 1Ω and a 2Ω resistor in series. The total resistance is:
Rtotal = 1 + 2 = 3Ω
Using Ohm’s law:
I = VRtotal = 6V3Ω = 2A
Power consumed by 2Ω resistor is given by:
P = I² × R
P = (2)² × 2 = 4 × 2 = 8W
Q2: (A) (i) Define electric power. Express it in terms of potential difference (V) and resistance (R). (4 to 5 Marks) (2024) (ii) An electric oven is designed to work on the mains voltage of 220 V. This oven consumes 11 units of electrical energy in 5 hours. Calculate : (a) power rating of the oven (b) current drawn by the oven (c) resistance of the oven when it is red hot OR (B) (i) Write the relation between resistance R and electrical resistivity p of the material of a conductor in the shape of cylinder of length / and area of cross-section A. Hence derive the SI unit of electrical resistivity. (ii) The resistance of a metal wire of length 3 m is 60 Ω. If the area of the cross-section of the wire is 4 x 10-7 m, calculate the electrical resistivity of the wire. (iii) State how would electrical resistivity be affected if the wire (of part ‘ii’) is stretched so that its length is doubled. Justify your answer.
Ans: (A) (i) Electric power: Rate at which electrical energy is dissipated or consumed / Rate of supplying energy to maintain the flow of current through a circuit.
(ii) (a) Energy consumed = 11 units
E = 11 kWh = 11 × 1000
P = 11000 W5 h = 2200 W
Thus, the power rating of the oven is 2200 W or 2.2 kW.
(b) I = PV = 2200220 = 10A
(c) R = V²P = (220)²2200 = 22 Ω
OR
(B)
(i) R = ρ lA
ρ = R × Al
= Ohm × (metre)²metre = ohm metre or Ωm
(ii)
Here l = 3 m, A = 4 × 10⁻⁷ m², R = 60 Ω
ρ = R × Al
= 60 × 4 × 10⁻⁷3 = 80 × 10⁻⁷ Ωm(iii) When a wire is stretched to double its length, its electrical resistivity (ρ) remains constant. A material’s electrical resistivity is intrinsic and does not alter its dimensions (length, cross-sectional area). Therefore, doubling the length of the wire will not affect its electrical resistance.
Q3: Study the I-V graph for three resistors of resistances R1, R2 and R3 and select the correct statement from the following: (1 Mark) (2024) (a) R1 = R2 = R3 (b) R1 > R2 > R3 (c) R3 > R2 > R1 (d) R2 > R3 > R1
This is a graph of current (I) versus voltage (V) for three resistors R1, R2, and R3. The slope of each line in an I-V graph is related to the conductance, which is the reciprocal of resistance (R). Thus, the steeper the slope, the lower the resistance.
From the graph:
R1 has the steepest slope, indicating the lowest resistance.
R3 has the least steep slope, indicating the highest resistance.
Correct interpretation:
The resistances follow the order:
R3 > R2 > R1
Q4: Use Ohm’s law to determine the potential difference across the 3 Ω resistor in the circuit shown in the following diagram when the key is closed. (2 Marks) (2024)
Q5: In the case of four wires of the same material, the resistance will be minimum if the diameter and length of the wire respectively are (1 Mark) (2024) (a) D/2 and L/4 (b) D/4 and 4L (c) 2D and L (d) 4D and 2L
Ans: (d) The resistance of a wire is inversely related to its cross-sectional area and directly related to its length. To minimize resistance, you should use a wire with a larger diameter and a shorter length. Therefore, if you choose a diameter of 4D (making the cross-sectional area much larger) and a length of 2L (shorter compared to the original), you will achieve the minimum resistance, making the correct answer (d) 4D and 2L.
Q6: Explain in brief the function of an electric fuse in a domestic circuit. An electric heater of current rating 3 kW; 220 V is to be operated in an electric circuit of rating 5 A. What is likely to happen when the heater is switched ‘ON’ ? Justify your answer with the necessary calculation. (3 Marks) (CBSE 2024)
An electric fuse is a safety device used in electrical circuits to prevent excessive current from flowing through the wires and damaging the circuit components.
It is designed to melt and break the circuit when the current exceeds a certain limit (which is its rated value).
This helps protect the wiring and connected appliances from damage due to overcurrent.
Here P = 3 kW = 3000 W, V = 220 V, I = ? P = V I
The current drawn by the heater is 13.64 A, but the current rating of the circuit is only 5 A.
This means the current drawn by the heater is significantly higher than the circuit’s rated current.
Since the current drawn by the heater (13.64 A) exceeds the current rating of the circuit (5 A), the circuit will be overloaded, and the fuse will blow to protect the circuit.
The fuse will disconnect the heater from the circuit, preventing further damage to the wires and avoiding potential hazards like overheating or fire.
Q7: (a) State Ohm’s law. Write the formula for the equivalent resistance R of the parallel combination of three resistors of values R1, R2, and R3. (b) Find the resistance of the following network of resistors: (3 Marks) (2024)
Ans: (a) Ohm’s Law: The potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same.
Formula:
1/Rp = 1/R1 + 1/R2 + 1/R3
(b) R + R2 = 3R2
Q8: Case/Source-based questions. (4 to 5 Marks) (CBSE 2024) In a domestic circuit, five LED bulbs are arranged as shown. The source voltage is 220 V and the power rating of each bulb is marked in the circuit diagram. Based on the following circuit diagram, answer the following questions:
(a) State what happens when (i) key K1 is closed. (ii) key K2 is closed. (b) Find the current drawn by the bulb B when it glows. (c) Calculate (i) the resistance of bulb B, and (ii) the total resistance of the combination of four bulbs B, C, D, and E. OR (c) What would happen to the glow of all the bulbs in the circuit when keys K1 and K2 both are closed and the bulb C suddenly gets fused? Give a reason to justify your answer.
Ans: (a) (i) Bulb A glows (ii) Bulbs B, C, D and E glow (b) P = V × I (c) (i) Resistance of bulb B, (alternative formula for calculation (ii) Total resistance of the series combination of four bulbs = 4 x 275 = 1100 W OR (c)
Bulb A will keep glowing with same brightness.
Other bulbs i.e., B, D and E will stop glowing.
Reason: As the bulbs B, D and E are connected in series with fused bulb C, so no current flows through them and thus they will not glow. The bulb A remains unaffected as it is connected in parallel combination.
Q9: Consider the following combinations of resistors: (1 Mark) (2024) The combinations having equivalent resistance 1 is/are: (a) I and IV (b) Only IV (c) I and II (d) I, II and III
Ans: (c) To determine which combinations of resistors have an equivalent resistance of 1 ohm, you need to analyze the given arrangements (I, II, III, and IV) based on how resistors are connected, either in series or parallel. Combinations that meet the criteria of having an equivalent resistance of 1 ohm are identified in options I and II. Therefore, the correct answer is (c) I and II.
Q10: An electric iron of resistance 20Ω draws a current of 5 A. The heat developed in the iron in 30 seconds is: (1 Mark) (2024) (a) 15000 J (b) 6000 J (c) 1500 J (d) 3000 J
Ans: (a) To calculate the heat developed in the electric iron, we can use the formula H = I2 Rt, where H is the heat produced, I is the current (5 A), R is the resistance (20 ohms), and t is the time (30 seconds). Plugging in the values: H = (52) × 20 × 30 = 25 × 20 × 30 = 15000J So, the heat developed in the iron is 15000 J, making the correct answer (a) 15000 J.
Q11: Two wires A and B of the same material, having the same lengths and diameters 0·2 mm and 0·3 mm respectively, are connected one by one in a circuit. Which one of these two wires will offer more resistance to the flow of current in the circuit? Justify your answer. (1 Mark) (2024)
Ans: Wire A will offer more resistance Justification:
Smaller diameter ⇒ Smaller cross-sectional area ⇒ Higher resistance.
Wire A (0.2 mm diameter) has higher resistance than wire B (0.3 mm diameter).
Q12: The following questions are source-based/case-based questions. Read the case carefully and answer the questions that follow. Study the following circuit: (4 to 5 Marks) (2024)
On the basis of this circuit, answer the following questions: (a) Find the value of total resistance between points A and B. (b) Find the resistance between the points B and C. (c) (i) Calculate the current drawn from the battery, when the key is closed. OR (c) (ii) In the above circuit, the 16 Ω resistor or the parallel combination of two resistors of 8 Ω, which one of the two will have more potential difference across its two ends? Justify your answer.
OR (c) (ii) 16 Ω Justification: According to Ohm’s law when same current flows, the potential difference across a higher resistance is always higher. Potential difference across 16 Ω = V = IR = 0.2 x 16 = 3.2V Potential difference across 8 Ω = V = IR(total) = 0.2 x 4 = 0.8V
Q13: An electric source can supply a charge of 500 coulomb. If the current drawn by a device is 25 mA, find the time in which the electric source will be discharged completely. (1 Mark) (2024)
Q14: (a) (i) The potential difference across the two ends of a circuit component is decreased to one-third of its initial value, while its resistance remains constant. What change will be observed in the current flowing through it? Name and state the law which helps us to answer this question. (ii) Draw a schematic diagram of a circuit consisting of a battery of four 1.5 V cells, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor, and a plug key, all connected in series. Now find (I) the electric current passing through the circuit, and (II) the potential difference across the 10 Ω resistor when the plug key is closed. OR (b) (i) When is the potential difference between two points said to be 1 volt? (ii) A copper wire has a diameter of 0.2 mm and resistivity of 1.6 x 10-8 Ω m. What will be the length of this wire to make its resistance 14 Ω? How much does the resistance change if the diameter of the wire is doubled? (4 to 5 Marks(2024)
The potential difference across the ends of a conductor is directly proportional to the current flowing through it, provided its temperature remains the same. (ii) Total Voltage = V = 4 x 1·5 V = 6 V Total resistance, R(s) = R1+ R2 + R3 = 5 Ω + 10 Ω + 15 Ω = 30 Ω
(I) Current, I = VR = 6 V30 Ω = 0.2 A
(II) V = IR = 0.2 A × 10 Ω = 2 V
OR
(b) (i) When 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
(ii) d = 0.2 mm = 2 × 10⁻⁴ m; R = 14 Ω
ρ = 1.6 × 10⁻⁸ Ω m; A = πd²4
R = ρ lA = 4 ρ lπd² or l = π d² R4 ρ
l = 227 × (2 × 10⁻⁴)²4 × 1.6 × 10⁻⁸ × 14
= 27.5 m
When the diameter is doubled, d’ = 2d A’ = 4A
R’R = AA’ or R’ = RAA’ = RA4A
R’14 = A4A
R’ = 3.5 Ω
Change (14.0 – 3.5) = 10.5 Ω
Q15: In the given circuit the total resistance between X and Y is:
Ans: (d) To find the total resistance between points X and Y, we need to analyze the circuit configuration.
The given resistances are: 2 Ω, 3 Ω, and 6 Ω.
Looking at the circuit: The 3 Ω and 6 Ω resistors are connected in parallel. This parallel combination is then in series with the 2 Ω resistor. Step 1: Calculate the equivalent resistance of the parallel combination of 3 Ω and 6 Ω. Using the formula for parallel resistance:
1/Rparallel = 1/3 + 1/6
1/Rparallel = 2 + 1/6 = 3/6 = 1/2
So, Rparallel = 2Ω. Step 2: Add the series resistance. Now, the 2 Ω (from the parallel combination) is in series with the 2 Ω resistor at the start: Rtotal = 2 + 2 = 4Ω Therefore, the total resistance between points X and Y is 4 Ω. The correct answer is (b) 4 Ω.
Q16: Draw a schematic diagram of a circuit consisting of a battery of four dry cells of 1.5 V each, a 2 Ω resistor, a 6 Ω resistor, a 16 Ω resistor and a plug key all connected in series. Put an ammeter to measure the current in the circuit and a voltmeter across the 16 Ω resistor to measure the potential difference across its two ends. Use Ohm’s law to determine: (A) ammeter reading, and (B) voltmeter reading when the key is closed. (3 Marks) (2024)
Equivalent resistance in the circuit: R = 2 + 6 + 16 = 24 Ω The voltage supplied by the battery: V = 4 × 1.5 = 6 V (A) Using Ohm’s law we can calculate ammeter’s reading I. I = V / R = 6 / 24 = = 0.25 A (B) Voltmeter’s reading is: V’ = IR = 0.25 ×16 = 4 V
Previous Year Questions 2023
Q1: The expressions that relate (i) Q, I and t and (ii) Q, V and W respectively are (here the symbols have their usual meanings): (1 Mark) (2023) (a) (i) I = Q/t (ii) W = V/Q (b) (i) Q = I × t (ii) W = V × Q (c) (i) Q = I/t (ii) V = W/Q (d) (i) I = Q/t (ii) Q = V/W
Ans: (b) The correct expressions that relate charge (Q), current (I), time (t), voltage (V), and work done (W) are as follows: For the relationship between charge, current, and time: Q = I × t (the total charge is the product of current and time). For the relationship between work, voltage, and charge: W = V × Q (the work done is the product of voltage and charge). Thus, the correct answer is (b) (i) Q = I × t and (ii) W = V × Q.
Q2: (i) How is electric current related to the potential difference across the terminals of a conductor? Draw the labelled circuit diagram to verify this relationship. (4 to 5 Marks) (2023) (ii) Why should an ammeter have low resistance? (iii) Two V-I graphs A and B for series and parallel combinations of two resistors are as shown. Giving reason state which graph shows (a) series, (b) parallel combination of the resistors.
Ans: (i) It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically, V ∝ I V = RI where R is resistance of the conductor. (ii) To measure the entire current passing through the circuit, the ammeter should have low resistance. (iii) Series Combination:
RS = R1 + R2 = Maximum resistance. Therefore, the V-I graph for a series combination will have a steeper slope, indicating that the current increases less with the potential difference. Parallel Combination:
Therefore, the V-I graph for a parallel combination will have a flatter slope, showing that the current increases more sharply with the potential difference.So, RA > RB
Graph A: The graph with a steeper slope corresponds to the series combination of resistors.
Graph B: The graph with a flatter slope corresponds to the parallel combination of resistors.
Q3: (a) Define electric power and state its SI unit. The commercial unit of electrical energy is known as ‘unit’. Write the relation between this ‘unit’ and joule. (5 Marks) (2023) (b) In a house, 2 bulbs of 50 W each are used for 6 hours daily and an electric geyser of 1 kW is used for 1 hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ 8.00 per kWh.
Ans: (a) The electric power of an electrical device is the rate at which electric energy is dissipated or consumed in an electric circuit. Power = Work/Time = Energy/Time The SI unit of electric power is watt (W). 1W = 1 volt x 1 ampere = 1 VA 1 unit = 1 kWh = 3.6 x 106 J (b) 2 bulbs: P = 50 W, t = 6 hr daily 1 geyser: P = 1 kW, t = 1hr E=P x t used by every 2 bulbs + energy used by geyser = 30 days, Rs. = 8/kWh Now, total energy consumed in one day = 2 × 50 × 6 + 1 x 1000 = 1600 Wh Total energy consumed in 30 days = 30 x 1600 Wh = 48 kWh Bill = 8 x 48 = Rs. 384.
Q4: If four identical resistors, of resistance 8 ohms, are first connected in series so as to give an effective resistance Rs, and then connected in parallel so as to give an effective resistance Rp, then the ratio RS / RP is: (a) 32 (b) 2 (c) 0.5 (d) 16 (1 Mark) (CBSE 2023)
Ans: (d) Given: Four identical resistors, each with a resistance of 8Ω. Calculate the total resistance in series ( RS):When resistors are connected in series, the total resistance is the sum of individual resistances: RS = 8 + 8 + 8 + 8 = 4 × 8 = 32Ω Calculate the total resistance in parallel (RP):When resistors are connected in parallel, the total resistance is given by:
So, RP = 2Ω Calculate the ratio RS / RP
Therefore, the correct answer is (d) 16.
Q5: In the following diagram, the position of the needle is shown on the scale of a voltmeter. The least count of the voltmeter and the reading shown by it respectively are:
(a) 0.15 V and 1.6 V (b) 0.05 V and 1.6 V (c) 0.15 V and 1.8 V (d) 0.05 V and 1.8 V (1 Mark) (CBSE 2023)
Ans: (c) To determine the least count of the voltmeter and the reading shown by the needle, let’s analyze the diagram.
Least Count Calculation:
The scale on the voltmeter ranges from 0 to 3 volts.
There are 15 divisions between 0.0 and 1.5 volts, and similarly 15 divisions between 1.5 and 3 volts.
Therefore, each division represents: 1.5V / 15 = 0.1V
So, the least count of the voltmeter is 0.1 V.
Reading of the Voltmeter:
The needle is positioned at 1.8 V on the scale.
Based on this analysis, the correct answer is:
Least count: 0.1 V
Reading shown by the voltmeter: 1.8 V
Thus, the correct option is (c) 0.15 V and 1.8 V.
Q6: (A) An electric iron consumes energy at a rate of 880 W when the heating is at the maximum rate and 330 W when the heating is at the minimum. If the source voltage is 220 V, calculate the current and resistance in each case. (B) What is the heating effect of electric current? (C) Find an expression for the amount of heat produced when a current passes through a resistor for some time. (4 to 5 Marks) (CBSE 2023)
Ans: (A) Power = VI = V2 / R, where V is voltage, I is current, R is resistance. First case, Power = 880 W Voltage = 220 V ⇒ 880 = 220 × I ⇒ I = 4 A ⇒ 880 = 2202 / R ⇒ R = 55 Ω Second Case, Power = 330 W P = VI = V2/R P = 220 × I ⇒ 330 = 2202 / R ⇒ R = 440 / 3 Ω (B) When a conductor provides resistance to current flow, the work done by the current in overcoming this resistance is converted into heat energy. This is known as the current heating effect. (C) The amount of work done W in carrying a charge Q through a wire of resistance R in time t is given by: W = Q × V Since Q = I × t Therefore. W = V × I × t where V is the potential difference across the wire. Since by Ohm’s law, V = IR Therefore, W = I2Rt The electric energy dissipated or consumed is directly proportional to the square of the current I, directly proportional to the resistance R and to the time t during which current flows. H = I2Rt
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Previous Year Questions 2022
Q1: (a) State Ohm’s Law. Represent it mathematically. (2022) (b) Define 1 ohm. (c) What is the resistance of a conductor through which a current of 0.5 A flows when a potential difference of 2V is applied across its ends?
Ans: (a) Ohm’s law states that the electric current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided physical conditions like temperature etc., are kept unchanged. Mathematically, V ∝ I or V/I = constant or V/I = R ⇒ V = IR where, R is called the resistance of the conductor. SI unit of resistance is ohm and is denoted by Ω. It is a constant of proportionality and its value depends upon the size, nature of material and temperature. (b) If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is said to be 1Ω. We know that, R = V/l
Therefore, 1 ohm = 1 volt1 ampere or 1Ω = 1 V1 A
(c) Given, potential difference V = 2V
Current, I = 0.5 A
Using Ohm’s law, V = IR
⇒ R = VI ⇒ R = 2V0.5 A = 4Ω
Q2: In the following figure, three cylindrical conductors A, B and C are shown along with their lengths and areas of cross-section.
If these three conductors are made of the same material and RA,RB, and RC be their respective resistances, then find (I) RA/RB, and (II)RA/RC . (2022)
Ans: (i) L We have three cylindrical conductors A, B and C- These three conductors are made of the same materials and their respective resistance are σ RA. RB and Rc.
∴ R = ρ LA
So, RA = ρ × LA = ρ LA
RB = ρ × L / 22A ⇒ ρ L4A
RC = ρ × LA / (2 × 2) ⇒ ρ LA
Then, (I) RARB = ρ LA / ρ L4A ⇒ 4; RA : RB = 4
(II) RARC = ρ LA / ρ LA ⇒ 1
Q3: (a) List the factors on which the resistance of a uniform cylindrical conductor of a given material depends. (b) The resistance of a wire of 0.01 cm radius is 10Ω. If the resistivity of the wire is 50 x 10-8 Ω m, find the length of this wire. (2022)
Ans: (a) The resistance of a uniform cylindrical conductor depends on (i) Length of the conductor (l) (ii) Area of cross-section of the conductor (A) (b) The formula of resistance, R = ρl/A,
Given, Radius, r = 0.01 cm = 0.0001 m
Resistance, R = 10 Ω
Resistivity, ρ = 50 × 10⁻⁸ Ω m
R = ρ (L / A)
L = (R × A) / ρ
= (10 × 3.14 × (0.0001)²) / (50 × 10⁻⁸)
= 0.628 m
Hence, the length of the wire is 0.628 m.
Q4: Calculate the equivalent resistance of the following electric circuit: (2022)
Ans: First, calculate the resistance of 2 series resistors inside the loop, i.e., = R1 +R2 =10Ω + 10Ω = 20Ω To calculate the equivalent resistance in the given electric circuit, let us find the parallel resistance. For that we use
= R1 × R2R1 + R2
= 20Ω × 20Ω20Ω + 20Ω
= 400Ω40Ω
= 10Ω
Now, again applying the series formula to add the resistors together =10Ω + 10Ω + 20Ω = 40Ω So the total resistance of the given 40Ω
Q5: (i) Write the formula for determining the equivalent resistance between A and B of the two combinations (I) and (II) of three resistors and arranged as follows:
(ii) If the equivalent resistance of the arrangements (I) and (II) are and respectively, then which one of the following graphs is correctly labelled? Justify your answer. (2022)
Ans: (i) We know that resistance R1, R2 and R3 are in series. So, RAB = R1 + R2 + R3 So,
1/R = 1/R1 + 1/R2 + 1/R3
1/R = R2R3 + R3R1 + R1R2/R1R2R3
(ii) The slope of V-I graph gives the resistance. The greater the slope greater will be the resistance. We know, Rs > Rp ∴ Graph (I) is correct
Q6: Study the following electric circuit in which the resistors are arranged in three arms A, B and C (2022) (a) Find the equivalent resistance of arm A. (b) Calculate the equivalent resistance of the parallel combination of the arms B and C. (c) (i) Determine the current that flows through the ammeter. OR (ii) Determine the current that flows in the ammeter when the arm B is withdrawn from the circuit
Ans: The equivalent resistance in the arm A = 5Ω + 15Ω + 20Ω =40 Ω (b) The equivalent resistance of the parallel combination of the arms B and C= ((1/10Ω + 1/20Ω + 1/30Ω) + (1/5Ω + 1/10Ω + 1/15Ω))-1 =20Ω (c) (i) Current flow flowing through the ammeter = 6/60 = 0.1 A OR (ii) Current that flows in the ammeter when the arm B is withdrawn from the circuit = 6/60 = 0.1A
Q7: (i) State Joule’s law of heating. Express it mathematically when an appliance of resistance R is connected to a source of voltage V and the current I flows through the appliance for a time t. (ii) Α 5 Ω resistor is connected across a battery of 6 volts. Calculate the energy that dissipates as heat in 10 seconds. (2022)
Ans: (i) According to Joule’s law of heating, when a current (I) is passed through a conductor of resistance (R) for a certain time (t), the conductor gets heated up and the amount of energy released is given by
Q8: (a) Calculate the resistance of a metal wire of length 2 m and area of cross-section 1.55 × 10-6 m2 (Resistivity of the metal is 2.8 × 10-8 Ωm) (b) Why are alloys preferred over pure metals to make the heating elements of electrical heating devices? (2022)
Ans: (a) Here, the length of the wire, l = 2m Area of cross-section, A = 1.55 × 10-6 m2 Resistivity of metal, ρ = 2.8 × 10-8 Ω m
∴
(b) Why Alloys Are Preferred Over Pure Metals for Heating Elements
Higher Resistivity: Alloys have higher resistivity than pure metals, which helps in generating more heat.
Less Oxidation & Corrosion: Alloys do not oxidize or corrode easily at high temperatures, increasing durability.
Better Heat Tolerance: Alloys can withstand high temperatures without melting or softening.
Stable Resistance: The resistance of alloys does not change significantly with temperature, ensuring consistent heating. Thus, alloys are preferred for making heating elements in electrical heating devices.
Q9: An electric heater rated 1100 W operates at 220 V. Calculate (i) its resistance, and (ii) the current drawn by it. (2022)
Ans: Power of electric heater, P = 1100 W Operating voltage, V = 220 V (i) Its resistance, R= V2/P R = 220 x 220/1100 Ω : R = 44 Ω operating voltage, V = 220 V So, current I = V/R = 220/44= 5A (ii) Using the formula: I = P / V = 1100 / 220 = 5 A
Q10: (a) What is the meaning of electric power of an electrical device? Write its SI unit. (b) An electric kettle of 2kW is used for 2 hours. Calculate the energy consumed in (i) kilowatt-hour and (ii) joules (2022)
Ans: (a) The electric power of an electrical device is the rate at which electric energy is dissipated or consumed in an electric circuit. Power = Work/Time = Energy/Time The SI unit of electric power is watt (W) 1 W =1 volt × 1 ampere = 1 V A (b) (i) Electrical energy is the product of power and time. E = P x t = 2 kW × 2h = 4kW h (ii) Electric energy consumed in joules 1 kWh =3.6× 106 J 4 kW h = 3.6 × 106 x 4 = 14.4 x 106 J.
Q11: (i) Define electric power and write its SI unit. (ii) Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to an electric mains supply. What current is drawn from the line if the supply voltage is 220 V? (2022)
Ans: (i) The rate at which electric energy is dissipated or consumed in an electric circuit is called electric power. SI unit of electric power is watt (W). (ii) Given, P1 = 100 W, V1 = 220 V P2 = 60 W, V2 = 220 V P = VI
∴ Total current = l1 + l2 = 0.45 + 0.27 = 0.72 A
Q12: Define the term electric power. An electric device of resistance R when connected across an electric source of voltage V draws a current I. Derive an expression for the power in terms of resistance R and voltage V. What is the power of a device of resistance 400 Ω operating at 200 V ? (2022)
Ans: Electric Power: The rate at which electrical energy is consumed or dissipated is called electrical power. P = VI, [∵ V = IR] P = (IR) I = I2 R
Given:
Resistance (R) = 400 Ω
Voltage (V) = 200 V
Power Calculation: Using the formula: P = V² / R = (200 × 200) / 400 = 40000 / 400 = 100 W
Previous Year Questions 2021
Q1: Two lamps one rated 100W at 220V, and the other 60W at 220V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220V? (2021)
Q1: If a person has five resistors each of value 5Ω, then the maximum resistance he can obtain by connecting them is (2020) (a) 1Ω (b) 5Ω (c) 10Ω (d) 25Ω
Ans: (a) Sol: The maximum resistance can be produced from a group of resistors by connecting them in series. Thus, Rtotal = 12 + 12 + 12 + 12 = 4 × 12 = 2Ω
Q4: A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph. (2020)
Ans: V-I graph: The nichrome cable’s V-I graph exhibits a straight line. It signifies that the resistance of the wire stays unchanged while the current supply has been varied. That is an ohmic conductor because it follows the ohm’s law. Mathematically, ohm’s law can be represented as: V = IR Here, R is the resistance. V is the voltage. I is the current. The resistance of a wire will be: ⇒ R = V/I = 0.4/0.1 = 4Ω Thus, the circuit diagram for the given case is, Therefore, because nichrome wire has a constant resistance of 4 but also obeys Ohm’s law, it is classified as an ohmic conductor.
Q5: (a) Write the mathematical expression for Joule’s law of heating. (b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V. (2020)
Ans: (a) As per mathematical expression for Joule.s law of heating Heat produced H = VIt = I2 Rt = V2/R t When a voltage V is applied across a resistance R for time t so that a current I flows through it. (b) Here charge Q = 96000 C, time t = 2 h and potential difference V = 40 V Heat generated H=VIt = VQ ⇒ H = 40 x 96000 = 3840000 J = 3.84 x 106 J
Q6: Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1A. The current through the 40 W bulb will be: (2020) (a) 0.4A (b) 0.6A (c) 0.8A (d) 1A
The same current will flow through both bulbs in series.
When the same current flows through all of the circuit’s components, the circuit is said to be connected in series.
The current has only one path in such circuits.
As an example of a series circuit, consider the household decorative string lights.
This is nothing more than a string of tiny bulbs connected in series.
If one of the bulbs in a series burns out, none of the others will light up.
Hence, Option (D) is correct.
Q7: (a) What is meant by the statement, “The resistance of a conductor is one ohm”? (b) Define electric power. Write an expression relating electric power, potential difference and resistance. (c) How many 132 Ω resistors in parallel are required to carry 5 A on a 220 V line? (2020)
Ans: (a) If a current of 1 ampere flows through it when the potential difference across it is 1 volt (b) The electric power is the electric work done per unit time [P=W/T]. the relation between electric power , potential difference and resistance is P = V2/R (c) R= V/I = 220/5 = 44 ohm R= 44 = 132/n hence, 132/44 = 3 Resistors
Q8: (a) Write the mathematical expression for Joule’s law of heating. (b) Compute the heat generated while transferring 96000 coulombs of charge in two hours through a potential difference of 40 V. (2020)
Ans: (a) The Joule’s law of heating implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor. i.e., H = I2 Rt (b) Given, charge q = 96000C time t = 2h = 7200s and potential difference V = 40V We know,
Q9: A cylindrical conductor of length ‘l‘ and uniform area of cross-section ‘A‘ has resistance ‘R‘. Another conductor of length 2.5l and resistance 0.5R and of the same material has area of cross-section: (a) 5A (b) 2.5A (c) 0.5A (d) 1 / 5A (CBSE 2020)
ρ is the resistivity of the material (constant for a given material),
l is the length of the conductor,
A is the area of cross-section.
Given:
A conductor with length l, area A, and resistance R.
Another conductor of the same material with length 2.5l and resistance 0.5R.
Let the area of cross-section of the second conductor be A′.
Since the two conductors are made of the same material, their resistivity ρ is the same. We can set up the resistance formulas for each conductor:
For the first conductor:
R = ρ ⋅ lA
For the second conductor:
0.5R = ρ ⋅ (2.5l)A’
Now, substitute R = ρ ⋅ lA into the equation for the second conductor:
0.5 ⋅ ρ ⋅ lA = ρ ⋅ (2.5l)A’
Canceling ρ and l from both sides:
0.5 ⋅ 1A = 2.5A’
Rearranging to solve for A’:
A’ = 2.5 × 0.5A = 5A
Therefore, the area of cross-section A′A′ of the second conductor is 5A.
Q10: The resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects in the resistor will become: (a) two times (b) half (c) one-fourth (d) four times (CBSE 2020)
The heating effect H in a resistor is given by Joule’s law of heating:
H = I2Rt
where:
I is the current,
R is the resistance,
t is the time.
If the resistance R is reduced to half of its initial value, and assuming the voltage V across the resistor remains unchanged, the current I through the resistor will increase.
Using Ohm’s law: I = V/R
When R is reduced to R/2, the new current I’ becomes:
I’ = VR/2 = 2 × VR = 2I
Now, substituting into the heating formula:
H’ = (I’)² R’ t = (2I)² × R2 × t
= 4I² × R2 × t = 2I² R t
Thus, the heating effect becomes twice the initial heating effect. Therefore, the correct answer is (a) two times.
Q11: Assertion (A): Alloys are commonly used in electrical heating devices like electric iron and heater. Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points than their constitutent metals. (a) Both (A) and (R) are true, and (R) is the correct explanation of (A). (b) Both (A) and (R) are true, and (R) is not the correct explanation of (A). (c) (A) is true but (R) is false. (d) (A) is false but (R) is true. (CBSE 2020)
Ans: (a) Assertion (A): Alloys are commonly used in electrical heating devices like electric irons and heaters. This is true because alloys have certain properties that make them suitable for these applications, such as high resistivity, which allows them to produce heat effectively. Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points than their constituent metals.” This statement is also generally true. Alloys like nichrome have a higher resistivity, which is beneficial for heating applications, and they also tend to have controlled melting points, which can sometimes be lower than certain individual metals but still appropriate for their purpose in heating devices. Therefore, both the assertion and reason are correct, and the reason correctly explains why alloys are used in heating devices. So, the correct answer is (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
Q12: Assertion (A): At high temperatures, metal wires have a greater chance of short circuiting. Reason (R): Both resistance and resistivity of a material vary with temperature. (a) Both (A) and (R) are true, and (R) is the correct explanation of (A). (b) Both (A) and (R) are true, and (R) is not the correct explanation of (A). (c) (A) is true but (R) is false. (d) (A) is false but (R) is true. (CBSE 2020)
Ans: (b) Assertion (A): “At high temperatures, metal wires have a greater chance of short circuiting.” This statement is true. At high temperatures, metal wires can overheat, which may lead to insulation breakdown, increasing the risk of short circuits. Reason (R): “Both resistance and resistivity of a material vary with temperature.” This statement is also true. The resistance and resistivity of metals generally increase with temperature due to increased atomic vibrations, which impede the flow of electrons. However, the reason is not the direct cause of the assertion. The increased risk of short circuiting at high temperatures is mainly due to the possibility of insulation breakdown and not solely because of changes in resistance or resistivity. Therefore, the correct answer is (b) Both (A) and (R) are true, and (R) is not the correct explanation of (A).
Q13: (A) An electric bulb is rated at 200 V; 100 W. What is its resistance? (B) Calculate the energy consumed by 3 such bulbs if they glow continuously for 10 hours for complete month of November. (C) Calculate the total cost if the rate is ₹ 6.50 per unit. (CBSE 2020)
(B) P = 100 W (given 3 such bulbs) = 100 / 1000 = 0.1 kW No. of bulbs = 3 time, t = 10 hours Number of days in the month of November = 30 E = P × t Total energy consumed by 3 bulbs in 30 days = 3 × 0.1 × 10 × 30 = 90 kWh Hence, the energy consumed by 3 such bulbs for the month of November will be 90 kWh. (C) Total cost = ? Rate of per unit = ₹ 6.50 per unit 1 kWh = 1 unit 90 kWh = 90 units Tost cost = 90 × 6.50 = ₹ 585.00 Hence, the total of operating 3 such bulbs will ₹ 585.
Previous Year Questions 2019
Q1: A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively. Which of the following is true? (2019) (a) R1 = R2 = R3 (b) R1 > R2 > R3 (c) R3 > R2 > R1 (d) R2 > R3 > R1
Ans: (c) Sol: Resistance is given by the slope of V-I graph. Here the graph is I-V graph. So, resistance is given by the inverse of the slope. From the graph, the order of the slope is slope of R1 > slope of R2 > slope of R3 Therefore order of resistance is R3 > R2 > R1. Hence, option (c) is correct.
Q2: When do we say that the potential difference between two points of a circuit in 1 volt? (2019)
Ans: Potential difference between two points of an electric circuit is said to be 1 volt, when a work of 1 J is to be done for moving a charge of 1 C between these two points.
Ans: Resistance of a conductor is the measure of opposition offered by it for the flow of electric charge through it. SI unit of resistance is ‘ohm’ (Ω).
Ans: According to Ohm’s law, temperature remaining constant, the current passing through a conductor is directly proportional to the potential difference across its ends, i.e., V ∝ I or V = IR Here, constant R is known as the resistance of given conductor.
Q5: In the circuit given below, the resistors and have the values and respectively, which have been connected to a battery of 12 V. Calculate
(a) the current through each resistor,(b) the total circuit resistance, and(c) the total current in the circuit. (2019)
(a) As all the resistances are in parallel, the voltage across each of them will be the same, which is 12V.
I1 = V / R1 = 12 / 5 = 2.4A
I2 = V / R2 = 12 / 10 = 1.2A
I3 = V / R3 = 12 / 30 = 0.4A
(b) Total current = I1 + I2 + I3 = 2.4 + 1.2 + 0.4 = 4A
(c) Total circuit resistance = V / Total current = 12 / 4 = 3Ω
Q6: On what factors does the resistance of a conductor depend? Or List the factors on which the resistance of a conductor in the shape of a wire depends. (2019)
Ans: Resistance is defined as the opposition to the flow of electrical current through a conductor. The resistance of an electric circuit can be measured numerically. Conductivity and resistivity are inversely proportional. The more conductive, the less resistance it will have. Resistance = Potential difference/ Current Factors on which the conductor depends
The temperature of the conductor
The cross-sectional area of the conductor
Length of the conductor
Nature of the material of the conductor
Electrical resistance is directly proportional to the length (L) of the conductor and inversely proportional to the cross-sectional area (A). It is given by the following relation. R = ρl/A where ρ is the resistivity of the material (measured in Ωm, ohm meter) Resistivity is a qualitative measurement of a material’s ability to resist flowing electric current. Obviously, insulators will have a higher value of resistivity than of conductors.
Q7: (a) A bulb is rated 40 W, 220 V. Find the current drawn by it when it is connected to a 220 V supply. Also, find its resistance. (b) If the given blub is replaced by a blub of rating 25 W, 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change. (2019)
Ans: (a) Here power of bulb P = 40 W and voltage V = 220 V
∴ Current drawn by the bulb I = PV = 40220 = 211 A
and resistance of the bulb R = VI = 220(2 / 11) = 220 × 112 = 1210 Ω
(b) On taking another bulb of power P’ = 25 W and voltage V = 220 V, there is a change in the value of current and resistance because their values depend on the power of the bulb.
New current I’ = P’V = 25220 = 544 A
and new resistance R’ = VI’ = 220(5 / 44) = 220 × 445 = 1936 Ω
Q8: Derive the expression for power P consumed by a device having resistance R and potential difference V. Or A device of resistance R is connected across a source of V voltage and draws a current I . Derive an expression for power in terms of voltage (or current) and resistance. (2019)
Ans: When a 2 Ω resistor is joined to a 6 V battery in series with 1 Ω and 2 Ω resistors, total resistance of the combination Rs = 2 + 1 + 2 = 5Ω
∴ Current in the circuit I1 = 6V5Ω = 1.2 A
∴ Power used in the 2Ω resistor P1 = I1² R = (1.2)² × 2 = 2.88 W.
(ii) When 2Ω resistor is joined to a 4V battery in parallel with 12Ω and 2Ω resistors, current flowing in 2Ω resistor is independent of the other resistors.
∴ Current flowing through 2Ω resistor I2 = 4V2Ω = 2 A
∴ Power used in the 2Ω resistor P2 = I2² R = (2)² × 2 = 8 W
P1P2 = 2.88 W8 W = 0.36 : 1.
Q10: Derive the relation R = R1 + R2 + R3 when resistors are joined in series. (2019)
In series combination, the same current flows in all the resistances but the potential difference across each of the resistance is different.
According to Ohm’s law, we have
V1 = IR1, V2 = IR2, V3 = IR3
If the total potential difference between A and B is V, then
V = V1 + V2 + V3
= IR1 + IR2 + IR3
= I(R1 + R2 + R3)
Let the equivalent resistance be R, then
V = IR
and hence IR = I(R1 + R2 + R3)
⇒ R = R1 + R2 + R3.
Q11: (a) Write the relation between resistance and electrical resistivity of the material of a conductor in the shape of a cylinder of length ‘l’ and area of cross-section ‘A’ Hence derive the SI unit of electrical resistivity. (b) Resistance of a metal wire of length 5 m is 100 Ω. If the area of cross-section of the wire is 3 x 10-7 m2, calculate the resistivity of the material. (2019)
Ans: (a) The resistance R of a conductor, the shape of a cylinder, of length l and area of cross-section A is given as: where ρ is a constant, which is known as the electrical resistivity of the material of conductor. Thus, resistivity ρ = RA/l ∴ SI unit of resistivity ρ shall be = Ω-m (b) Here l = 5 m, R = 100 Ω and A = 3 x 10-7 m2 ∴ Resistivity of the material of wire ρ = RA/l =
Q12: (a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistors joined in parallel is equal to the sum of the reciprocals of the individual resistances. (b) In an electric circuit two resistors of 12Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery. (2019)
In parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 I2 and I3. Thus, I = I1 + I2 + I3. The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
I1 = V/R1, I2 = V/R2, I3 = V/R3
If R is the equivalent resistance then,
I = V/R
∴ V/R = V/R1 + V/R2 + V/R3
and
∴ 1/R = 1/R1 + 1/R2 + 1/R3
(b) Here R1 = R2 = 12Ω and V = 6V
Net resistance R of the parallel grouping is:
1/R = 1/R1 + 1/R2 = 1/12 + 1/12 = 1/6 ⇒ R = 6Ω
∴ Current drawn by the circuit from the battery I = V/R = 6V/6Ω = 1 A.
Q13: Study the circuit of Fig. and find out : (i) Current in 12 Ω, resistor (ii) difference in the readings of A1 and A2 if any. (2019)
Ans: (i) In the circuit resistor of R1 = 12 Ω is connected in series with parallel combination of two resistors R2 and R3 of 24 Ω each. The effective resistance of parallel combination of R2 and R3 is given as :
∴ Net resistance of the circuit R = R1 + R23 = 12 + 12 = 24Ω
∴ Current through 12Ω resistor = Circuit current I = V/R = 6V/24Ω = 0.25 A
(ii) Both ammeters A1 and A2 give the same reading of 0.25 A and there is no difference in their readings.
Q14: (i) Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor and a plug key all connected in series. (ii) Calculate the electric current passing through the above circuit when the key is closed. (iii) Potential difference across 15 Ω resistor. (2019)
(i) The schematic diagram is given in Fig. 12.25. (ii) Here total voltage V = 5 x 2 = 10 V and total resistance R = R1 + R2 + R3 = 5 + 10 + 15 = 30 Ω ∴ Current passing through the circuit when the key is closed (iii) Potential difference across resistor R3 of 15 Ω
Q15: An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate (а) the total resistance of the circuit, (b) the current through the circuit, (c) the potential difference across the (i) electric, lamp and (ii) conductor, and (d) power of the lamp. (2019)
Ans: Here voltage of battery V = 6 V, resistance of electric lamp = R1 = 20 Ω and resistance of conductor R2 = 4 Ω. (a) Since R1 and R2 are connected in series, the total resistance of the circuit R = R1 + R2 = 20 + 4 = 24 Ω (b) The current through the circuit I = V/R = 6/24 = 0.25 A (c) (i) Potential difference across the electric lamp V1 = IR1 = 0.25 x 20 = 5 V (ii) Potential difference across the conductor V2 = IR2 = 0.25 x 4 = 1 V (d) Power of the lamp P = I2R1 = (0.25)2 x 20 = 1.25 W
Q16: (a) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected to a battery, ammeter, voltmeter and key. Draw suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors. (b) Calculate the equivalent resistance of the network shown in Fig. (2019)
(a) The arrangement is shown in circuit diagram of Fig. In parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 I2 and I3. Thus, I = I1 + I2 + I3. The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
I1 = V/R1, I2 = V/R2, I3 = V/R3
If R is the equivalent resistance then,
I = V/R
∴ V/R = V/R1 + V/R2 + V/R3
and
∴ 1/R = 1/R1 + 1/R2 + 1/R3
(b) In the network, resistors of 20Ω and 20Ω are joined in parallel and make a resistance R1, where
1/R1 = 1/20 + 1/20 = 1/10 or R1 = 10Ω.
This combined resistance R1 = 10Ω is joined in series with a given 10Ω resistance. Hence, the equivalent resistance of the network will be
R = 10 + 10 = 20Ω.
Q17: (a) Three resistors o f resistances R1, R2 and R3 are connected in (i) series, and (ii) parallel. Write expression for the equivalent resistance of the combination in each case. (b) Two identical resistances of 12 Ω each are connected to a battery of 3 V. Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance. (2019)
Ans: (a) (i) In series arrangement, equivalent resistance Rs = R1 + R2 + R3 (ii) In parallel arrangement, equivalent resistance Rp is given as:
1/Rp = 1/R1 + 1/R2 + 1/R3
(b) Here R1 = R2 = 12Ω and V = 3V.
For minimum resistance, two resistors must be connected in parallel so that
1/Rp = 1/R1 + 1/R2 = 1/12 + 1/12 = 1/6 ⇒ Rp = 6Ω
Hence power
Pp = V²/Rp = (3)²/6 = 1.5 W
For maximum resistance, two resistors must be connected in series so that
Rs = R1 + R2 = 12 + 12 = 24Ω
So, the power
Ps = V²/Rs = (3)²/24 = 0.375 W
⇒ Pp/Ps = 1.5/0.375 = 4
Q18: Experimentally prove that in series combination of three resistances: (а) current flowing through each resistance is same, and (b) total potential difference is equal to the sum of potential differences across individual resistors. (2019)
Ans: Series combination o f resistors : We take three resistors R1 R2 and R3 and join them in series between the points X and Y in an electric circuit as shown in Fig. 12.42. (a) Plug the key and note the ammeter reading. Then change the position of ammeter to anywhere in between the resistors and again note the ammeter reading. We find that ammeter reading remains unchanged. It shows that in series arrangement same current flows through each resistor. (b)
Insert a voltmeter across the ends X and Y of the series combination of resistors.
Plug the key so as to complete the circuit and note the voltmeter reading V across the series combination of resistors.
Take out plug from key K and disconnect the voltmeter. Now insert the voltmeter across the ends of first resistor R1 as shown in Fig. 12.43.
Plug the key and note the voltmeter reading V1. Similarly, measure the potential difference across the other two resistors R2 and R3 separately.
Let these potential differences be V2 and V3, respectively. Experimentally we find that V = V1 + V2 + V3
It shows that in series arrangement of resistors total potential difference is equal to the sum of potential differences across individual resistors.
Q19: Unit of electric power may also be expressed as: (a) volt-ampere (b) kilowatt-hour (c) watt-second (d) joule-second (CBSE 2019)
Ans: (a) The unit of electric power can be expressed in terms of volt-ampere (V·A), which represents the power calculated as the product of voltage (in volts) and current (in amperes). This is particularly common in electrical engineering, where 1 watt = 1 volt × 1 ampere. Here’s an explanation of the other options: (b) kilowatt-hour: This is a unit of energy, not power. It represents the amount of energy used over time (1 kilowatt of power used for 1 hour). (c) watt-second: This is also a unit of energy, as it represents the power used over a second (1 watt of power used for 1 second). (d) joule-second: This is a unit related to angular momentum or action in physics, not specifically for electric power. Therefore, the correct answer is (a) volt-ampere.
Q20: When a 4 V battery is connected across an unknown resistor there is a current of 100 mA in the circuit. The value of the resistance of the resistor is: (a) 4 Ω (b) 40 Ω (c) 400 Ω (d) 0.4 Ω (CBSE 2019)
Ans: (b) Using Ohm’s law: V = I × R where: V = 4V (voltage), I = 100mA = 0.1A (current). We can rearrange the formula to solve for R :
Therefore, the resistance of the resistor is 40 Ω.
Q21: (A) In a given ammeter, a student saw that needle indicates 12th division in ammeter while performing an experiment to verify Ohm’s law. If ammeter has 10 divisions between 0 to 0·5 A, then what is the ammeter reading corresponding to 12th division? (B) How do you connect an ammeter and a voltmeter in an electric circuit? (CBSE 2019)
Ans: (A) Least count of ammeter = 0.5/10 = 0.05 A Thus, value corresponding to 12 divisions = 0.05 ×12 = 0.6 A (B) An ammeter is connected in series and a voltmeter is connected in parallel in an electric circuit.
Previous Year Questions 2018
Q1: Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω? (2018)
Ans: Here resistances R1 – R2 = R3 = 9Ω (i) To obtain an equivalent resistance Req = 13.5Ω, we connect one resistor R1 in series to the parallel com bination o f R2 and R3 as shown in figure (i). Then (ii) To obtain equivalent resistance Req = 6 Ω, we connect resistor R1 in parallel to the series combination of R2 and R3 as shown in figure (ii). Then
Ans: As per Joule’s law the heat produced in a resistor is (i) directly proportional to square of current flowing through it, (ii) directly proportional to resistance, and (iii) directly proportional to time. Mathematically, Heat H = I2Rt
Q3: Why are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason. (2018)
The electrons in the metal are loosely bound while the electrons in the glass are tightly bound together.
Glass has one of the lowest possible heat conduction a solid. A good electrical conductivity is the same as a small electrical resistance.
Silver is the best conductor because its electrons are freer to move than those of the other elements.
When electricity is applied to the metal, ions start to migrate from one end to the other end of the metal, but it is not possible for electrons to travel freely in glass in which electrons are tightly bound.
Glass is a bad conductor of electricity as it has high resistivity and has no free electrons.
In parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 I2 and I3. Thus, I = I1 + I2 + I3. The potential difference across each of these resistances is the same. Thus, from Ohm’s law If R is the equivalent resistance then, I = V/R ∴ and
Previous Year Questions 2016
Q1: Why do electricians wear rubber hand gloves while working? (2016)
Ans: Rubber is an electrical insulator. Hence electrician can work safely while working on an electric circuit without a risk of getting any electric shock.
Q2: How are two resistors with resistances R1Ω and R2Ω are to be connected to a battery of emf 3 volts to obtain maximum current flowing through it? (2016)
Ans: We know that the power can be stated as the amount of electric energy consumed per unit time. On solving For Energy, we get Energy, E = P/T Here; P = 60W t = 1s E = 60W x 1sec ∴ E = 60J Thus, the energy in joules is 60J.
Q5: (a) What do you mean by the resistance of a conductor? Define its unit. (b) In an electric circuit with a resistance wire and a cell, the current flowing is I. What would happen to this current if the wire is replaced by another thicker wire of the same material and the same length? Give reason. (2016)
Ans: (a) The resistance of a conductor is a property of the conductor, which affects the flow of current through it on maintaining a potential difference across its ends. Unit of resistance is ohm. Resistance of a conductor is said to be 1 ohm, if a potential difference of 1 V is to be applied across its ends for maintaining flow of 1 A current. (b) If given resistance wire is replaced by another thicker wire o f same material and same length then cross-section area of wire is increased and consequently its resistance decreases So the current flowing in the circuit increases.
Q6: (a) Why are electric bulbs filled with chemically inactive nitrogen or argon gas? (b) The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 x 10-8 Ω-m, find the length of the wire. (2016)
Ans: (a) Electric bulbs are generally filled with some inert gas like nitrogen or argon. This enables to prolong the life of the filament of electric bulb. (b) Here radius of wire r = 0.01 cm = 0.01 x 10-2 m, resistance R = 10 Ω and resistivity p = 50 x 10-8 Ω-m.
As R = ρLA = ρLπr², hence length L = Rπr²ρ
⇒ L = 10 × 22 × (0.01 × 10⁻²)²7 × 50 × 10⁻⁸ = 2235 m = 0.629 m or 62.9 cm
Q7: (a)Define electric power. Express it in terms of V, I and R where V stands for potential difference, R for resistance and I for current. (b) V -I graphs for two wires A and B are shown in the Fig. 12.34. Both of them are connected in series to a battery. Which of the two will produce more heat per unit time? Give justification for your answer. (2016)
Ans: (a) Electric power is defined as the rate of supplying electrical energy for maintaining current flow through a circuit. Electric power P = VI = I2R = V2/R. (b) We know that slope of V-I graph for a given wire gives its resistance and in given figure slope of graph is more for wire A. It means that RA > RB. In series arrangement same current I flows through both the resistance. As heat produced per unit time is given by I2R, hence it is obvious that more heat will be produced per unit time in wire A.
Q8: (a) Establish a relationship to determine the equivalent resistance IS of a combination of three resistors having resistances R1, R2 and R3 connected in parallel. (b) Three resistors are connected in an electrical circuit as shown. Calculate the resistance between A and B. (2016)
Ans: (a) The arrangement is shown in circuit diagram of Fig. 12.39. In parallel combination of three resistances R1 R2 and R3, the current in each of the resistances is different. If I is the current drawn from the cell then it is divided into branches I1 I2 and I3. Thus, I = I1 + I2 + I3. The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
I1 = V/R1, I2 = V/R2, I3 = V/R3
If R is the equivalent resistance then,
I = V/R
∴ V/R = V/R1 + V/R2 + V/R3
and
∴ 1/R = 1/R1 + 1/R2 + 1/R3
(b) Here series combination of R1 and R2 is joined to R3 in parallel arrangement. Hence the net resistance R between points A and B is given as
Q9: (a) Establish a relationship to determine the equivalent resistance R of a combination of three resistors having resistances R1, R2 and R3 connected in series. (b) Calculate the equivalent resistance R of a combination of three resistors of 2 Ω, 3 Ω and 6 Ω joined in parallel. (2016)
Q8: Define an electric circuit. Draw a labelled, schematic diagram of an electric circuit comprising of a cell, a resistor, an ammeter, a voltmeter and a closed switch. Distinguish between an open and a closed circuit. (2015)
A continuous and closed path of an electric current is called an electric circuit.
A labelled, schematic diagram of an electric circuit showing a cell E, a resistor R, an ammeter A, a voltmeter V and a closed switch S is shown here.
An electric circuit is said to be an open circuit when the switch is in ‘off’ mode (or key is unplugged) and no current flows in the circuit.
The circuit is said to be a closed circuit when the switch is in ‘on’ mode (or key is plugged) and a current flows in the circuit.
Q9: (a) n electrons, each carrying a charge -e, are flowing across a unit cross section of a metallic wire in unit time from east to west. Write an expression for electric current and also give its direction of flow. Give reason for your answer. (b) The charge possessed by an electron is 1.6 x 10-19 coulomb. Find the number of electrons that will flow per second to constitute a current of 1 ampere. (2015)
Ans:(a) Electric current in a circuit is defined as the time rate of flow of electric charge through any cross-section and its direction is opposite to that of flow of electrons. Hence in present case electric current I = As electrons are flowing from east to west, the direction of electric current is from west to east. (b) Here current I = 1 A, time t = 1 s and charge on each electron e = 1.6 x 10-19 C. Hence, number of electrons flowing n =
Q10: (a) List the factors on which the resistance of a cylindrical conductor depends and hence write an expression for its resistance. (b) How will the resistivity of a conductor change when its length is tripled by stretching it? (2015)
Ans: (a) The resistance of a cylindrical conductor i.e., a wire (R) is (i) directly proportional to its length L, (ii) inversely proportional to its cross-section area A and (iii) depends on the nature of material of wire. Mathematically, Here, ρ is known as the resistivity of given material. It is defined as the resistance offered by a unit cube of given material when current flows perpendicular to the opposite faces. (b) The resistivity of the conductor remains unchanged.
Q11: (a) V-I graphs for two wires A and B are shown in the Fig. If both the wires are made of same material and are of same length, which of the two is thicker? Give justification for your answer. (b) A wire of length L and resistance R is stretched so that the length is doubled and area of cross-section halved. How will (i) resistance change, and (ii) resistivity change? (2015)
Ans: (a) Resistance R of wire A is more than that of B (RA > RB) because slope of V-I graph for A is more. The resistance of a wire is inversely proportional to its cross-section area. Hence area of cross-section of wire B, of smaller resistance, must be more. Thus, wire B is thicker. (b) (i) The new resistance of wire Thus, resistance increases to four times its original value. (ii) The resistivity remains unchanged because it does not depend on the dimensions of a conductor of given material.
Q12: Study the electric circuit of Fig. and find (i) the current flowing in the circuit, and (ii) the potential difference across 10 Ω resistor. (2015)
Ans: (i) Here V = 3 V, R1 = 10 Ω and R 2 = 20 Ω. Since R1 and R2 are connected in series, the effective resistance of the circuit R = R1 + R2 = 10 + 20 – 30 Ω ∴ Current flowing in the circuit (ii) The potential difference across R 1 = 10 Ω resistor is V1 = IR1 = 0.1 x 10 = 1.0 V.
Q13: Three resistors of 3 Ω each are connected to a battery of 3 V as shown in Fig Calculate the current drawn from the battery. (2015)
Ans: In the electric circuit shown the series combination of R1 and R2 is joined in parallel to the resistance R3. Hence equivalent resistance R of the circuit is
∴ Current drawn from the battery I = V/R = 3/2 = 1.5 A
Q14: A 5 Ω resistor is connected across a battery of 6 volts. Calculate: (i) the current flowing through the resistor. (ii) the energy that dissipates as heat in 10 s. (2015)
Ans: Here V = 6 V and R = 5 Ω (i) The current flowing through the resistor I = (ii) Energy dissipated as heat in time t = 10 s is ∴ H = I2Rt = (1.2)2 x 5 x 10 = 72 J
Q15: Calculate the amount of heat generated while transferring 90000 coulombs of charge between the two terminals of a battery of 40 V in one hour. Also determine the power expended in the process. (2015)
Ans: Here charge transferred Q = 90000 C, potential difference b etw een the terminals o f battery V = 40 V and time t = 1 h = 3600 s. Current = Amount of heat generated H = Vlt = 40 x 25 x 3600 = 3600000 J = 3.6 x 106 J and power expended
Q16: How many 40 W; 220 V lamps can be safely connected to a 220 V, 5 A line? Justify your answer. (2015)
Ans: The current drawn by a 40 W, 220 V electric lamp As the electric line is o f rating 220 V, 5 A, hence we can connect n lamps in parallel where Thus, we can safely connect 27 lamps of 40 W, 220 V rating to a 220 V, 5 A line.
Q17: What is meant by electric current ? Name and define its SI unit. In a conductor electrons are flowing from B to A. What is the direction of conventional current ? Give justification for your answer. A steady current of 1 ampere flows through a conductor. Calculate the number of electrons that flow through any section of the conductor in 1 second. (Charge on electron — 1.6 x 10-19 coulomb) (2015)
Electric current is defined as the rate of flow of electric charge through a cross- section of a conductor.
If Q charge passes through a section of a conductor in time t, then-current I = Q/t.
SI unit of electric current is an ampere (A). Current is said to be one ampere, if rate of flow of charge through a cross-section of conductor be 1 coulomb per second.
Direction of conventional current is taken as the direction of flow of positive charge or opposite to the direction of flow of negative charge. If negatively charged in a conductor flow from B to A then the direction of conventional current will be from A to B.
Here current I = 1 A, time t = 1 s and charge on electron e = 1.6 x 10-19 C.
Let n electrons flow through a section of conductor so that charge passing through the section is Q= ne. ∴
Q18: What is heating effect of electric current ? Find an expression for amount of heat produced. Name some appliances based on heating effect of current. (2015)
Ans: When a current flows through a conducting wire (resistance wire), heat is developed, and the temperature of the wire rises. It is known as the heating effect of electric current. If V is the potential difference maintained across the ends of a wire then, by definition, the amount of work done for flow of 1 C charge through the wire is V. ∴ Work done for flow of Q charge W = V/Q = V/It [∵ Q = It] where I is the current flowing in time t. As V = IR, hence W= V/It = (IR)It = I2Rt This work done (i.e., electrical energy dissipated) is converted into heat. Hence, the amount of heat produced, Q = I2Rt J This is known as Joule’s law of heating. Incandescent lamps, electric irons, electric stoves, toasters, geysers, electric room heaters, etc., are appliances based on the heating effect of electric current.
Q19: What is the minimum resistance which can be made using five resistors, each of 1 / 5Ω? (a) 1 / 5Ω (b) 1 / 25Ω (c) 1 / 10Ω (d) 25Ω (CBSE 2015, 13, 12)
Ans: (b) To achieve the minimum resistance, all resistors should be connected in parallel because the equivalent resistance of resistors in parallel is always less than any individual resistor. Given: Each resistor has a resistance of 1/5Ω. There are five resistors. When resistors are connected in parallel, the equivalent resistance Req is given by:
Therefore, the minimum resistance that can be made using five resistors, each of 1/5Ω, is 1/25Ω.
Previous Year Questions 2014
Q1: Should the resistance of an ammeter be low or high? Give reason. (CBSE 2014)
An ammeter is used to measure the current flowing through a circuit. To ensure accurate measurement of current, the ammeter should have very low resistance. This is because if the ammeter had high resistance, it would impede the flow of current, and as a result, the current in the circuit would decrease, leading to an incorrect reading.
By having low resistance, the ammeter minimizes the potential drop across it, ensuring that it does not alter the current in the circuit significantly and provides an accurate measurement.
In summary, low resistance in an ammeter ensures that it doesn’t affect the current flow in the circuit, allowing for precise measurement.
∴ By distance formula,
Given, diameter 
and the required ratio is 5 : 1.
⇒ 
⇒ x = 4, y = – 8∴ Coordinates of A are (4, – 8).
Since, the perimeter the right triangle is 60 cm, 
So, AQ + CY= 1 + 2 = 3 cm
By Basic Proportionality Theorem, 
Given: In ΔABC, ∵ IIBC, which intersects AB and AC at ‘D’ and ‘E’ respectively.
Given, in triangles ΔDEF and ΔPQR 
Given: ∠APC = ∠BAC
∴ DF/BF = AF/EF [Corresponding sides of similar triangles] 
Given, In ΔABC, AD ⊥ BC
The total area of the lawn and the walkway is 360 square metres. The width of the walkway is same on all sides. The dimensions of the lawn itself are 12 metres by 10 metres. Based on the information given above, answer the following questions: 
OR
Hence, (6, 0) is the solution of given system of equations.
is an irrational number. 
is an irrational number.
is an irrational number. 3
be a rational number.
where b ≠ 0 and a, b are co-prime integers.
which is a rational number.
is an irrational number.
is an irrational number.
be a rational number.
is an irrational number.
…[b ≠ 0; a and b are integers]
is a rational number.
Biomagnification of DDT in an Aquatic Food Chain
Types of Natural Ecosystem
Hence, B depends on:
(For a positive charge it would be into the page.)
(b) Permanent magnet / Current carrying solenoid/ Electromagnet is used to magnetise a piece of magnetic material. 
Hence, the total equivalent resistance becomes
= 0.04 Ω.
We can rearrange formula to solve for V:
Substitute the values:
⇒ The potential difference across the 4 Ω resistor is 20 volts.
(I) Draw a circuit diagram showing the above two resistors X and Y joined in parallel with the same battery and same ammeter and voltmeter. (1 Mark)
where:
(a) R1 = R2 = R3
= Ω-m
So the current flowing in the circuit increases.
Therefore, the minimum resistance that can be made using five resistors, each of 1/5Ω, is 1/25Ω.