Q1: A ball thrown up is caught by the thrower after 4 s. With what velocity was it thrown up? How high did it go? Where was it after 3 s ? (g = 9.8 ms–2) Ans: In case of upward motion of the ball from A to B
Initial velocity, u = ?
Final velocity, v = 0 (at maximum height)
Time taken by the ball to reach the highest point = 2s (time of ascent = time of descent)
Acceleration due to gravity g = -9.8 m/s2 (upward motion)
Finding the initial velocity of the ball
Using the first equation of motion, v = u + gt:
v = u – gt
0 = u – 9.8 × 2
u = 19.6 m/s
The initial velocity of the ball is 19.6 m/s.
Finding the maximum height (h) attained by the ball
Using the second equation of motion, h = ut + 1/2 gt2:
h = 19.6 × 2 – 1/2 × 9.8 × (2)2
h = 39.2 – 19.6
h = 19.6 m
Let the ball be at C after t = 3 sec
Consider motion from A to C
u = 19.6 m/s
t = 3 s
g = -9.8 m/s2
s = h’
s = ut + 1/2 gt2
h’ = 19.6 × 3 – 1/2 × 9.8 × (3)2
h’ = 58.8 – 44.1 = 14.7 m
Distance from top
x = h – h’
x = 19.6 – 14.7 = 4.9 m
Hence, the ball goes to the maximum height of 19.6 m, the velocity at which it was thrown is 19.6 m/s, and the distance below its highest point after 3 sec is 4.9 m.
Q2: Calculate the force of gravitation due to a child of mass 25 kg on his mother of mass 75 kg if the distance between their centres is 1 m from each other. Given G = (20/3) × 10–11 Nm2 kg–2 Ans: Here m1 = 25 kg; m2 = 75 kg; d = 1 m; Using,
F = 12,500 × 10–11 or F = 1.25 × 10–7 N
Q3: A sealed tin of Coca-Cola of 400 g has a volume of 300 cm3. Calculate the density of the tin. Ans: Here, mass of tin, M = 400 g Volume of tin, V = 300 cm3 Density of tin,
Q4: A sealed can of mass 600 g has a volume of 500 cm3. Will this can sink in water? Density of water is 1 g cm–3. Ans: Here, mass of can, M = 600 g Volume of can, V = 500 cm3 Density of can, Since, density of the can is greater than the density of water, so the can will sink in water.
Q5: The gravitational force between two objects is 49 N. How much distance between these objects be decreased so that the force between them becomes double? Ans: Let ‘r’ be the distance between the object of mass m1 and m2
Now, the distance is reduced to ‘x’ so that the force become twice, then
Dividing eq. (i) by (ii)
So, the distance must decrease by times the original distance.
Q6: A force of 200 N is applied perpendicular to its surface having area 4 square metres. Calculate the pressure. Ans: Thrust = 200 N Area = 4 m2 Pressure = ? Pressure = Thrust / Area = = 50 Nm–2 = 50 Pa
Q7: The density of water is 1000 kg m3. If relative density of iron is 7.874, then calculate the density of iron. Ans: Density of water = 1000 kg/m3 Relative density (R.D.) of iron = 7.874 Using, R.D. of iron we get Density of iron = R.D. of iron × density of water = 7.874 × 1000 kg/m3 = 7874 kg/m3.
Q8: What is the force of gravitation between two point masses of 1 kg and 2 kg kept 1 m apart? Ans: m = 1 kg, m2 = 2 kg, r = 1 m
= 13.34 x 10-11N This is an extremely small force.
Q9: A particle is thrown up vertically with a velocity of 50 m/s. (a) What will be its velocity at the highest point of its journey? (b) How high would the particle rise? (c) What time would it take to reach the highest point? Ans: At the highest point the velocity will be zero. Considering activity A to B Using = u + at 0 = 50 – 9.8 × t t = 5.1 sec Also v2 – u2 = 2as s = 127.5 m
Q10: Weight of a girl is 294 N. Find her mass. Ans: W = mg 294 = m × 9.8 m Q11: How much force should be applied on an area of 1 cm2 to get a pressure of 15 Pa? Ans: Here, Area, A = 1 cm2 = 10–4 m2 Pressure (P) = 15 Pa = 15 N/m2 As F = P × A = (15 N/m2) × (10–4 m2) = 1.5 × 10–3 N
Q12: A force of 20N acts upon a body whose weight is 9.8N. What is the mass of the body and how much is its acceleration? Given, Force = 20 N, Weight W = 9.8 N. We know, W = mg; 9.8 = m × 9.8 m = 1 kg Ans: We know F = ma 20 = 1 × a a = 20 m/s2
Q13: An object is thrown vertically upwards and reaches a height of 78.4 m. Calculate the velocity at which the object was thrown? (g = 9.8 m/s2) Ans: Given, h = 78.4 m v = 0 g = –9.8 m/s2 Now, v2 = u2 – 2ghu = 39.2 m/s2
Q14: Two bodies A and B having masses 2 kg and 4 kg respectively are separated by 2 m. Where should a body of mass 1 kg be placed so that the gravitational force on this body due to bodies A and B is zero? Ans: Mass of body a is Ma = 2 kg Mass of body b is Mb = 4 kg Mass of body c is Mc = 1 kg Separation between a and b = 2 m Let the body C be placed at a distance d from body A Gravitational force between A and C
Gravitational force between B and C is
For body C the gravitational force is 0.
Hence, FAC = FBC d = 0.83
Q15: Let us find force of attraction between two blocks lying 1 m apart. Let the mass of each block is 40 kg. Ans: F = ? m1 = 40 kg m2 = 40 kg d = 1 m G = 6.67 × 10–11 Nm2kg–2 = 1.0672 x 10-7N
Newton’s I law of motion : An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.
Newton’s II law of motion : The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Newton’s III law of motion : To every action, there is an equal and opposite reaction and they act on two different bodies.
Q2: Explain inertia and momentum. Ans :
Inertia : The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia. For example : A book lying on a table will remain there until an external force is applied on it to remove or displace it from that position.
Momentum : Momentum of body is the quantity of motion possessed by the body. It is equal to the product of the mass and velocity of the body and is denoted by p. p = mv Momentum is a vector quantity and its direction is same as the direction of velocity of the object. Its SI unit is kilogram metre per second (kg ms–1).
Q3: Define force. What are different types forces? Ans:
Force : It is a push or pull on an object that produces acceleration in the body on which it acts. The S.I. unit of force is Newton.
Types of forces : Balanced force : When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
Unbalanced force : When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced forces.
Frictional force : Force of friction is the force that always opposes the motion of object.
Q4: What is inertia? Explain different types of inertia. Ans:
Inertia : The natural tendency of an object to resist change in their state of rest or of motion is called inertia. The mass of an object is a measure of its inertia. Its S.I. unit is kg.
Types of inertia :
Inertia of rest : The object remain in rest unless acted upon by an external unbalanced force.
Inertia of motion : The object in the state of uniform motion will continue to remain in motion with same speed and direction unless external force is not applied on it.
Q5: Give example to show the effects of force. Ans: (i) Place a ball in the ground. Kick it with your foot. The ball starts moving. The ball moves because of the force applied to it. (ii) If ball is coming towards you, you can kick it in any direction. The direction of motion of the ball changes because of the force applied to it. (iii) Place a rubber on the ground. Press it with your foot. It is found that the ball is no longer round but takes the shape of an egg, i.e. it is oblong. The shape of the ball has changed because of the force applied on the ball.
Q6: (a) What is friction? (b) What are the advantages of friction? or Why is friction necessary? Ans: (a) The force which opposes the motion of one body over the surface of another is called friction or the force of friction. (b) Friction is very desirable and an important force in our daily life. Some advantages of friction are :
The nails and screws hold the wooden boards together due to friction.
It is due to the friction between the ground/road and the soles of our shoes that we are able to walk.
The friction between the road and the surface of the tyres permits safe driving.
W e are able to write on paper because of the friction between the pen/pencil and the paper. That is why, it is very difficult to write on a glazed/waxed paper.
The application of brakes to stop a cycle, scooter or car, etc., is possible due to friction between the brake-lining and the rim of the wheel.
Q7: (i) What are the causes of friction? (ii) How can friction between any two surfaces can be reduced? or Describe some method for reducing friction. Ans: The friction rises due to the following factors :
Due to the force of attraction between the molecules of the two surfaces in contact. This is called the force of adhesion.
Due to the interlocking of the surface irregularities.
Thus, friction is due to the roughness of the two surfaces in contact.
The friction between two surfaces can be reduced by following methods :
By polishing surfaces : Rough surfaces can be made smooth by polishing. Therefore, polishing reduces friction.
By applying oil or grease on the surfaces : Oil/ grease forms a thin layer between the two surfaces and reduces friction.
Long Question Answer: Force and Law of Motion
Q1: State all 3 Newton’s law of motion. Ans:
Newton’s I law of motion : An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.
Newton’s II law of motion : The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Newton’s III law of motion : To every action, there is an equal and opposite reaction and they act on two different bodies.
Q2: Explain inertia and momentum. Ans :
Inertia : The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia. For example : A book lying on a table will remain there until an external force is applied on it to remove or displace it from that position.
Momentum : Momentum of body is the quantity of motion possessed by the body. It is equal to the product of the mass and velocity of the body and is denoted by p. p = mv Momentum is a vector quantity and its direction is same as the direction of velocity of the object. Its SI unit is kilogram metre per second (kg ms–1).
Q3: Define force. What are different types forces? Ans:
Force : It is a push or pull on an object that produces acceleration in the body on which it acts. The S.I. unit of force is Newton.
Types of forces : Balanced force : When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
Unbalanced force : When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced forces.
Frictional force : Force of friction is the force that always opposes the motion of object.
Q4: What is inertia? Explain different types of inertia. Ans:
Inertia : The natural tendency of an object to resist change in their state of rest or of motion is called inertia. The mass of an object is a measure of its inertia. Its S.I. unit is kg.
Types of inertia :
Inertia of rest : The object remain in rest unless acted upon by an external unbalanced force.
Inertia of motion : The object in the state of uniform motion will continue to remain in motion with same speed and direction unless external force is not applied on it.
Q5: Give example to show the effects of force. Ans: (i) Place a ball in the ground. Kick it with your foot. The ball starts moving. The ball moves because of the force applied to it. (ii) If ball is coming towards you, you can kick it in any direction. The direction of motion of the ball changes because of the force applied to it. (iii) Place a rubber on the ground. Press it with your foot. It is found that the ball is no longer round but takes the shape of an egg, i.e. it is oblong. The shape of the ball has changed because of the force applied on the ball.
Q6: (a) What is friction? (b) What are the advantages of friction? or Why is friction necessary? Ans: (a) The force which opposes the motion of one body over the surface of another is called friction or the force of friction. (b) Friction is very desirable and an important force in our daily life. Some advantages of friction are :
The nails and screws hold the wooden boards together due to friction.
It is due to the friction between the ground/road and the soles of our shoes that we are able to walk.
The friction between the road and the surface of the tyres permits safe driving.
W e are able to write on paper because of the friction between the pen/pencil and the paper. That is why, it is very difficult to write on a glazed/waxed paper.
The application of brakes to stop a cycle, scooter or car, etc., is possible due to friction between the brake-lining and the rim of the wheel.
Q7: (i) What are the causes of friction? (ii) How can friction between any two surfaces can be reduced? or Describe some method for reducing friction. Ans: The friction rises due to the following factors :
Due to the force of attraction between the molecules of the two surfaces in contact. This is called the force of adhesion.
Due to the interlocking of the surface irregularities.
Thus, friction is due to the roughness of the two surfaces in contact.
The friction between two surfaces can be reduced by following methods :
By polishing surfaces : Rough surfaces can be made smooth by polishing. Therefore, polishing reduces friction.
By applying oil or grease on the surfaces : Oil/ grease forms a thin layer between the two surfaces and reduces friction.
Q8: When a force of 40 N is applied on a body it moves with an acceleration of 5 ms2. Calculate the mass of the body. Ans: Let m be the mass of the body. Given : F = 40 N, a = 5 ms2 From the relation F = m a, we have 40 = m × 5 m = 40/5 = 8 kg
Q9: An object undergoes an acceleration of 8 ms–2 starting from rest. Find the distance travelled in 1 second. Ans: Given, Acceleration, a = 8 ms–2 Initial velocity, u = 0 Time interval, t = 1 s Distance travelled, s = ? Using the equation of motion, s = ut + 1/2 at2, one gets s = 0 × 1 + 1/2 × 8 × 12 = 4 m The object travels a distance of 4 m.
Q10: Calculate the force required to impact to a car, a velocity of 30 ms–1 in 10 seconds. The mass of the car is 1,500 kg. Ans: Here u = 0 ms–1; v = 30 ms–1; t = 10 s; a = ? Using v = u + at, we have 30 = 0 + a (10) a = 3 ms–2 Now F = ma = 1,500 × 3 or F = 4,500 NLong Question Answer: Force and Law of Motion
Q1: State all 3 Newton’s law of motion. Ans:
Newton’s I law of motion : An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.
Newton’s II law of motion : The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Newton’s III law of motion : To every action, there is an equal and opposite reaction and they act on two different bodies.
Q2: Explain inertia and momentum. Ans :
Inertia : The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia. For example : A book lying on a table will remain there until an external force is applied on it to remove or displace it from that position.
Momentum : Momentum of body is the quantity of motion possessed by the body. It is equal to the product of the mass and velocity of the body and is denoted by p. p = mv Momentum is a vector quantity and its direction is same as the direction of velocity of the object. Its SI unit is kilogram metre per second (kg ms–1).
Q3: Define force. What are different types forces? Ans:
Force : It is a push or pull on an object that produces acceleration in the body on which it acts. The S.I. unit of force is Newton.
Types of forces : Balanced force : When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
Unbalanced force : When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced forces.
Frictional force : Force of friction is the force that always opposes the motion of object.
Q4: What is inertia? Explain different types of inertia. Ans:
Inertia : The natural tendency of an object to resist change in their state of rest or of motion is called inertia. The mass of an object is a measure of its inertia. Its S.I. unit is kg.
Types of inertia :
Inertia of rest : The object remain in rest unless acted upon by an external unbalanced force.
Inertia of motion : The object in the state of uniform motion will continue to remain in motion with same speed and direction unless external force is not applied on it.
Q5: Give example to show the effects of force. Ans: (i) Place a ball in the ground. Kick it with your foot. The ball starts moving. The ball moves because of the force applied to it. (ii) If ball is coming towards you, you can kick it in any direction. The direction of motion of the ball changes because of the force applied to it. (iii) Place a rubber on the ground. Press it with your foot. It is found that the ball is no longer round but takes the shape of an egg, i.e. it is oblong. The shape of the ball has changed because of the force applied on the ball.
Q6: (a) What is friction? (b) What are the advantages of friction? or Why is friction necessary? Ans: (a) The force which opposes the motion of one body over the surface of another is called friction or the force of friction. (b) Friction is very desirable and an important force in our daily life. Some advantages of friction are :
The nails and screws hold the wooden boards together due to friction.
It is due to the friction between the ground/road and the soles of our shoes that we are able to walk.
The friction between the road and the surface of the tyres permits safe driving.
W e are able to write on paper because of the friction between the pen/pencil and the paper. That is why, it is very difficult to write on a glazed/waxed paper.
The application of brakes to stop a cycle, scooter or car, etc., is possible due to friction between the brake-lining and the rim of the wheel.
Q7: (i) What are the causes of friction? (ii) How can friction between any two surfaces can be reduced? or Describe some method for reducing friction. Ans: The friction rises due to the following factors :
Due to the force of attraction between the molecules of the two surfaces in contact. This is called the force of adhesion.
Due to the interlocking of the surface irregularities.
Thus, friction is due to the roughness of the two surfaces in contact.
The friction between two surfaces can be reduced by following methods :
By polishing surfaces : Rough surfaces can be made smooth by polishing. Therefore, polishing reduces friction.
By applying oil or grease on the surfaces : Oil/ grease forms a thin layer between the two surfaces and reduces friction.
Q8: When a force of 40 N is applied on a body it moves with an acceleration of 5 ms2. Calculate the mass of the body. Ans: Let m be the mass of the body. Given : F = 40 N, a = 5 ms2 From the relation F = m a, we have 40 = m × 5 m = 40/5 = 8 kg
Q9: An object undergoes an acceleration of 8 ms–2 starting from rest. Find the distance travelled in 1 second. Ans: Given, Acceleration, a = 8 ms–2 Initial velocity, u = 0 Time interval, t = 1 s Distance travelled, s = ? Using the equation of motion, s = ut + 1/2 at2, one gets s = 0 × 1 + 1/2 × 8 × 12 = 4 m The object travels a distance of 4 m.
Q10: Calculate the force required to impact to a car, a velocity of 30 ms–1 in 10 seconds. The mass of the car is 1,500 kg. Ans: Here u = 0 ms–1; v = 30 ms–1; t = 10 s; a = ? Using v = u + at, we have 30 = 0 + a (10) a = 3 ms–2 Now F = ma = 1,500 × 3 or F = 4,500 NLong Question Answer: Force and Law of Motion
Q1: State all 3 Newton’s law of motion. Ans:
Newton’s I law of motion : An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.
Newton’s II law of motion : The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Newton’s III law of motion : To every action, there is an equal and opposite reaction and they act on two different bodies.
Q2: Explain inertia and momentum. Ans :
Inertia : The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia. For example : A book lying on a table will remain there until an external force is applied on it to remove or displace it from that position.
Momentum : Momentum of body is the quantity of motion possessed by the body. It is equal to the product of the mass and velocity of the body and is denoted by p. p = mv Momentum is a vector quantity and its direction is same as the direction of velocity of the object. Its SI unit is kilogram metre per second (kg ms–1).
Q3: Define force. What are different types forces? Ans:
Force : It is a push or pull on an object that produces acceleration in the body on which it acts. The S.I. unit of force is Newton.
Types of forces : Balanced force : When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
Unbalanced force : When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced forces.
Frictional force : Force of friction is the force that always opposes the motion of object.
Q4: What is inertia? Explain different types of inertia. Ans:
Inertia : The natural tendency of an object to resist change in their state of rest or of motion is called inertia. The mass of an object is a measure of its inertia. Its S.I. unit is kg.
Types of inertia :
Inertia of rest : The object remain in rest unless acted upon by an external unbalanced force.
Inertia of motion : The object in the state of uniform motion will continue to remain in motion with same speed and direction unless external force is not applied on it.
Q5: Give example to show the effects of force. Ans: (i) Place a ball in the ground. Kick it with your foot. The ball starts moving. The ball moves because of the force applied to it. (ii) If ball is coming towards you, you can kick it in any direction. The direction of motion of the ball changes because of the force applied to it. (iii) Place a rubber on the ground. Press it with your foot. It is found that the ball is no longer round but takes the shape of an egg, i.e. it is oblong. The shape of the ball has changed because of the force applied on the ball.
Q6: (a) What is friction? (b) What are the advantages of friction? or Why is friction necessary? Ans: (a) The force which opposes the motion of one body over the surface of another is called friction or the force of friction. (b) Friction is very desirable and an important force in our daily life. Some advantages of friction are :
The nails and screws hold the wooden boards together due to friction.
It is due to the friction between the ground/road and the soles of our shoes that we are able to walk.
The friction between the road and the surface of the tyres permits safe driving.
W e are able to write on paper because of the friction between the pen/pencil and the paper. That is why, it is very difficult to write on a glazed/waxed paper.
The application of brakes to stop a cycle, scooter or car, etc., is possible due to friction between the brake-lining and the rim of the wheel.
Q7: (i) What are the causes of friction? (ii) How can friction between any two surfaces can be reduced? or Describe some method for reducing friction. Ans: The friction rises due to the following factors :
Due to the force of attraction between the molecules of the two surfaces in contact. This is called the force of adhesion.
Due to the interlocking of the surface irregularities.
Thus, friction is due to the roughness of the two surfaces in contact.
The friction between two surfaces can be reduced by following methods :
By polishing surfaces : Rough surfaces can be made smooth by polishing. Therefore, polishing reduces friction.
By applying oil or grease on the surfaces : Oil/ grease forms a thin layer between the two surfaces and reduces friction.
Q8: When a force of 40 N is applied on a body it moves with an acceleration of 5 ms2. Calculate the mass of the body. Ans: Let m be the mass of the body. Given : F = 40 N, a = 5 ms2 From the relation F = m a, we have 40 = m × 5 m = 40/5 = 8 kg
Q9: An object undergoes an acceleration of 8 ms–2 starting from rest. Find the distance travelled in 1 second. Ans: Given, Acceleration, a = 8 ms–2 Initial velocity, u = 0 Time interval, t = 1 s Distance travelled, s = ? Using the equation of motion, s = ut + 1/2 at2, one gets s = 0 × 1 + 1/2 × 8 × 12 = 4 m The object travels a distance of 4 m.
Q10: Calculate the force required to impact to a car, a velocity of 30 ms–1 in 10 seconds. The mass of the car is 1,500 kg. Ans: Here u = 0 ms–1; v = 30 ms–1; t = 10 s; a = ? Using v = u + at, we have 30 = 0 + a (10) a = 3 ms–2 Now F = ma = 1,500 × 3 or F = 4,500 NLong Question Answer: Force and Law of Motion
Q1: State all 3 Newton’s law of motion. Ans:
Newton’s I law of motion : An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.
Newton’s II law of motion : The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Newton’s III law of motion : To every action, there is an equal and opposite reaction and they act on two different bodies.
Q2: Explain inertia and momentum. Ans :
Inertia : The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia. For example : A book lying on a table will remain there until an external force is applied on it to remove or displace it from that position.
Momentum : Momentum of body is the quantity of motion possessed by the body. It is equal to the product of the mass and velocity of the body and is denoted by p. p = mv Momentum is a vector quantity and its direction is same as the direction of velocity of the object. Its SI unit is kilogram metre per second (kg ms–1).
Q3: Define force. What are different types forces? Ans:
Force : It is a push or pull on an object that produces acceleration in the body on which it acts. The S.I. unit of force is Newton.
Types of forces : Balanced force : When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
Unbalanced force : When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced forces.
Frictional force : Force of friction is the force that always opposes the motion of object.
Q4: What is inertia? Explain different types of inertia. Ans:
Inertia : The natural tendency of an object to resist change in their state of rest or of motion is called inertia. The mass of an object is a measure of its inertia. Its S.I. unit is kg.
Types of inertia :
Inertia of rest : The object remain in rest unless acted upon by an external unbalanced force.
Inertia of motion : The object in the state of uniform motion will continue to remain in motion with same speed and direction unless external force is not applied on it.
Q5: Give example to show the effects of force. Ans: (i) Place a ball in the ground. Kick it with your foot. The ball starts moving. The ball moves because of the force applied to it. (ii) If ball is coming towards you, you can kick it in any direction. The direction of motion of the ball changes because of the force applied to it. (iii) Place a rubber on the ground. Press it with your foot. It is found that the ball is no longer round but takes the shape of an egg, i.e. it is oblong. The shape of the ball has changed because of the force applied on the ball.
Q6: (a) What is friction? (b) What are the advantages of friction? or Why is friction necessary? Ans: (a) The force which opposes the motion of one body over the surface of another is called friction or the force of friction. (b) Friction is very desirable and an important force in our daily life. Some advantages of friction are :
The nails and screws hold the wooden boards together due to friction.
It is due to the friction between the ground/road and the soles of our shoes that we are able to walk.
The friction between the road and the surface of the tyres permits safe driving.
W e are able to write on paper because of the friction between the pen/pencil and the paper. That is why, it is very difficult to write on a glazed/waxed paper.
The application of brakes to stop a cycle, scooter or car, etc., is possible due to friction between the brake-lining and the rim of the wheel.
Q7: (i) What are the causes of friction? (ii) How can friction between any two surfaces can be reduced? or Describe some method for reducing friction. Ans: The friction rises due to the following factors :
Due to the force of attraction between the molecules of the two surfaces in contact. This is called the force of adhesion.
Due to the interlocking of the surface irregularities.
Thus, friction is due to the roughness of the two surfaces in contact.
The friction between two surfaces can be reduced by following methods :
By polishing surfaces : Rough surfaces can be made smooth by polishing. Therefore, polishing reduces friction.
By applying oil or grease on the surfaces : Oil/ grease forms a thin layer between the two surfaces and reduces friction.
Q8: When a force of 40 N is applied on a body it moves with an acceleration of 5 ms2. Calculate the mass of the body. Ans: Let m be the mass of the body. Given : F = 40 N, a = 5 ms2 From the relation F = m a, we have 40 = m × 5 m = 40/5 = 8 kg
Q9: An object undergoes an acceleration of 8 ms–2 starting from rest. Find the distance travelled in 1 second. Ans: Given, Acceleration, a = 8 ms–2 Initial velocity, u = 0 Time interval, t = 1 s Distance travelled, s = ? Using the equation of motion, s = ut + 1/2 at2, one gets s = 0 × 1 + 1/2 × 8 × 12 = 4 m The object travels a distance of 4 m.
Q10: Calculate the force required to impact to a car, a velocity of 30 ms–1 in 10 seconds. The mass of the car is 1,500 kg. Ans: Here u = 0 ms–1; v = 30 ms–1; t = 10 s; a = ? Using v = u + at, we have 30 = 0 + a (10) a = 3 ms–2 Now F = ma = 1,500 × 3 or F = 4,500 N Q8: When a force of 40 N is applied on a body it moves with an acceleration of 5 ms2. Calculate the mass of the body. Ans: Let m be the mass of the body. Given : F = 40 N, a = 5 ms2 From the relation F = m a, we have 40 = m × 5 m = 40/5 = 8 kg
Q9: An object undergoes an acceleration of 8 ms–2 starting from rest. Find the distance travelled in 1 second. Ans: Given, Acceleration, a = 8 ms–2 Initial velocity, u = 0 Time interval, t = 1 s Distance travelled, s = ? Using the equation of motion, s = ut + 1/2 at2, one gets s = 0 × 1 + 1/2 × 8 × 12 = 4 m The object travels a distance of 4 m.
Q10: Calculate the force required to impact to a car, a velocity of 30 ms–1 in 10 seconds. The mass of the car is 1,500 kg. Ans: Here u = 0 ms–1; v = 30 ms–1; t = 10 s; a = ? Using v = u + at, we have 30 = 0 + a (10) a = 3 ms–2 Now F = ma = 1,500 × 3 or F = 4,500 N
Q1: A person moves along the boundary of a square field of side 10 m in 20 s. What will be the magnitude of displacement of that person at the end of 2 minutes 20 seconds? Ans: Q2: Neha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Neha. Ans: Total distance = 180 m Total displacement = 0 Time taken t = 1 min = 60 s
Q3: A car decreases its speed from 80 km/h to 60 km/h is 5 seconds. Find the acceleration of the car. Ans:
Time (t) = 5 s Acceleration (a) =
Q4: A train starting from a railway station and moving with uniform acceleration, attains a speed 40 km/h in 10 minutes. Find its acceleration. Ans: u = 0 (starting from rest) v = 40 km/h= 11.11 m/s Time t = 10 minutes = 600 s Acceleration (a) =
Q5: The average speed of a bicycle, an athlete and car are 18 km/h, 7 m/s and 2 km/min respectively. Which of the three is the fastest and which is the slowest Ans:
Thus, the average speeds of the bicycle, the athlete and the car are 5 m/s, 7 m/s and 33.3 m/s respectively. So, the car is the fastest and the bicycle is the slowest.
Q6: Figure shows distance-time graph of two objects A and B, which object is moving with greater speed when both are moving?
Ans: The line for object B makes a longer angle with the time-axis. Its slope is longer than the slope of the line for object A. Thus, the speed of B is greater than that of A.
Q7: Figure represents the speed time graph for a particle. Find the distance covered by the particle between t = 10 min and t = 30 min.
Ans: We draw perpendicular lines from the 10-minute point and the 30-minute point to the time-axis (fig.). The distance covered is equal to the area of the rectangle ABCD, its value is ABCD = (30 min – 10 min) x (10 km/h) = 20 min × 10 km/h
Q8: Find the distance covered by a particle during the time interval t = 0 to t = 20 s for which the speed time graph is shown in figure. Ans: The distance covered in the time interval 0 to 20 s is equal to the area of the shaded triangle. So, Distance = 1/2 x Base x Height = 1/2 x (20 s) x (20 m/s) = 200 m.
Q9: A bus moves 30 km in 30 min and the next 30 km. in 40 min. Calculate the average speed for the entire journey. Ans: Given, the total time taken is 30 min + 40 min = 70 min and the total distance travelled is 30 km + 30 km = 60 km The average speed is = 51.4 km/h
Q10: It is estimated that the radio signal takes 1.27 seconds to reach the Earth from the surface of the Moon. Calculate the distance of the Moon from the Earth. Speed of radio signal = 3 × 108 ms–1 (speed of light in air). Ans: Here, time = 1.27 s Speed = 3 × 108 ms–1 Distance = ? Using distance = speed × time, we get Distance = 3 × 108 ms–1 × 1.27 s = 3.81 × 108 m = 3.81 × 105 km
Q11: Divya walked 2 km on a straight road and then walked back 1 km. Which of the two quantities involved in her walking is greater- the scalar or vector? Ans: Distance travelled by Divya = 2 km + 1 km = 3 km Displacement = 2 km – 1 km = 1 km Hence, distance which is a scalar quantity is greater than the displacement which is a vector quantity.
Q12: Two satellites A and B revolve around a planet C. The time taken by satellite B to go around the planet is twice the time taken by A. Which of the two satellites will have a greater magnitude of velocity? Ans: Satellite A will have greater magnitude of velocity since velocity is inversely proportional to time.
Q13: A child moving on a circular track of radius 40 m completes one revolution in 5 minutes. What is his (i) average speed (ii) average velocity in one full revolution? Ans: Distance travelled in one revolution, 2p r =2p × 40m Displacement in one revolution = 0 Time taken = 5 minutes = 5 × 60s (i) Average speed = Distance / Time = (ii) Average velocity = Displacement / Time = Zero
Q1: Explain the structure of a nervous tissue with details about its location and function. Ans:
Structure: Nervous tissue is composed of specialized cells called neurons, which consist of a cell body containing the nucleus and cytoplasm. Neurons have long, slender extensions known as axons, which transmit electrical signals, and shorter, branching extensions called dendrites, which receive signals from other neurons. These neurons are interconnected in complex networks.
Location: Nervous tissue is primarily found in the brain, spinal cord, and nerves throughout the body.
Function: Nervous tissue plays a crucial role in transmitting and processing information within the body. It facilitates communication between different parts of the body, coordinates responses to stimuli, and regulates various bodily functions, including movement, sensation, and cognition.
Q2: Explain the structure of parenchyma. What are its major modifications? Ans: Parenchyma: It is the basic or fundamental tissue found in plants. Cells of this tissue are thin walled, circular or polygonal. They are living with a nucleus and a vacuole. Intercellular spaces are present between the cells of this tissue. Two modifications of parenchyma are chlorenchyma and aerenchyma.
Chlorenchyma : Sometimes cells of the parenchyma contain chlorophyll and perform photosynthesis. This kind of parenchyma is known as chlorenchyma.
Aerenchyma : In aquatic plants, parenchyma contains big air spaces in between them. Such a parenchyma tissue is known as aerenchyma.
Q3: Explain plant tissue in detail. Ans: Plant tissue is mainly divided into two categories :
Meristematic tissue
Permanent tissue
Meristematic tissue: The cells divide very fast. It helps in the growth of the plants. The shape of the cell is oval, round and polygonal. There is no intercellular space. There are three types of meristematic tissues : (i) Apical meristem : Growth in length (ii) Lateral meristem : Growth in breadth or thickness (iii) Intercalary meristem : Growth in inter-nodes Permanent tissue: When meristematic tissue stops dividing and gets mature, then it forms permanent tissue. There are two types of permanent tissues : (i) Simple t issue (ii) Complex tissue (i) Simple Tissue : Simple tissues are same in structure and perform the same functions. There are three types of simple tissues : (a) Parenchyma : It is present in soft parts. (b) Collenchyma : It provides mechanical strength to plants and is found in stalks. (c) Sclerenchyma : It provides support as well as flexibility to plants. (ii) Complex Tissue : They are different in structure but perform the same function in group. There are two types of complex tissues : (a) X ylem : It transports water from roots to shoot in plants. (b) P hloem : Transports foods to all parts of the plant.
Q4: Explain connective tissue along with its types. Ans: Connective tissue consists various types of cells which perform the same function. These are of three types : (i) Proper connective tissue (ii) Fluid tissue (iii) Skeletal tissue
Proper connective tissue : These are of four types– Areolar and ligament connective tissue : It is present between muscles and skin and in the bone marrow. It is also present around nerves and blood vessels. They fill the space inside the organ. They also provide strength to internal organs and helps in repair of tissues.
Adipose tissue: It is found below the skin and also between internal organs. It stores fat and due to this fat storage, it behaves as an insulator.
Tendon: It is fibrous, strong and flexible and joins muscles with bone.
Ligament: It is elastic and strong and joins bone with bone.
Fluid tissue consists of: Blood: It is a liquid tissue called plasma which has RBCs, WBCs, plasma and blood platelets. It helps to transport substances like gases, hormones, digested food and waste material.
Lymph: It transports digested fat and white blood cells in plasma.
Skeletal tissue is made up of : Bone : It is a hard tissue which helps in the movement and support of our body.
Cartilage: It softens the bone surface at joints. It is found in our ear, nose, trachea and larynx.
Q5: Explain epidermis in plants. Ans: It forms the outermost layer of the plant. It is comprised of a single cell layer. This tissue forms a protective layer for plants and that helps to protect the internal parts of plants. It helps in protection against loss of water, attack by parasitic fungi and mechanical injury. Epidermis has small pores known as stomata. They are small holes or pores on the surface of leaves which help in exchange of gases and also in transpiration. Epidermis has long parts like hair that provide greater surface area for water absorption in roots. In plants found in deserts, epidermis consists of a thick waxy coating called cutin which makes the outer layer water resistant.
Q6: Give the difference between the types of muscle fibres diagrammatically Ans:
Q7: Explain complex tissue in plants. Ans: Generally, complex tissues consist of more than one type of cell. They are different in structure but together perform the same function. There are two types of complex tissues :
X ylem : It transports water from roots to shoot in plants. The movement is only in one direction that is, from roots to shoot. It provides mechanical support to the plant. It has mostly dead elements. Its elements are tracheids, xylem sclerenchyma, xylem parenchyma, vessel elements.
P hloem : Conduct foods to all parts of the plant. The movement of phloem is bidirectional that is, movement in both the directions is possible. Its elements are mostly living. Its elements are sieve tubes, phloem parenchyma, companion cells and intermediary cells.
Above given both complex tissues are conductive tissues and form the vascular bundle.
Q8: Explain the structure of three types of muscle fibres. Also write the locations where they are found in the body. Ans: The followings are the three types of muscle cells :
Unstriated muscles (also known as smooth, involuntary muscles) : This type of muscular tissue consists of spindle-shaped, long uninucleated cells. This type of muscles are present in alimentary canal, blood vessels, iris of eye, in ureters and bronchi of lungs, etc.
Striated muscles (also known as voluntary muscles because oftheir function being in our control or will) : This type ofmuscular cells are long multinucleated and enclosed in a membrane called sarcolemma. Each fibre has several longitudinal filaments embedded in cytoplasm. These filaments give these muscles striated appearance. These muscles are attached to the skeleton; so they are called skeletal muscles.
Cardiac muscles : These muscles are found in heart. They are not under the control of the will. They contract rhythmically and involuntarily throughout life without the sign of fatigue. Structurally they show the characters of both unstriated and striated muscles. They are made up of branched fibres. These fibres are uninucleated and show alternate light and dark bands (striation).
Q9: Draw a labelled diagram of section of a phloem. Ans:
Q10: What is the difference between meristematic cells and permanent cells? Ans: Difference between meristematic cells and permanent cells : Meristematic cells:
They have dense cytoplasm and a large centrally placed nucleus.
These cells are capable of dividing to _produce new cells.
Permanent cells:
They have a large central vacuole and normal nucleus.
They attain permanent shape and are not capable of producing new cells.
Q11: How many types of meristems are present in plants, on the basis of position? Ans: On the basis of location of meristem, it is classified into three types :
A pical meristem is present at the tip of stem, roots and their branches.
In tercalary meristem is found at the leaf base, above the nodes (i.e. at the base of internodes as in grasses) or below the nodes (i.e. at the uppermost region of internode as in mint).
La teral meristem
V ascular cambium and cork cambium are the examples of lateral meristem.
V ascular cambium is found in vascular bundles while cork cambium is found underneath the bark of trees. Both of these cause increase in girth of plants.
Q12: Differentiate between parenchyma and collenchyma. Ans: Difference between parenchyma and collenchyma : Q13: Differentiate between collenchyma and sclerenchyma. Ans: Difference between collenchyma and sclerenchyma :
Q14: Describe the structure of phloem. Ans: Phloem is a complex tissue responsible for transporting organic nutrients, primarily sugars, throughout the plant. It consists of several cell types, including:
Sieve tube elements: Elongated cells with perforated end walls called sieve plates, through which nutrients flow. These cells lack nuclei and are supported by companion cells.
Companion cells: Adjacent to sieve tube elements, companion cells are metabolically active and provide support and assistance in nutrient transport.
Phloem fibers: Long, slender cells that provide structural support to the phloem tissue.
Phloem parenchyma: Thin-walled cells that store reserves of organic compounds and provide structural support.
The interconnectedness of these cell types facilitates the efficient transport of nutrients, such as sugars produced during photosynthesis, from sources (such as leaves) to sinks (such as roots, fruits, and developing tissues) throughout the plant.
Q15: Differentiate between chlorenchyma and arenchyma. Ans: Difference between chlorenchyma and arenchyma: Q16: What is xylem? Explain its structure. Which one of its component is very important and why? Ans: Xylem is a complex plant tissue which transports water and dissolved minerals from roots to all other plant parts.
Structure: Xylem consists of four kinds of cells (also known as elements).
Tracheids: A tracheid is an elongated, hollow cell with its both ends tapering. The walls of these cells are thick by the deposition of lignin. At certain spots lignin is not present. These spots are termed as pits. The tracheids are dead cells.
Vessels: These are tube-like structures formed by a number of cells placed end to end with their transverse walls dissolved. The side walls of these tubes also have deposition of lignin. The thickening of the walls show various kinds of patterns. They are also dead cells.
Xylem Parenchyma: They are prenchymatous, thin walled, living cells. They help in lateral conduction of water and sap. They also store food.
Xylem Fibres: They are lignified dead fibres which provide mechanical support to plant.
The most important element of xylem is vessel because most of the water and minerals are carried upward through this component of xylem.
Q17: Differentiate between : (i) Xylem and phloem (ii) Vessel and sieve tube (iii) Tracheid and vessel Ans: (i) Xylem and phloem
(ii) Vessel and sieve tube
(iii) Tracheids and vessel
Q18: What are three main categories of connective tissue? Ans: Categories of connective tissue are : Connective tissue proper: There is a matrix in which generally two types of (white and yellow) fibres are present. In between these fibres some connective tissue cells are present. Example of this kind of connective tissues are aerolar tissue and adipose tissue.
Skeletal tissue: This type of tissues form the skeleton of an organism. It is of two types : Cartilage and bone.
C artilage has solid matrix called chondrin, in which fibres and cells known as chondrocytes are present. Usually cells are present in clusters of 2-3 cells in small spaces called lacunae. Cartilage is found in the regions of pinna, nose, trachea and larynx.
In bones, matrix is formed of a protein called ossein impregnated with phosphate and carbonates of calcium and magnesium.
Fluid tissue: Blood and lymph are examples of fluid connective tissues. These are specialized connective tissues. It consists of liquid matrix with no fibres. In liquid matrix called plasma corpuscles remain suspended. Blood transports food material, gases and other substances to the various parts of the body.
Q19: Explain the structure of a fluid connective tissue. Ans: Blood is a fluid connective tissue. Blood consists of: (i) Blood plasma, (ii) Blood cells.
Blood plasma: It is the fluid matrix which contains 85 to 95% water, 7% different types of proteins, 0.9% of salts, about 0.1% glucose and a very small amount of hormones, wastes, etc. In the plasma, blood corpuscles (cells) are suspended.
Blood cells: Three kinds of blood cells are found suspended in the blood plasma. These are : (i) Red blood corpuscles (Erythrocytes) or RBCs (ii) White blood corpuscles (leucocytes) or WBCs and (iii) Blood platelets. Red blood corpuscles (Erythrocytes) or RBCs : The red blood corpuscles are biconcave, disc-like cells which are devoid of nucleus. They contain a substance called haemoglobin because of this they appear red in colour. The most important function of the RBCs is the transport of oxygen and carbon dioxide. White blood corpuscles (Leucocytes) or WBCs : These cells are comparatively large in size, colourless and irregular in appearance. They are devoid of haemoglobin. They protect our body from diseases by destroying germs. Blood platelets : These are small, 2-4 m in diameter. They are without nucleus. Their main function is to liberate some substances which helps in blood clotting.
Q20:Differentiate between bone and cartilage. Ans: Q21: Give one function of each of the following : (a) Stomata, (b) Root nodules, (c) Cardiac muscle fibres. Ans: (a) Exchange of gases in plants. (b) Root nodules are found in leguminous plants. They harbour bacteria which can fix free atmospheric nitrogen into nitrates and nitrites which plants like pulses can use for protein synthesis. (c) Cardiac muscles show rhythmic contraction and relaxation throughout life. Because of this heart can pump the blood.
Q22: Describe the structure of cartilage and bone. Ans: Cartilage: Cartilage is a flexible and resilient connective tissue found in various parts of the body, including the nose, ears, trachea, and joints. It consists of chondrocytes (cartilage cells) embedded in a matrix of collagen and elastic fibers. Chondrocytes reside within small spaces called lacunae and maintain the integrity of the cartilage matrix. Cartilage provides support and cushioning to joints and other structures, allowing for smooth movement and shock absorption.
Bone: Bone is a rigid connective tissue that forms the skeletal framework of the body. It consists of osteocytes (bone cells) embedded in a matrix of collagen fibers and mineral salts, primarily calcium phosphate. Osteocytes are housed within small cavities called lacunae and are interconnected by tiny channels called canaliculi. Bone tissue is highly vascularized, allowing for the exchange of nutrients and waste products. It provides structural support, protection of internal organs, and serves as a reservoir for minerals such as calcium and phosphorus.
Q23: What are the two main components of blood? Why is blood considered a type of connective tissue? Ans:
Blood has two main components : (a) Fluid (liquid) matrix called plasma. (b) Suspended red blood cells (RBCs), white blood cells (WBCs) and platelets.
Blood is considered as connective tissue because : (a) It has the same origin as the other connective tissues. (b) It flows to different parts of the body and thus connects different parts of the body with one another to exchange materials and gases.
Q1: Give difference between hypotonic solution, isotonic solution and hypertonic solution. Ans:
Q2: (a) Name the organelle which provides turgidity and rigidity to the plant cell. Name any two substances which are present in it. (b) How are they useful in unicellular organisms? Ans: (a) Plant cells have big vacuoles that provide them turgidity and rigidity. Plant vacuoles store amino acids, sugars, various organic acids and some proteins. (b) I n unicellular organism they can serve the following works :
Forming food vacuoles : In single celled organisms like amoeba, the food vacuole contains the food items that the amoeba has engulfed. After that the food items are digested by the enzymes.
Removal of excess water and wastes : In some unicellular organisms, vacuoles play important roles in egesting excess water and some wastes from the cell.
Q3: Write a note on the structure of cell. Ans : (a) Cell is the basic unit of all living organisms. It is surrounded by an outer selectively permeable Plasma Membrane. Plant cells have an additional covering called “cell wall” outer to the Plasma Membrane. (b) Inside the plasma membrane there is a translucent viscous substance the cytoplasm in which the organelles are embedded. The control centre of the cell is the nucleus; it contains all the information necessary for the cell to function and to reproduce. Surrounding the nucleus is the endoplasmic reticulum (ER) on which ribosomes may be embedded. Ribosomes are granular structures which are the site of protein synthesis. (c) The powerhouse of cell is the mitochondria. It helps in releasing energy by the oxidation of food in cell. There are flat membranous secretory structures in the cell called the Golgi bodies. In plant cells, an additional structure located near the nucleus called the chloroplast, is also present. They are the site of photosynthesis. (d) Cells also contain lysosomes which are also called suicide bags. They digest and remove the unwanted debris of the cell. Centriole (present in animal cells) located near the nucleus helps in cell division. Cytoplasm also contains vacuoles filled with the cell sap. In plant cells, vacuole is large and centrally placed.
Q4: Give the difference between plant cell and animal cell. Ans:
No.
Plant Cell
Animal Cell
1.
Usually larger in size.
Comparatively smaller in size.
2.
Enclosed by a rigid cellulose cell wall in addition to plasma membrane.
Enclosed by a plasma membrane only.
3.
Has a fixed shape.
Shape is usually variable.
4.
Plastids present (chloroplasts in photosynthetic cells).
Plastids absent.
5.
A mature plant cell has a large central vacuole.
Vacuoles are small, temporary or absent.
6.
Nucleus usually lies at the periphery due to large vacuole.
Nucleus usually lies in the centre.
7.
Centrioles usually absent.
Centrioles present.
8.
Lysosomes rare.
Lysosomes common.
9.
Plasmodesmata present.
Plasmodesmata absent.
10.
Reserve food is starch.
Reserve food is glycogen.
11.
Can synthesize all amino acids, vitamins and coenzymes required.
Cannot synthesize all amino acids, vitamins and coenzymes.
12.
Does not burst in hypotonic solution due to rigid cell wall (becomes turgid).
May burst in hypotonic solution due to absence of cell wall.
Q5: Draw a neat labelled diagram of plant cell. Ans: Plant cell
Q6: (i) Name the organelle which provides turgidity and rigidity to the plant cell. Name any two substances which are present in it. (ii) How are they useful in unicellular organisms? Ans : (i) Plant cells have big vacuoles full of cell sap that provide them turgidity and rigidity. Plant vacuoles store amino acids, sugars, various organic acids and some proteins. (ii) I n unicellular organism they may serve the following purposes :
Forming food vacuoles: In single celled organisms like amoeba, the food vacuole contains the food items that the amoeba has ingested. The food items are digested by the enzymes later on.
Removal of excess water and wastes: In some unicellular organisms, specialized vacuoles play important roles in expelling excess water and some wastes from the cell.
Q7: Describe the role played by the lysosomes. Why are these termed as suicidal bags? How do they perform their functions? Ans : Functions of lysosomes :
Extracellular digestion. Sometimes lysosome enzymes are released outside the cell to break down extracellular material.
Digestion of foreign material. Lysosome also destroys any foreign material which enters inside the cell such as bacteria.
Cellular digestion. In damaged cells, ageing cells or dead cells lysosomes get ruptured and enzymes are released. These enzymes digest their own cell.
Lysosomes contain about 40 hydrolytic enzymes. When the cell gets damaged, lysosomes burst and their enzymes digest their own cell. So, lysosomes are called ‘suicide bags’.
Q8: Describe an activity to demonstrate endosmosis and exosmosis. Draw a diagram also. Ans: 1. Endosmosis : The movement of water in the cell or a body through a semipermeable membrane is called endosmosis. It can be demonstrated as follows : (i) Take some raisins with stalks and put them in plain water in a beaker. (ii) Observation : Raisins absorb water and swell. Raisins have high concentration of sugar than surrounding plain water. Because of this, water from the outside passing through semipermeable membrane enters into the cell. This is endosmosis. 2. Exosmosis : The movement of water out from a cell or a body through a semipermeable membrane is called exosmosis. This can be demonstrated as follows : (i) We place the swollen raisins (from above activity) into a beaker containing a concentrated solution of sugar or salt. (ii) Observation : When swollen raisins are placed in concentrated sugar or salt solution, they shrink because the solution surrounding the raisins is having low water concentration. Thus, raisins loose water by osmosis, this process is called exosmosis. Q9: Draw a neat labelled diagram of Animal cell. Ans: Animal Cell
Q10: Explain the structure and function of Golgi bodies. Ans: Golgi bodies consist of a system of membrane-bound vesicles arranged in stacks parallel to each other called cisterns. These membranes are connected with the membrane of endoplasmic reticulum (ER). Golgi apparatus
Functions of Golgi apparatus : (i) Golgi apparatus packages and dispatches the material synthesized in the cell. (ii) Golgi complex is also involved in the formation of lysosomes. (iii) Golgi apparatus is also involved in the synthesis of many substances such as polysaccharides, glycoprotein, etc. Q11: Give difference between plasma membrane and cell wall. Ans:
Feature
Plasma Membrane
Cell Wall
Composition
Made up of lipids and proteins
Made up of cellulose (in plants)
Nature
Living
Dead / non-living
Presence
Present in both plant and animal cells
Present in plant cells only (absent in animal cells)
Permeability
Semi-permeable
Freely permeable
Physical nature
Soft and flexible
Hard and rigid
Function
Regulates entry and exit of substances
Provides shape, support, and protection
Q12: Explain the structure of nucleus. Give a neat labelled diagram of a nucleus of cell. Give brief information about nucleus. Ans: Nucleus is the control centre of the cell. It is covered by a double layered envelope called nuclear membrane. The nuclear membrane has some pores which allow the transfer of material from inside the nucleus to cytoplasm. Inside the nuclear membrane some thread like structures are present. This is known as chromatin material. Structure of a nucleus
The chromatin material mainly formed through DNA (deoxyribonucleic acid) and proteins. When a cell starts to divide, chromatin material condenses into rod-shaped structures called chromosomes. The chromosomes contain DNA which are called genes. The nucleus is a large, centrally located spherical cellular component. It is bounded by two nuclear membranes, both forming a nuclear envelope. The nuclear envelope separates the nucleus from the cytoplasm. Within nucleoplasm two types of nuclear structures are embedded : the nucleolus and chromatin material. The nucleolus may be one or more in number and is not bounded by any membrane. It is rich in protein and RNA molecules and acts as the site for ribosome formation. Nucleus
Q13: Explain the following terms : (a) Plasma membrane, (b) Cytoplasm, (c) Nucleus. Ans: (a) Plasma membrane: It is a thin membrane which controls the passage of materials in and out of the cell. It is also called selectively permeable membrane. It makes the outer boundary of the cell and is made up of lipids and proteins (phospholipid bilayer). (b) Cytoplasm: It is transparent jelly-like thick substance present in the cell. It makes the ground of the cell in which all the cell organelles are suspended. (c) Nucleus: It is a double-layered membrane structure which contains chromosomes required for the inheritance of characteristics from one generation to the other.
Q14: Give difference between diffusion and osmosis. Write any two examples where a living organism uses osmosis to absorb water. Ans: Example of Osmosis : (i) Plant roots absorb water. (ii) Unicellular organisms such as amoeba absorb water from freshwater.
Q15: What would happen if when we put an animal cell into a solution of sugar or salt in water? Ans: The following three things could happen :
If the solution surrounding the cell is very dilute than cytoplasm, the water will move into the cell, i.e., the cell will gain water.
I f the solution has exactly similar water concentration as that of cytoplasm of cell, there will be no net movement of water across the cell membrane, i.e., no gain or loss of water from the cell.
If the medium (solution) has a lower concentration of water than the cell, i.e., the solution is concentrated, the cell will lose water by osmosis. How do all cells look alike in terms of shape and size?
Cells vary in shapes and sizes according to the fusion. Generally, cells are spherical but they may be long and branched as in nerve cell, Kidney shaped as guard cell in plant’s leaves, discoid as RBC, spindle shaped as muscle cell, etc. Size of cell varies from 0.2 mm to 18 cm in diameter. Some are microscopic while some are visible with naked eyes. For example : (i) Size of a typical cell in a multicellular organism ranges from 20-30 mn. (ii) The largest cell is ostrich egg (15 cm in diameter with shell and 8 cm in diameter without shell). (iii) The longest cell is nerve cell (up to 1 m. or more) and red blood cells are the smallest cell in our body. (iv) Smallest cells so far known are PPLOs, e.g. mycoplasma (0.1 µm in diameter). (v) Human egg is 0.1 mm in diameter.
Q16: How do lysosomes perform their function? Ans: Functions of lysosomes: (i) Extracellular digestion: Sometimes lysosome enzymes are released outside the cell to break down extracellular material. (ii) Destruction of foreign material: Lysosome also destroys any foreign material which enters inside the cell such as bacteria. (iii) Cellular digestion : Enzymes are released in damaged cells, ageing cells or dead cells. These enzymes digest their own cell.
Lysosomes contain about 40 hydrolytic enzymes. Lysosomes burst and their enzymes digest their own cell when the cell gets damaged. So, lysosomes are called ‘suicide bags’. Foreign materials entering the cell, such as bacteria or food, as well as dead old organelles in the lysosomes break up into small pieces.
Q17: What types of enzymes are present in the lysosomes? What is their function? Which organelle membranes manufacture these enzymes? Ans: Lysosomes contain powerful digestive enzymes capable of breaking down all organic material. Lysosomes help to keep the cell clean by digesting worn out cell organelles and foreign material such as bacteria or food. RER (Rough Endoplasmic Reticulum) makes the digestive enzymes present in the lysosomes.
Q18: Give brief information about the mitochondria. Describe the structure of mitochondria. Ans: The mitochondria are tiny bodies of varying shapes and size. Each mitochondria is bounded by a double membrane envelope. Outer membrane is porous. The inner membrane is thrown into folds. These folds are called cristae and are studded with small rounded bodies known as oxysomes. The interior cavity of the mitochondria is filled with a protein matrix which contains a few small-sized ribosomes, a circular DNA molecule and phosphate granules. Mitochondria are sites of cellular respiration. Mitochondria are membrane bound cell organelle found in the cytoplasm. Each mitochondria is a double membrane bounded structure. The outer membrane of mitochondrion is smooth. But, the inner membrane of the mitochondrion is folded inwardly, into the matrix of mitochondrion forming finger like projections. The inward finger like projections of inner membrane is called cristae. Cristae greatly increase the surface area of inner membrane. Mitochondria contain extra nuclear DNA. Mitochondria
Q1: Number of electrons, protons and neutrons in chemical species A, B, C and D is given below:
Now, answer the following questions : (a) What is the mass number of A and B? (b) What is the atomic number of B? (c) Which two elements represent a pair of isotopes and why? (d) What is the valency of element C? Also, justify your answers. Ans: (a) Mass number = Number of protons + Number of neutrons Mass number of A = 3 + 4 = 7 Mass number of B = 9 + 8 = 17 (b) Atomic Number = Number of protons; therefore, atomic number of B = 9. (c) Elements C and D represent a pair of isotopes. Explanation: Isotopes are atoms of the same element that have the same atomic number (same number of protons) but different mass numbers (different numbers of neutrons). Since C and D have the same number of protons but different numbers of neutrons, they are isotopes of each other. (d) To find the valency of element C, look at its electronic configuration. Given that C has 8 electrons, its configuration is 2, 6. It needs 2 more electrons to complete its octet and attain a stable noble gas configuration. Therefore, its valency = 2.
Q2: Describe in brief Rutherford’s alpha-particle scattering experiment with the help of a labelled diagram. Write any three important conclusions drawn from the experiment. Ans: Rutherford’s alpha-particle scattering experiment:
Ernest Rutherford directed a beam of fast-moving alpha particles (positively charged) at a very thin gold foil and observed their scattering using a fluorescent screen. The alpha particles were produced by a radioactive source and the scattered particles produced tiny flashes of light on the screen, which were counted to record their paths.
Observations:
Most alpha particles passed straight through the foil without deflection.
Some alpha particles were deflected by small angles.
A very few alpha particles (about 1 in 12,000) were deflected back, i.e., scattered through large angles.
Conclusions drawn:
Most of the atom is empty space, because the majority of alpha particles passed through the foil without any deflection.
There is a small, dense, positively charged nucleus at the centre of the atom; this concentrated positive charge repelled and deflected the positively charged alpha particles, causing the observed deflections.
The size of the nucleus is very small compared to the overall size of the atom, since only a tiny fraction of alpha particles experienced large deflections.Scattering of alpha particles by a gold foil
Q3: Give the number of electrons, protons and neutrons in 59CO27 and 108Ag47. Ans : For 59Co27: Atomic number = 27 → 27 protons
Number of electrons = 27 (since the atom is neutral)
Number of neutrons = Mass number − Atomic number = 59 − 27 = 32
For 108Ag47:
Atomic number = 47 → 47 protons
Number of electrons = 47 (neutral atom)
Number of neutrons = 108 − 47 = 61
Q4: Give the difference between isotopes and isobars. Ans:
Difference between isotopes and isobars:
Isotopes: Atoms of the same element that have the same atomic number (same number of protons) but different mass numbers (different numbers of neutrons). Example: 35Cl and 37Cl.
Isobars: Atoms of different elements that have the same mass number but different atomic numbers (different numbers of protons). Example: 40Ca (Z = 20) and 40Ar (Z = 18) are isobars because both have mass number 40.
Q5: Chlorine occurs in nature in two isotopic forms with masses 35u and 37u in the ratio of 3 : 1. What should be the mass of a chlorine atom? Ans: The average atomic mass of chlorine is calculated using the masses of its isotopes and their relative abundances.
Average atomic mass = (fractional abundance × mass of isotope) summed over isotopes.
Using the given ratio 3 : 1, the fractional abundances are 3/4 and 1/4, respectively.
Average mass = (3 × 35u + 1 × 37u) / 4 = 142 / 4 = 35.5 u. Hence, the average atomic mass of chlorine is 35.5 u.
Q6: An element 12X24 loses two electrons to form a cation, which combines with the anion of element 17Y35 formed by gaining an electron. (i) Write the electronic configuration of element X. (ii) Write the electronic configuration of the anion of element Y. (iii) Write the formula for the compound formed by the combination of X and Y. Ans: (i) Atomic number of X = 12 → electronic configuration = 2, 8, 2. On losing 2 electrons, it forms X²⁺ whose configuration becomes 2, 8 (like a noble gas).
(ii) Atomic number of Y = 17 → electronic configuration = 2, 8, 7. On gaining 1 electron, Y forms the anion Y⁻ with configuration 2, 8, 8 (stable octet).
(iii) To balance charges: one X²⁺ ion will combine with two Y⁻ ions. X²⁺ + 2 Y⁻ → XY₂. So, the formula of the compound is XY₂.
Q7: Give reasons : (i) The mass number of an atom excludes the mass of an electron. (ii) The nucleus of an atom is charged. (iii) Alpha-particle scattering experiment was possible by using a gold foil only and not by a foil of any other metal. Ans:
(i) The mass of an electron is negligible compared to that of a proton or neutron (about 1/1836 of a proton). Therefore, the mass number, which counts the major contributors to atomic mass, includes only protons and neutrons and excludes electrons.
(ii) The nucleus contains protons, which are positively charged, while neutrons are neutral. As a result, the overall charge of the nucleus is positive.
(iii)Gold is highly malleable and can be hammered into extremely thin foils (a few atoms thick) without breaking. Such a thin foil was essential to allow alpha particles to pass through and to observe their scattering. This property made gold especially suitable for Rutherford’s experiment.
Q8: Give the postulates of Dalton’s atomic theory. Ans: Dalton proposed an atomic theory in 1808 to explain the nature of matter. The main postulates of Dalton’s atomic theory are:
All matter is composed of tiny indivisible particles called atoms.
Atoms of a given element are identical in mass and properties.
Atoms of different elements have different masses and properties.
Atoms combine in simple whole-number ratios to form compounds.
Atoms are indivisible and indestructible in chemical reactions (they are not created or destroyed in ordinary chemical changes).
The relative number and kinds of atoms in a compound remain constant (fixed composition).
Note: Some parts of Dalton’s theory were later revised after the discovery of subatomic particles and isotopes, but his ideas laid the foundation of modern atomic theory.
Q9: Explain Bohr’s Model of the atom? Ans: Bohr’s model was proposed to address limitations of Rutherford’s model. Its key postulates are:
Quantised orbits: Electrons move in certain fixed circular orbits or energy levels around the nucleus without radiating energy. These orbits are labelled K, L, M, N or by principal quantum number n = 1, 2, 3, 4,…
No radiation in stationary orbits: While an electron remains in a permitted orbit, it does not emit energy.
Energy emission or absorption: An electron can move from one allowed orbit to another by absorbing energy (moving to a higher orbit) or emitting energy (moving to a lower orbit). The energy absorbed or emitted equals the difference between the two energy levels and is given by E = hν, where ν is the frequency of radiation.
Quantised energy levels explain spectral lines: Since only specific energy differences are possible, atoms emit or absorb radiation of specific frequencies, producing discrete spectral lines.
Bohr’s model thus introduced the idea of quantised energy levels, explaining atomic spectra and the stability of atoms in a simple, conceptual way appropriate at the class 9 level.
Q10: How are electrons disturbed in different shells or orbits?
Ans: Electron distribution in different shells:
The distribution of electrons in orbits (or energy levels) is explained by the following rules:
Maximum electrons in a shell: The maximum number of electrons that can be accommodated in the nth shell is given by 2n².
K-shell (n = 1): 2 × 1² = 2 electrons
L-shell (n = 2): 2 × 2² = 8 electrons
M-shell (n = 3): 2 × 3² = 18 electrons
N-shell (n = 4): 2 × 4² = 32 electrons
Maximum electrons in outermost shell: The outermost shell of an atom can have a maximum of 8 electrons (for the common main-group elements) to attain a stable configuration.
Filling order of shells: Electrons occupy the inner shells first; an outer shell is not filled unless all inner shells are complete. Thus electrons are filled in a step-wise manner starting from the innermost shell.
Example: For sodium (atomic number 11), the electron distribution is 2, 8, 1, showing inner shells filled first and one electron in the outermost shell.
Q1: Glucose has the molecular formula C6H12O6. Calculate : (a) Its molecular mass. (b) The number of atoms in one molecule of glucose. (c) The number of gram molecules in 18 g of glucose. Ans: The molecular mass of a compound is the sum of the atomic masses of all the atoms present in its molecular formula. (a) Molecular mass of C6H12O6 = (6 × 12u) + (12 × 1u) + (6 × 16u) = 72u + 12u + 96u = 180u (b) The number of atoms in one molecule of C6H12O6 = 6 atoms of C + 12 atoms of H + 6 atoms of O = 6 + 12 + 6 = 24 atoms (c) Number of gram molecules = Mass of glucose (g) / Molecular mass of glucose (g) = 18 g / 180 g/mol = 0.1 mol
Q2:What is the use of the mole concept in chemistry?
Ans: The mole concept is extremely useful in chemistry for quantitative analysis:
It allows chemists to count atoms, molecules, or ions using mass.
One mole of any substance contains 6.022 × 1023 particles (Avogadro’s number).
One mole of any gas occupies 22.4 litres at standard temperature and pressure (STP).
It helps in calculating reactants and products in chemical reactions through balanced equations.
It is used to determine molar mass and establish relationships between mass, volume, and number of particles.
Q3: Give symbol and valency of: Potassium, Barium, Aluminium, Calcium, Cobalt, Fluorine, Lead, Zinc, Iodine, Sulphide. Ans:
Q4: In a reaction, 6.83 g of lead chloride precipitated. The initial solution had 5 g each of lead nitrate and sodium chloride. What is the amount of sodium nitrate formed?
Ans: Initially, we have:
50 g of 10% lead nitrate solution = 5 g Pb(NO3)2 + 45 g water
50 g of 10% sodium chloride solution = 5 g NaCl + 45 g water Total mass before reaction = 5 g + 5 g + 90 g water = 100 g
After reaction, 6.83 g of lead chloride (PbCl2) precipitates out. Assuming no mass is lost, Total mass after = 100 g Remaining components (water and NaNO3) = 100 – 6.83 = 93.17 g
Water is 90 g, so: Mass of sodium nitrate formed = 93.17 – 90 = 3.17 g
Q5: Write a formula for the following : (a) Zinc sulphate, (b) Methane, (c) Ammonium carbonate. Ans:
(a) Zinc sulphate
Thus, Zn2(SO4)2 and finally = ZnSO4
(b) Methane
Thus, finally = CH4
(c) Ammonium carbonate
Thus, finally = (NH4)2CO3
Q6: Explain the law of multiple proportions. Ans: According to the law of multiple proportions, when two elements combine to make one or more compounds, then the ratio of weights of these elements remains in a fixed ratio to one another.
For example: Hydrogen and oxygen combine to form water (H2O) and hydrogen peroxide (H2O2) under different conditions. 2 grams of hydrogen combine with 16 grams of oxygen in the case of water, while 2 grams of hydrogen combine with 32 grams of oxygen to form hydrogen peroxide. Now, the weights of oxygen combine with a fixed weight of hydrogen in water and hydrogen peroxide, respectively ,are 16 and 32, which are in a simple ratio of 16: 32 or 1 : 2.
Q7: What are molecules? Give a brief explanation of the arrangement of the constituent atoms in the molecules. Ans: A molecule is the smallest particle of an element or compound which is stable under normal conditions. And it can freely show all the properties of that element or compound. It may be made up of one, two or more atoms. A molecule with one atom is called a monoatomic molecule. E.g. helium, neon, etc.
A molecule with two atoms is called a diatomic molecule. E.g. Cl2, O2. Similarly, there are molecules containing three atoms (CO2), four atoms (P4) and so on.
Q8: The mass of one molecule of a substance is 4.65 × 10–23 grams. What is its molecular mass? Ans: Mass of 1 molecule of a substance = 4.65 × 10–23 grams
Mass of 6.023 × 1023 molecules of a substance
= 4.65 × 10–23 × 6.023 × 1023
= 28 g/mol
Molecular mass of the substance = 28 g/mol
Q9: An element 12X24 loses two electrons to form a cation, which combines with the anion of element 17Y35 formed by gaining an electron. (a) Write the electronic configuration of element X. (b) Write the electronic configuration of the anion of element Y. (c) Write the formula for the compound formed by the combination of X and Y. Ans:
(a) X = 2, 8, 2
(b) Y– = 2, 8, 8
(c) XY2
Q10: Calculate the formula unit masses of ZnO, Na2O, K2CO3 given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and 0 = 16u. Ans: Formula unit mass of ZnO = 1 × 65u + 1 × 16u = 81u
Formula unit mass of Na2O = 2 × 23u + 1 × 16u = 62u
Formula unit mass of K2CO3 = 2 × 39u + 1 × 12u + 3 × 16u = 138u
Q11: What is the mass of : (a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms (atomic mass of aluminium = 27)? (c) 10 moles of sodium sulphite (Na2SO3)? Ans:
(a) 1 mole of nitrogen atoms
= 1 × gram atomic mass of nitrogen atom
= 1 × 14 g = 14 g
(b) 4 moles of aluminium atoms
= 4 × gram atomic mass of aluminium atoms
= 4 × 27 g = 108 g
(c) 10 moles of sodium sulphite (Na2SO3)
= 10 (2 × gram atomic mass of Na + 1 × gram atomic mass of sulphur + 3 × gram atomic mass of oxygen)
= 10 (2 × 23 g + 1 × 32 g + 3 × 16g)
= 10 (46 g + 32 g + 48 g)
= 10 × 126 g = 1260 g
Q12: Give the postulates of Dalton’s atomic theory. Ans: John Dalton proposed his atomic theory in 1808. It explains the nature of matter in terms of atoms. The main postulates of Dalton’s atomic theory are:
All matter is made up of indivisible particles called atoms.
Atoms of a given element are identical in mass and properties.
Atoms of different elements have different masses and chemical properties.
Atoms combine in simple whole-number ratios to form compounds.
Atoms can neither be created nor destroyed in a chemical reaction.
The relative number and types of atoms in a given compound remain constant.
These postulates laid the foundation for modern chemistry, although some parts were later modified with the discovery of subatomic particles.
Q13: (a) Give one point of difference between an atom and an ion. (b) Give one example each of a polyatomic cation and an anion. (c) Identify the correct chemical name of FeSO3: Ferrous sulphate, Ferrous sulphide, Ferrous sulphite. (d) Write the chemical formula for the chloride of magnesium.
Ans:
(a) An atom is a neutral particle containing equal numbers of protons and electrons. An ion is a charged particle formed when an atom loses or gains electrons. Positive ions are called cations, and negative ions are called anions.
(b) (i) Polyatomic cation: Ammonium ion, (NH4)+
(ii) Polyatomic anion: Sulphate ion, (SO4)2–
(c) FeSO3 contains Fe²⁺ and SO3²⁻ (sulphite ion), so the correct name is Ferrous sulphite.
(d) Magnesium has a valency of 2, and chlorine has a valency of 1. Therefore, the formula is MgCl2 (Magnesium chloride).
Q14: When 3.0 g of magnesium is burnt in 2.00 g of oxygen, 5.00 g of magnesium oxide is produced. What mass of magnesium oxide will be formed when 3.00 g of magnesium is burnt in 5.00 g of oxygen? Which law of chemical combination will govern your answer? State the law. Ans: According to the given reaction: 3.00 g magnesium + 2.00 g oxygen → 5.00 g magnesium oxide. This shows that 3 g of magnesium reacts completely with 2 g of oxygen.
Now, when 3.00 g magnesium is burnt in 5.00 g oxygen, only 2 g of oxygen will be used (as per the fixed ratio), and 3 g of oxygen will remain unused.
So, mass of magnesium oxide formed = 3 + 2 = 5.00 g
Law involved: This illustrates the Law of Definite Proportions, which states that a chemical compound always contains the same elements in the same fixed ratio by mass.
Long Question Answer: Atoms and Molecules
Q1: Glucose has the molecular formula C6H12O6. Calculate : (a) Its molecular mass. (b) The number of atoms in one molecule of glucose. (c) The number of gram molecules in 18 g of glucose. Ans: The molecular mass of a compound is the sum of the atomic masses of all the atoms present in its molecular formula. (a) Molecular mass of C6H12O6 = (6 × 12u) + (12 × 1u) + (6 × 16u) = 72u + 12u + 96u = 180u (b) The number of atoms in one molecule of C6H12O6 = 6 atoms of C + 12 atoms of H + 6 atoms of O = 6 + 12 + 6 = 24 atoms (c) Number of gram molecules = Mass of glucose (g) / Molecular mass of glucose (g) = 18 g / 180 g/mol = 0.1 mol
Q2:What is the use of the mole concept in chemistry?
Ans: The mole concept is extremely useful in chemistry for quantitative analysis:
It allows chemists to count atoms, molecules, or ions using mass.
One mole of any substance contains 6.022 × 1023 particles (Avogadro’s number).
One mole of any gas occupies 22.4 litres at standard temperature and pressure (STP).
It helps in calculating reactants and products in chemical reactions through balanced equations.
It is used to determine molar mass and establish relationships between mass, volume, and number of particles.
Q3: Give symbol and valency of: Potassium, Barium, Aluminium, Calcium, Cobalt, Fluorine, Lead, Zinc, Iodine, Sulphide. Ans:
Q4: In a reaction, 6.83 g of lead chloride precipitated. The initial solution had 5 g each of lead nitrate and sodium chloride. What is the amount of sodium nitrate formed?
Ans: Initially, we have:
50 g of 10% lead nitrate solution = 5 g Pb(NO3)2 + 45 g water
50 g of 10% sodium chloride solution = 5 g NaCl + 45 g water Total mass before reaction = 5 g + 5 g + 90 g water = 100 g
After reaction, 6.83 g of lead chloride (PbCl2) precipitates out. Assuming no mass is lost, Total mass after = 100 g Remaining components (water and NaNO3) = 100 – 6.83 = 93.17 g
Water is 90 g, so: Mass of sodium nitrate formed = 93.17 – 90 = 3.17 g
Q5: Write a formula for the following : (a) Zinc sulphate, (b) Methane, (c) Ammonium carbonate. Ans:
(a) Zinc sulphate
Thus, Zn2(SO4)2 and finally = ZnSO4
(b) Methane
Thus, finally = CH4
(c) Ammonium carbonate
Thus, finally = (NH4)2CO3
Q6: Explain the law of multiple proportions. Ans: According to the law of multiple proportions, when two elements combine to make one or more compounds, then the ratio of weights of these elements remains in a fixed ratio to one another.
For example: Hydrogen and oxygen combine to form water (H2O) and hydrogen peroxide (H2O2) under different conditions. 2 grams of hydrogen combine with 16 grams of oxygen in the case of water, while 2 grams of hydrogen combine with 32 grams of oxygen to form hydrogen peroxide. Now, the weights of oxygen combine with a fixed weight of hydrogen in water and hydrogen peroxide, respectively ,are 16 and 32, which are in a simple ratio of 16: 32 or 1 : 2.
Q7: What are molecules? Give a brief explanation of the arrangement of the constituent atoms in the molecules. Ans: A molecule is the smallest particle of an element or compound which is stable under normal conditions. And it can freely show all the properties of that element or compound. It may be made up of one, two or more atoms. A molecule with one atom is called a monoatomic molecule. E.g. helium, neon, etc.
A molecule with two atoms is called a diatomic molecule. E.g. Cl2, O2. Similarly, there are molecules containing three atoms (CO2), four atoms (P4) and so on.
Q8: The mass of one molecule of a substance is 4.65 × 10–23 grams. What is its molecular mass? Ans: Mass of 1 molecule of a substance = 4.65 × 10–23 grams
Mass of 6.023 × 1023 molecules of a substance
= 4.65 × 10–23 × 6.023 × 1023
= 28 g/mol
Molecular mass of the substance = 28 g/mol
Q9: An element 12X24 loses two electrons to form a cation, which combines with the anion of element 17Y35 formed by gaining an electron. (a) Write the electronic configuration of element X. (b) Write the electronic configuration of the anion of element Y. (c) Write the formula for the compound formed by the combination of X and Y. Ans:
(a) X = 2, 8, 2
(b) Y– = 2, 8, 8
(c) XY2
Q10: Calculate the formula unit masses of ZnO, Na2O, K2CO3 given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and 0 = 16u. Ans: Formula unit mass of ZnO = 1 × 65u + 1 × 16u = 81u
Formula unit mass of Na2O = 2 × 23u + 1 × 16u = 62u
Formula unit mass of K2CO3 = 2 × 39u + 1 × 12u + 3 × 16u = 138u
Q11: What is the mass of : (a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms (atomic mass of aluminium = 27)? (c) 10 moles of sodium sulphite (Na2SO3)? Ans:
(a) 1 mole of nitrogen atoms
= 1 × gram atomic mass of nitrogen atom
= 1 × 14 g = 14 g
(b) 4 moles of aluminium atoms
= 4 × gram atomic mass of aluminium atoms
= 4 × 27 g = 108 g
(c) 10 moles of sodium sulphite (Na2SO3)
= 10 (2 × gram atomic mass of Na + 1 × gram atomic mass of sulphur + 3 × gram atomic mass of oxygen)
= 10 (2 × 23 g + 1 × 32 g + 3 × 16g)
= 10 (46 g + 32 g + 48 g)
= 10 × 126 g = 1260 g
Q12: Give the postulates of Dalton’s atomic theory. Ans: John Dalton proposed his atomic theory in 1808. It explains the nature of matter in terms of atoms. The main postulates of Dalton’s atomic theory are:
All matter is made up of indivisible particles called atoms.
Atoms of a given element are identical in mass and properties.
Atoms of different elements have different masses and chemical properties.
Atoms combine in simple whole-number ratios to form compounds.
Atoms can neither be created nor destroyed in a chemical reaction.
The relative number and types of atoms in a given compound remain constant.
These postulates laid the foundation for modern chemistry, although some parts were later modified with the discovery of subatomic particles.
Q13: (a) Give one point of difference between an atom and an ion. (b) Give one example each of a polyatomic cation and an anion. (c) Identify the correct chemical name of FeSO3: Ferrous sulphate, Ferrous sulphide, Ferrous sulphite. (d) Write the chemical formula for the chloride of magnesium.
Ans:
(a) An atom is a neutral particle containing equal numbers of protons and electrons. An ion is a charged particle formed when an atom loses or gains electrons. Positive ions are called cations, and negative ions are called anions.
(b) (i) Polyatomic cation: Ammonium ion, (NH4)+
(ii) Polyatomic anion: Sulphate ion, (SO4)2–
(c) FeSO3 contains Fe²⁺ and SO3²⁻ (sulphite ion), so the correct name is Ferrous sulphite.
(d) Magnesium has a valency of 2, and chlorine has a valency of 1. Therefore, the formula is MgCl2 (Magnesium chloride).
Q14: When 3.0 g of magnesium is burnt in 2.00 g of oxygen, 5.00 g of magnesium oxide is produced. What mass of magnesium oxide will be formed when 3.00 g of magnesium is burnt in 5.00 g of oxygen? Which law of chemical combination will govern your answer? State the law. Ans: According to the given reaction: 3.00 g magnesium + 2.00 g oxygen → 5.00 g magnesium oxide. This shows that 3 g of magnesium reacts completely with 2 g of oxygen.
Now, when 3.00 g magnesium is burnt in 5.00 g oxygen, only 2 g of oxygen will be used (as per the fixed ratio), and 3 g of oxygen will remain unused.
So, mass of magnesium oxide formed = 3 + 2 = 5.00 g
Law involved: This illustrates the Law of Definite Proportions, which states that a chemical compound always contains the same elements in the same fixed ratio by mass.
Q15: (a) Calculate the number of molecules of SO2 present in 44 g of it. (b) If one mole of oxygen atoms weighs 16 grams, find the mass of one atom of oxygen in grams. Ans: (a) Molecular mass of SO2 = Atomic mass of S + 2 × Atomic mass of O = 32 + 2 × 16 = 64u Molar mass = 64 g Number of molecules, N = = 44/64 x 6.022 x 1023 = 4.14 × 1023 molecules (b) One mole of oxygen contains 6.022 × 1023 atoms of oxygen Mass of one atom of oxygen = = 2.66 × 10–23 g
Q16: Sodium is represented as 23Na11. (a) What is its atomic mass? (b) Write its gram atomic mass. (c) How many atoms of Na will be there in 11.5 g of the sample? Ans: (a) Atomic mass = 23u (b) Gram atomic mass = 23 g (c) Given mass = 11.5 g Molar mass = 23 g Number of atoms (N) = = 3.011 × 1023 atomsLong Question Answer: Atoms and Molecules
Q1: Glucose has the molecular formula C6H12O6. Calculate : (a) Its molecular mass. (b) The number of atoms in one molecule of glucose. (c) The number of gram molecules in 18 g of glucose. Ans: The molecular mass of a compound is the sum of the atomic masses of all the atoms present in its molecular formula. (a) Molecular mass of C6H12O6 = (6 × 12u) + (12 × 1u) + (6 × 16u) = 72u + 12u + 96u = 180u (b) The number of atoms in one molecule of C6H12O6 = 6 atoms of C + 12 atoms of H + 6 atoms of O = 6 + 12 + 6 = 24 atoms (c) Number of gram molecules = Mass of glucose (g) / Molecular mass of glucose (g) = 18 g / 180 g/mol = 0.1 mol
Q2:What is the use of the mole concept in chemistry?
Ans: The mole concept is extremely useful in chemistry for quantitative analysis:
It allows chemists to count atoms, molecules, or ions using mass.
One mole of any substance contains 6.022 × 1023 particles (Avogadro’s number).
One mole of any gas occupies 22.4 litres at standard temperature and pressure (STP).
It helps in calculating reactants and products in chemical reactions through balanced equations.
It is used to determine molar mass and establish relationships between mass, volume, and number of particles.
Q3: Give symbol and valency of: Potassium, Barium, Aluminium, Calcium, Cobalt, Fluorine, Lead, Zinc, Iodine, Sulphide. Ans:
Q4: In a reaction, 6.83 g of lead chloride precipitated. The initial solution had 5 g each of lead nitrate and sodium chloride. What is the amount of sodium nitrate formed?
Ans: Initially, we have:
50 g of 10% lead nitrate solution = 5 g Pb(NO3)2 + 45 g water
50 g of 10% sodium chloride solution = 5 g NaCl + 45 g water Total mass before reaction = 5 g + 5 g + 90 g water = 100 g
After reaction, 6.83 g of lead chloride (PbCl2) precipitates out. Assuming no mass is lost, Total mass after = 100 g Remaining components (water and NaNO3) = 100 – 6.83 = 93.17 g
Water is 90 g, so: Mass of sodium nitrate formed = 93.17 – 90 = 3.17 g
Q5: Write a formula for the following : (a) Zinc sulphate, (b) Methane, (c) Ammonium carbonate. Ans:
(a) Zinc sulphate
Thus, Zn2(SO4)2 and finally = ZnSO4
(b) Methane
Thus, finally = CH4
(c) Ammonium carbonate
Thus, finally = (NH4)2CO3
Q6: Explain the law of multiple proportions. Ans: According to the law of multiple proportions, when two elements combine to make one or more compounds, then the ratio of weights of these elements remains in a fixed ratio to one another.
For example: Hydrogen and oxygen combine to form water (H2O) and hydrogen peroxide (H2O2) under different conditions. 2 grams of hydrogen combine with 16 grams of oxygen in the case of water, while 2 grams of hydrogen combine with 32 grams of oxygen to form hydrogen peroxide. Now, the weights of oxygen combine with a fixed weight of hydrogen in water and hydrogen peroxide, respectively ,are 16 and 32, which are in a simple ratio of 16: 32 or 1 : 2.
Q7: What are molecules? Give a brief explanation of the arrangement of the constituent atoms in the molecules. Ans: A molecule is the smallest particle of an element or compound which is stable under normal conditions. And it can freely show all the properties of that element or compound. It may be made up of one, two or more atoms. A molecule with one atom is called a monoatomic molecule. E.g. helium, neon, etc.
A molecule with two atoms is called a diatomic molecule. E.g. Cl2, O2. Similarly, there are molecules containing three atoms (CO2), four atoms (P4) and so on.
Q8: The mass of one molecule of a substance is 4.65 × 10–23 grams. What is its molecular mass? Ans: Mass of 1 molecule of a substance = 4.65 × 10–23 grams
Mass of 6.023 × 1023 molecules of a substance
= 4.65 × 10–23 × 6.023 × 1023
= 28 g/mol
Molecular mass of the substance = 28 g/mol
Q9: An element 12X24 loses two electrons to form a cation, which combines with the anion of element 17Y35 formed by gaining an electron. (a) Write the electronic configuration of element X. (b) Write the electronic configuration of the anion of element Y. (c) Write the formula for the compound formed by the combination of X and Y. Ans:
(a) X = 2, 8, 2
(b) Y– = 2, 8, 8
(c) XY2
Q10: Calculate the formula unit masses of ZnO, Na2O, K2CO3 given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and 0 = 16u. Ans: Formula unit mass of ZnO = 1 × 65u + 1 × 16u = 81u
Formula unit mass of Na2O = 2 × 23u + 1 × 16u = 62u
Formula unit mass of K2CO3 = 2 × 39u + 1 × 12u + 3 × 16u = 138u
Q11: What is the mass of : (a) 1 mole of nitrogen atoms? (b) 4 moles of aluminium atoms (atomic mass of aluminium = 27)? (c) 10 moles of sodium sulphite (Na2SO3)? Ans:
(a) 1 mole of nitrogen atoms
= 1 × gram atomic mass of nitrogen atom
= 1 × 14 g = 14 g
(b) 4 moles of aluminium atoms
= 4 × gram atomic mass of aluminium atoms
= 4 × 27 g = 108 g
(c) 10 moles of sodium sulphite (Na2SO3)
= 10 (2 × gram atomic mass of Na + 1 × gram atomic mass of sulphur + 3 × gram atomic mass of oxygen)
= 10 (2 × 23 g + 1 × 32 g + 3 × 16g)
= 10 (46 g + 32 g + 48 g)
= 10 × 126 g = 1260 g
Q12: Give the postulates of Dalton’s atomic theory. Ans: John Dalton proposed his atomic theory in 1808. It explains the nature of matter in terms of atoms. The main postulates of Dalton’s atomic theory are:
All matter is made up of indivisible particles called atoms.
Atoms of a given element are identical in mass and properties.
Atoms of different elements have different masses and chemical properties.
Atoms combine in simple whole-number ratios to form compounds.
Atoms can neither be created nor destroyed in a chemical reaction.
The relative number and types of atoms in a given compound remain constant.
These postulates laid the foundation for modern chemistry, although some parts were later modified with the discovery of subatomic particles.
Q13: (a) Give one point of difference between an atom and an ion. (b) Give one example each of a polyatomic cation and an anion. (c) Identify the correct chemical name of FeSO3: Ferrous sulphate, Ferrous sulphide, Ferrous sulphite. (d) Write the chemical formula for the chloride of magnesium.
Ans:
(a) An atom is a neutral particle containing equal numbers of protons and electrons. An ion is a charged particle formed when an atom loses or gains electrons. Positive ions are called cations, and negative ions are called anions.
(b) (i) Polyatomic cation: Ammonium ion, (NH4)+
(ii) Polyatomic anion: Sulphate ion, (SO4)2–
(c) FeSO3 contains Fe²⁺ and SO3²⁻ (sulphite ion), so the correct name is Ferrous sulphite.
(d) Magnesium has a valency of 2, and chlorine has a valency of 1. Therefore, the formula is MgCl2 (Magnesium chloride).
Q14: When 3.0 g of magnesium is burnt in 2.00 g of oxygen, 5.00 g of magnesium oxide is produced. What mass of magnesium oxide will be formed when 3.00 g of magnesium is burnt in 5.00 g of oxygen? Which law of chemical combination will govern your answer? State the law. Ans: According to the given reaction: 3.00 g magnesium + 2.00 g oxygen → 5.00 g magnesium oxide. This shows that 3 g of magnesium reacts completely with 2 g of oxygen.
Now, when 3.00 g magnesium is burnt in 5.00 g oxygen, only 2 g of oxygen will be used (as per the fixed ratio), and 3 g of oxygen will remain unused.
So, mass of magnesium oxide formed = 3 + 2 = 5.00 g
Law involved: This illustrates the Law of Definite Proportions, which states that a chemical compound always contains the same elements in the same fixed ratio by mass.
Q15: (a) Calculate the number of molecules of SO2 present in 44 g of it. (b) If one mole of oxygen atoms weighs 16 grams, find the mass of one atom of oxygen in grams. Ans: (a) Molecular mass of SO2 = Atomic mass of S + 2 × Atomic mass of O = 32 + 2 × 16 = 64u Molar mass = 64 g Number of molecules, N = = 44/64 x 6.022 x 1023 = 4.14 × 1023 molecules (b) One mole of oxygen contains 6.022 × 1023 atoms of oxygen Mass of one atom of oxygen = = 2.66 × 10–23 g
Q16: Sodium is represented as 23Na11. (a) What is its atomic mass? (b) Write its gram atomic mass. (c) How many atoms of Na will be there in 11.5 g of the sample? Ans: (a) Atomic mass = 23u (b) Gram atomic mass = 23 g (c) Given mass = 11.5 g Molar mass = 23 g Number of atoms (N) = = 3.011 × 1023 atoms Q15: (a) Calculate the number of molecules of SO2 present in 44 g of it. (b) If one mole of oxygen atoms weighs 16 grams, find the mass of one atom of oxygen in grams. Ans: (a) Molecular mass of SO2 = Atomic mass of S + 2 × Atomic mass of O = 32 + 2 × 16 = 64u Molar mass = 64 g Number of molecules, N = = 44/64 x 6.022 x 1023 = 4.14 × 1023 molecules (b) One mole of oxygen contains 6.022 × 1023 atoms of oxygen Mass of one atom of oxygen = = 2.66 × 10–23 g
Q16: Sodium is represented as 23Na11. (a) What is its atomic mass? (b) Write its gram atomic mass. (c) How many atoms of Na will be there in 11.5 g of the sample? Ans: (a) Atomic mass = 23u (b) Gram atomic mass = 23 g (c) Given mass = 11.5 g Molar mass = 23 g Number of atoms (N) = = 3.011 × 1023 atoms
Q1. Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25 K or less. What part of fractional distillation apparatus makes it efficient and possess an advantage over a simple distillation process. Explain using a diagram. Answer:
The apparatus used for fractional distillation includes a fractionating column. This column is designed to enhance the separation of liquids with similar boiling points. Here’s how it works:
The fractionating column contains glass beads that increase the surface area.
As vapours rise, they encounter these beads, allowing them to cool and condense more effectively.
This process occurs in multiple cycles, improving the separation of components.
In contrast, a simple distillation lacks this feature, making it less efficient for separating miscible liquids with small boiling point differences.
Q2. (a) Under which category of mixtures will you classify alloys and why? Answer:When constituent particles of a combination of two or more element or compound retains their properties, then it is called mixture. In an alloy the constituent particles, hence alloys are classified as mixture. For example; steel is an alloy of carbon and iron.
(b) A solution is always a liquid. Comment. Answer: Since, a solution is the homogeneous mixture of two or more substances, thus it is not necessary that a solution would always a liquid.A solution can be in all the three states of matter. A solution is a homogeneous mixture and can be in all the three states of matter.
Example:
Solution of alcohol in water is a liquid. Air is a solution of different gas. Alloy is a solution which is in the form of solid.
(c) Can a solution be heterogeneous? Answer: Solution is defined as the homogeneous mixture, hence a solution cannot be heterogeneous. But when a mixture becomes heterogeneous, it cannot be fall under the definition of solution.
Q3. Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the Parts and evolution of gas was seen in both the cases. How will you identify the gases evolved? Answer: When iron filings and sulphur are mixed, Part B (unheated) reacts with dilute hydrochloric acid to produce hydrogen gas (Fe + 2HCl → FeCl2 + H2), while Part A (heated to form iron sulphide, FeS) reacts to produce hydrogen sulphide gas (FeS + 2HCl → FeCl2 + H2S). The gases can be identified by simple tests: hydrogen from Part B gives a pop sound with a burning splinter, whereas hydrogen sulphide from Part A turns moist lead acetate paper black. This confirms that heating transforms the mixture into a compound, changing the nature of the gas evolved.
Q4. A child wanted to separate the mixture of dyes constituting a sample of in 36. A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in Fig. The filter paper was removed when the water moved near the top of the filter paper.
(i) What would you expect to see, if the ink contains three different coloured components?
Answer: Streaks of different colours can be seen on the filter paper.
(ii) Name the technique used by the child. Answer: Chromatography
(iii) Suggest one more application of this technique. Answer: Chromatography is used for separating pigments from colours, for the separation drugs from blood sample, etc.
Q5. A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the lig37. A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the Fig. They were amazed to see that milk taken in the tumbler was Fig.illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?
(a) Explain why the milk sample was illuminated. Name the phenomenon involved. Answer: Since, milk is a colloid and when light scattered from the particles of colloids, it is illuminated, thus light was illuminated when passed through the milk. This is known as Tyndall Effect.
(b) Same results were not observed with a salt solution. Explain. Answer: For scattering of light the size of particles should be large enough. Since the particles of solution are not enough to scattered the beam of light, hence same result were not observed.
(c) Can you suggest two more solutions which would show the same effect as shown by the milk solution? Answer: Soap bubbles and fog are the colloids, hence same effect, i.e. scattering of light is shown by these. This is known as Tyndall effect.
Q6. Classify each of the following, as a physical or a chemical change. Give reasons. (a) Drying of a shirt in the sun. Answer: Drying of shirt in the sun is a Physical change. Since in this change no new substance is formed.
(b) Rising of hot air over a radiator. Answer: Since, in rising of hot air over a radiator no new substance is formed, hence it is a Physical change.
(c) Burning of kerosene in a lantern. Answer: While burning of kerosene in a lantern carbon dioxide, and water vapour is formed, hence it is a Chemical change.
(d) Change in the colour of black tea on adding lemon juice to it. Answer: In this change a new substance is formed, hence it is a Chemical change.
(e) Churning of milk cream to get butter. Answer: While churning of milk cream to get butter, no new substance is formed, hence it is a Physical change.
Q7. During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water while Sarika prepared it by dissolving 10g of sugar in water to make 100g of the solution.
(a) Are the two solutions of the same concentration Answer: No, the two solutions have different concentrations.
(b) Compare the mass % of the two solutions. Answer:We know;
For first solution:
Mass of solute = 10 gram
Mass of solution = 100 gram + 10 gram = 110 gram
Hence;
Mass % of solution = 10/110 x 100 = 9.09 %
For second solution:
Mass of solute = 10 gram
Mass of solution = 100 gram
Hence;
Mass percent of first solution: Mass percent of second solution = 9.09 : 10
Q8. You are provided with a mixture containing sand, iron filings, ammonium chloride and sodium chloride. Describe the procedures you would use to separate these constituents from the mixture? Answer:
The given mixture can be separated using the following methods:
Magnetic Separation: Use a magnet to attract the iron filings from the mixture. The iron filings will stick to the magnet, allowing for easy separation.
Sublimation: After removing the iron filings, heat the remaining mixture. Ammonium chloride will sublimate, turning into vapour without becoming liquid. This vapour will condense on the inner wall of a funnel, leaving sodium chloride and sand behind.
Filtration: Mix the remaining sand and sodium chloride with water. Stir the mixture; the sodium chloride will dissolve. Use filter paper to separate the undissolved sand from the solution.
Vapourisation: Heat the solution obtained from filtration to evaporate the water. This will leave behind sodium chloride crystals.
In summary, the components of the mixture—sand, iron filings, ammonium chloride, and sodium chloride—can be effectively separated using magnetic separation, sublimation, filtration, and vapourisation.
Q9. Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g of NaCl + 100g of water (b) 0.11g of NaCl + 100g of water (c) 0.01 g of NaCl + 99.99g of water (d) 0.10 g of NaCl + 99.90g of water
Answer: (c) 0.01 g of NaCl + 99.99 g of water The correct composition for a 0.01% (by mass) NaCl solution is 0.01 g of NaCl in 99.99 g of water. The other options either miscalculate the mass of NaCl or the water, making them incorrect.
Q10. Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100g of water? Answer: In a 20% solution containing 100 g water; the mass percentage of water = 100 – 20 = 80%
∴ 80% of solution is 100 gm
∴ 100% of solution is 100/80 gm
Hence; to prepare 20% (w/w) solution in 100 gram of water 25 gram of sodium sulphate is needed.
Q1: Discuss the factors which affect evaporation. Ans: Evaporation is the process by which liquid changes into vapour at temperatures below its boiling point. Several factors influence the rate of evaporation. These include:
1. Surface Area: The rate of evaporation increases with the increase in surface area. This is because more molecules are exposed to the surface and can escape into the atmosphere. For example, wet clothes dry faster when spread out.
2. Temperature: When the temperature rises, the kinetic energy of the particles increases. This allows more molecules to overcome the force of attraction and escape from the liquid surface as vapour.
3. Wind Speed: A higher wind speed removes the evaporated particles from the surface quickly, allowing more particles to escape. This increases the rate of evaporation.
4. Humidity: Humidity refers to the amount of water vapour present in the air. If the air is already humid, it cannot take in more vapour, so the rate of evaporation decreases.
Thus, evaporation is faster when the surface area and temperature are high, the wind is strong, and the humidity is low.
Q2: Explain the interconversion of three states in terms of force of attraction and kinetic energy of the molecules. Ans: Matter exists in three main states—solid, liquid, and gas. These states can change into one another by varying temperature and pressure. These changes are based on two key properties of particles:
Intermolecular Force of Attraction
Kinetic Energy
1. Solids: In solids, the particles are tightly packed due to strong forces of attraction. The kinetic energy is very low, so particles only vibrate at fixed positions.
2. Liquids: When a solid is heated, its particles gain energy and vibrate more vigorously. At a certain temperature (melting point), the force of attraction is weakened, and the solid becomes a liquid. Particles in liquids have more kinetic energy and can slide past each other.
3. Gases: Further heating of the liquid increases kinetic energy so much that the particles overcome almost all the attraction and move freely. The liquid changes into gas (boiling point). Gases have the weakest forces of attraction and the highest kinetic energy.
Hence, increasing temperature or decreasing pressure leads to a change of state by altering the kinetic energy and intermolecular attraction among particles.Interconversion of three states
Changes in temperature or pressure can cause these states to convert into one another:
Sublimation: Solid to gas without becoming liquid.
Deposition: Gas to solid without becoming liquid.
Boiling: Liquid to gas throughout the liquid.
Evaporation: Liquid to gas from the surface.
Q3: How is the high compressibility property of gas useful to us? Ans: Gases have a unique property called compressibility, which means they can be compressed easily because the particles are far apart. This property is extremely useful in our daily life and in industrial applications.
Uses of Compressibility:
LPG (Liquefied Petroleum Gas): LPG is a fuel made by compressing petroleum gas into a liquid and storing it in cylinders. When the pressure is released during use, it returns to gaseous form and is used for cooking.
Oxygen Cylinders: Oxygen used in hospitals is stored in compressed form in metal cylinders. This allows a large quantity of oxygen to be carried in a small container.
CNG (Compressed Natural Gas): CNG is methane gas compressed into high-pressure containers. It is used as a fuel for vehicles because it is cleaner and occupies less space in compressed form.
Thus, the compressibility of gases makes it possible to store and transport them efficiently for domestic, medical, and industrial use.
Q4: Pressure and temperature determine the state of a substance. Explain this in detail. Ans: The state of any matter (solid, liquid, or gas) depends on its temperature and the pressure applied to it. Both of these factors affect the particle movement and the intermolecular forces of attraction.
Effect of Temperature:
When the temperature increases, particles gain kinetic energy and start moving faster.
For example, if ice (solid) is heated, it melts into water (liquid). On further heating, the water boils and becomes steam (gas).
Similarly, cooling removes energy. Steam condenses into water, and upon further cooling, it forms ice.
Effect of Pressure:
An increase in pressure brings particles closer, which strengthens intermolecular forces and may change a gas into a liquid or a liquid into a solid
For instance, carbon dioxide gas changes into dry ice (solid) when high pressure is applied and the temperature is lowered.
In LPG cylinders, the gas is compressed under pressure to store it in liquid form. When the gas is used, the pressure is released and it becomes gaseous.
Both temperature and pressure play important roles in changing the state of a substance by affecting molecular motion and spacing between particles.
Q5: Give the difference between Evaporation and Boiling. Ans: Q6: The melting point of ice is 273.15 K. What does this mean? Explain in detail. Ans: The melting point of a solid is the temperature at which it changes into a liquid at normal atmospheric pressure. For ice, this melting point is 273.15 K (0°C).
Explanation:
At 273.15 K, ice starts to melt into water.
The temperature remains constant during this process, even though heat is continuously supplied.
The heat energy is used to overcome the strong intermolecular forces holding the solid particles together.
This energy is called the latent heat of fusion.
Once all the ice melts, the temperature of the resulting liquid water starts to rise if more heat is supplied.
The melting point of ice signifies the temperature at which it turns into liquid water without any change in temperature, by absorbing latent heat.
Q7: With the help of an example, explain how the diffusion of gases in water is essential. Ans: Diffusion is the process by which particles move from an area of higher concentration to an area of lower concentration. Gases from the atmosphere, such as oxygen and carbon dioxide, diffuse into water, and this diffusion is vital for aquatic life.
Importance of Diffusion in Water:
Oxygen from the air dissolves in water through diffusion. Aquatic animals like fish absorb this oxygen through their gills for respiration.
Carbon Dioxide also diffuses into water and is used by aquatic plants during photosynthesis to make food.
Example: Fish survive in water because they take in the dissolved oxygen, which has diffused from the air into the water. Similarly, underwater plants use dissolved carbon dioxide. Thus, the diffusion of gases like oxygen and carbon dioxide in water is crucial for the survival of aquatic organisms.
Q1: How do biotic and abiotic factors affect crop production? Ans: Living organisms such as honey bees and earthworms aid improve crop output, whereas pests (insects and rodents) and bacteria have a negative impact on crop production. Climate conditions and nonliving natural resources such as soil, water, and air are examples of abiotic factors.They also have an impact on crop productivity, as favourable temperature, humidity, and mineral nutrition boost crop yield.
Q2: What are the desirable agronomic characteristics for crop improvements? Ans: The following are desired agronomic traits for crop improvement: (i)For cereal crops, dwarfness is a beneficial trait since it allows the plants to use fewer nutrients. (ii)Tallness and profuse branching are ideal traits for fodder crops so that we can get more leaves to feed our animals.
Q3: What are macronutrients and why are they called macronutrients? Ans: There are sixteen nutrients that are required for plant growth. Six of these thirteen nutrients are considered macronutrients since they are required in high amounts. Nitrogen, phosphorus, potassium, calcium, magnesium, and sulphur are all macronutrients.
Q4: How do plants get nutrients? Ans: Plants get their nutrients from the air, water, and soil. Nutrients supplied as a source of Carbon in the air, oxygen in the air Hydrogen and oxygen are found in water. Soil nitrogen, phosphorus, potassium, calcium, magnesium, sulphur, iron, manganese, boron, zinc, copper, molybdenum, chlorine, zinc, manganese,
Q5: Why should preventive measures and biological control methods be preferred for protecting crops? Ans: It is likewise true for plants that prevention is preferable to treatment. Herbicides, weedicides, insecticides, pesticides, fungicides, and other chemicals are sprayed on the crop. Because their excessive usage can injure crop plants and pollute the environment, careful seed bed preparation, timely crop sowing, intercropping, and crop rotation are also recommended.
Q6: Which method is commonly used for improving cattle breeds and why? Ans: Cross-breeding is a typical technique for developing cow breeds. For instance, in dairy animals Exotic or foreign breeds (such as Jersey and Brown Swiss) are bred for lengthy lactation durations, whilst native breeds (such as RedSindhi and Sahiwal) are bred for disease resistance. The two can be crossed to produce animals with both desirable traits.
Q7: Discuss the implications of the following statement: “It is interesting to note that poultry is India’s most efficient converter of low fibre food stuff (which is unfit for human consumption) into highly nutritious animal protein food.” Ans: Under poultry the birds kept are fed on agricultural waste material and broken grains etc which are not useful for humans but those birds consuming such waste provide us with eggs and meat. It is a highly nutritious animal protein food hence the statement made is quite appropriate.
Q8: What management practices are common in dairy and poultry farming? Ans: Food requirements Proper cleaning and shelter facilities Protection from unfavourable climatic conditions and diseases are frequent management strategies in dairy and poultry farms. Protection against pests
Q9: How are fish obtained? Ans: Fishes can be obtained in two ways. (a) Capture fishing: This is a method of collecting fish from natural sources (rivers, lakes, oceans). (b) Cultural fishery: This is also known as fish farming and involves the rearing and breeding of selected fish.
Q10: What are the advantages of composite fish culture? Ans: The following are some of the benefits of composite fish culture:
In such systems, both indigenous and imported fish species can be utilised.
Food accessible in all regions of the water reservoir is exploited due to the non-competitive nature of selected species.
Increases the amount of fish in the water reservoir (intensive fish farming).
Q11: What are the desirable characters of bee varieties suitable for honey production? Ans: Characteristics of bee types appropriate for honey production include:
Honey harvesting capacity is high. They have to be less stingy.
They should stay in a beehive for a long time and breed prolifically.
Q12: What is pasturage and how is it related to honey production? Ans: Pasturage refers to the blooms that bees can collect nectar and pollen from. The pasturage determines the value or quality of honey. In addition, the type of flowers present will influence the honey’s flavour.
Q13: For increasing production, what is common in poultry, fisheries and bee-keeping? Ans: The following are steps that are commonly used in poultry, fisheries, and beekeeping to increase production:
The best types and breeds are used. Food is supplied that is both nutritional and tasty.
Cleanliness and hygienic conditions are maintained.
Q14: What are the benefits of cattle farming? Ans: Cattle farming provides two advantages: (i)Draught animals for farm labour (males), i.e. tilling, irrigation, and carting. (ii)Milch animals (dairy animals) are females who produce milk.
Q15: How do storage grain losses occur? Ans: Storage grain losses are caused by a variety of biotic and abiotic factors: biotic factors include insects, rodents, bacteria, fungi, and other organisms that feed on grains. Unfavorable humidity and temperature conditions are abiotic variables.
Q16: Why are manure and fertilizers used in fields? Ans: Manure contributes to soil fertility by supplementing it with nutrients and organic materials. The majority of organic matter in manure aids in soil structure improvement. Fertilizers are used to promote healthy vegetative growth (leaves, branches, and flowers) by supplying specific nutrients such as nitrogen, phosphorus, and potassium.
Q17: Define (a) Pisciculture Ans: Pisciculture is the large-scale rearing and management of fish. (b) hatcheries Ans: Hatcheries are nurseries where fish eggs or fish seed are placed in freshwater fisheries. (c) swarming Ans: Swarming is the process by which the new queen leaves the old hives and seeks out a new home for reproduction.
Q18: What is green manuring? Give examples of green manures. Ans: Green manure is made from herbaceous plants that are cultivated, ploughed beneath, and mixed with the soil while they are still green. Green manuring is the process of ploughing green plants and combining them with the soil. Sun hemp, cluster bean (guar), lentil (Masur), and cowpea are some of the plants utilised as green manure (Lobia).
Q19: Discuss the preventive measures for the storage of grains. Ans: The preventive measures for the storage of grains are: (a) Drying – For grain storage, the moisture content of the grains should be decreased to less than 14 percent. This can be accomplished by drying in the sun and then drying in the shade. (b) Hygiene should be maintained –Godowns and stores should be cleaned regularly. Remove any dirt, trash, webs, or debris from the previously kept grains. Waterproofing and sealing cracks and gaps in the walls, floor, and ceiling are essential. For storing food grains, new gunny bags should be used. The mouth of the gunny bag should be tightly sewn once it has been filled.
Q20: Name three basic scientific approaches for increasing yield of a crop. Ans: Three scientific approaches for increasing yields of a crop are – (i) Crop production management, which involves correct irrigation and fertiliser management, is one of three scientific methodologies for enhancing crop yields. Manure and fertilisers can be used to accomplish this. Crop rotation, intercropping, and mixed-cropping can all help with nutrient management. Plants require protection from weeds, insects, pests, and pathogens. (ii) Crop protection management It can be accomplished through biological, chemical, or cultural methods. (iii) Crop variety management: Crop variety can be enhanced by hybridization or transgenic techniques. It is possible to do so in order to get desired plant traits
Q21: What are the advantages of bee-keeping? Ans: The following are some of the benefits of beekeeping: (a)It takes minimal investment and offers the farmer additional income. (b)In addition to honey, beekeeping produces wax, royal jelly, and bee venom, among other things. (c) Cross pollination is aided by bees.
Q22: Differentiate between capture fishing, aquaculture and mariculture. Ans: (a) Capture fishing –Capture fishing is the process of obtaining fish from bodies of water such as rivers, seas, and oceans: (b) Aquaculture – Aquaculture is the cultivation of aquatic creatures in fresh or saltwater. (c) Mariculture – Mariculture is the cultivation of marine fish.
Q23: List the steps to be taken to prevent and control diseases in animals. Ans: The following steps should be followed to control diseases:
Providing adequate shelter.
Maintaining animal hygiene and disposing of dead animals and animal wastes properly.
Disease screening of animals on a regular basis, with unhealthy animals being isolated immediately.
Following the advice of a veterinary practitioner, providing a proper diet and appropriate medications.
Handling of all animal products and by-products in a sanitary manner.
Q24: What are the components of cattle feed? Ans: Roughage and concentrates are present in cattle feed in the form of hay and grain, as well as a large amount of water.
Roughage — Roughage is made up of coarse and fibrous components with poor nutrient content; animals acquire roughage from hay (cereal straw) and grain, respectively, as well as a lot of water.
Cotton seeds, oilseeds, oilcakes, and cereal grains like gramme and bajra are high in one or more nutrients (such as carbohydrates, lipids, proteins, minerals, and vitamins) and low in fibres. Cattle are fed green fodder in the winter, primarily Berseem and Lucerne, and maize, bajra, jowar, and dry fodder in the other seasons.
Q25: Define the following (i) White revolution Ans: The term “white revolution” refers to the increased production of milk. It required the utilisation of upgraded high-milk-yielding mulch animal crossbreeds. (ii) silver revolution Ans: The term “silver revolution” refers to a massive increase in egg output. (iii) blue revolution. Ans: The term “blue revolution” refers to an increase in fish production.
Q26: What is green manuring? Give an example of green manures. Ans: Green manure is made from herbaceous plants that have been cultivated, ploughed under, and mixed with the soil while they are still green. Green manuring is the term for this process. Sun hemp, cluster bean (guar), lentil (maser), and cowpea are some of the plants utilised as green manure (Berseem)
Q27: What are the main practices involved in keeping animals or animal husbandry? Ans: Animal husbandry day involves the following main practises.
Breeding – This is done to get animals with specific traits. Animals with high milk yields and meat yields can be developed through breeding.
Feeding – This is the study of the right food (called feed), as well as the mode and timing of feeding of various animals.
Weeding –This is the process of eliminating animals that are not economically viable.
Q28: Name the abiotic and biotic factors which affect stored grains and how? Ans: Insects, birds, rodents, mites, fungi, and bacteria are examples of biotic forces. (a) Moisture, temperature and the storage container’s material are all abiotic variables. As a result of the aforementioned conditions, cereal grains become infested with insects and microorganisms. (b) Quality deterioration (c) Weight reduction. (b) Poor grain germination potential (e) Produce discoloration
Q29: What is the need for crop improvement? What are the desirable agronomic characteristics for crop improvement? Ans: Crop enhancement entails creating superior plants with the following characteristics:
High-yielding
Varieties with higher-quality produce.
Disease-resistant cultivars
Plants have beneficial agronomic traits, such as
Dwarfness is required in cereals, which consumes less nutrients.
Plants for fodder crops that are tall and have a lot of branching.
Q30: Define (i) Draught breeds Ans: Draught cattle, sometimes known as bullocks, are cattle that are employed for labour. (ii) Dual purpose breeds Ans: Dual-purpose breeds are those that have females for milk and men for work. (iii) Dairy breeds Ans: Dairy animals are breeds that are solely used for milk production.
Q31: What are the symptoms of diseased animals? Ans:
The animal stops eating and becomes lethargic, looks exhausted, and remains isolated as a result of the condition.
The animal shivers as its body temperature rises.
The animal produces an excessive amount of saliva, which hangs from its mouth at times.
The animal excretes a mixture of loose dung and colourful urine.
The animal’s mouth and ears droop.
Q32: What do we get from cereals, pulses, fruits and vegetables? Ans: Cereals provide carbohydrates, pulses provide proteins, while fruits and vegetables provide vitamins and minerals.
Q33: What factors may be responsible for losses of grains during storage? Ans: Both biotic and abiotic factors may be responsible for losses of grains during storage are :
Humidity and temperature of the environment and moisture content of grains are the abiotic factors.
Biotic factors include organisms such as rodents, bacteria, fungi, and some insects that feed on grains.
Q34: What are weeds? Give two examples. Ans: Weeds are undesired plants that grow in fields. Common weeds include Amaranthus and Chenopodium.
Q35: What is crop rotation? Ans: Crop rotation is the technique of alternately cultivating various crops in the same land in a pre-planned succession.
Q36: What are drones? Ans: Drones are airborne devices that are used in agriculture to improve crop output and to track crop growth. They assist farmers in developing agricultural field systems for using water, fertilisers, herbicides, and seeds. These tools have revolutionised agriculture by allowing farmers to save significant amounts of money while also increasing efficiency and profitability.
Q37: What is pasturage and how is it important? Ans: Crop rotation is the practice of farming a variety of crops in a pre-determined order on the same piece of land.
Q38: What is a layer and a broiler? What are the differences between the two? Ans: The egg-laying poultry bird is known as an egg layer, whereas the meat-producing poultry bird is known as a chicken or broiler. Housing (shelter), food, and environmental requirements differ from layer requirements. Broiler feed is protein- and vitamin-rich, with an acceptable fat content.
Q39: What are the characteristic features of ideal shelters for cattle? Ans: The following are characteristics of a shelter:
The animals are protected from rain, heat, and cold by a suitably roofed shed.
The shed’s floor is slanted to make cleaning easier and to keep their sitting area dry.
A plan for safe drinking water is put in place.
Excreta disposal is properly arranged in the sheds.
Q40: What are the hazards of using fertilizers? Ans: Effects of fertiliser application – (a) Impact on soil quality – fertiliser application leads to a loss of organic matter and a deterioration of soil structure. (b) Eutrophication – Excessive fertiliser application causes nitrate buildup in the soil. Rain washes nitrates and phosphates into lakes, ponds, and rivers, where they encourage algae to develop excessively.
In case of upward motion of the ball from A to B
times the original distance.
s = 127.5 m
u = 39.2 m/s2
Ans:
Q21: Give one function of each of the following : (a) Stomata, (b) Root nodules, (c) Cardiac muscle fibres.
Plant cell
Animal Cell
Golgi apparatus
Structure of a nucleus
Nucleus
Mitochondria
Scattering of alpha particles by a gold foil
Thus, finally = (NH4)2CO3
Interconversion of three states
