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Page No. 102Q1. State the universal law of gravitation.
Ans: Everybody in this universe attracts every other body with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.Universal Law of GravitationQ2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Ans: Consider F as the force of attraction between an object on the surface of the earth and the earth.
Let the mass of earth = M
Mass of object = m.
If the radius of the object is comparatively negligible, r ≈ radius of earth = R
F = GMm/R² 

Page No. 104

Q1. What do you mean by free fall?
Ans: When an object falls with a constant acceleration, under the influence of the force of gravitation of the Earth, the object is said to have a free fall.

Q2. What do you mean by acceleration due to gravity?
Ans: During the course of its free fall, a body accelerates due to the force of gravity acting on it. This acceleration is known as acceleration due to gravity.

Page No. 106

Q1. What are the differences between the mass of an object and its weight?
Ans:

Q2. Why is the weight of an object on the moon 1/6th its weight on the Earth?
Ans: 
Mass of the moon (M) = 7.4 × 10²² kg
The radius of the moon (R) = 1.74 × 10⁶ m
Gravitational constant (G) = 6.7 × 10¹¹ Nm²kg²
We know that acceleration due to gravity(g) = GM/R²
Also, weight, W = mg
W = GMm/R²
Let the mass of the object be m, let its weight of the moon be M_m and its radius be R_m
By applying the universal law of gravitation, the weight of the object on the moon will be:

  (1)

Let the weight of the same object on the earth be We. The mass of the earth is M and its radius is R.

 (2)

Substituting the values in equations (1) and (2) we get:

W_m= 2.431 x 10¹⁰G x m

and W_e = 1.474 x 10¹¹ G x m

On dividing W_m and W_e we get:

∴ The weight of the object on the moonits weight on Earth

Page No. 109

Q1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Ans: 

• It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. 
• This is because the pressure is inversely proportional to the surface area on which the force acts. 
• The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

Q2. What do you mean by buoyancy? 
Ans: 

• The upward force exerted by a liquid on an object immersed in it is known as buoyancy.
  
• When you try to immerse an object in water, then you can feel an upward force exerted on the object, which increases as you push the object deeper into water.

Q3. Why does an object float or sink when placed on the surface of water?
Ans: 

• If the density of an object is more than the density of the liquid, then it sinks in the liquid.
  
• This is because the buoyant force acting on the object is less than the force of gravity. 
• On the other hand, if the density of the object is less than the density of the liquid, then it floats on the surface of the liquid. 
• This is because the buoyant force acting on the object is greater than the force of gravity. 

Page No. 110

Q1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg? 
Ans: Mass is always a constant quantity. Therefore, it cannot be more or less than 42 kg.

Q2. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why? 
Ans: The correct answer is cotton bag is heavier than an iron bar.
The reason is:

True weight = (apparent weight + up thrust)

• The density of the cotton bag is less than that of the iron bar, so the volume of the cotton bag is greater than the iron bar. So the cotton bag experiences more upthrust due to the presence of air.
• Therefore in the presence of air, the true weight of a cotton bag is greater than the true weight of an iron bar.

Page No. 111

Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Ans: The force of gravitation between two objects of masses m₁ and m₂ at separation r is given by:
F = (G m₁m₂)/r²
When the distance between them is reduced to half, the gravitational force.
F’ = (G m₁m₂)/r²
= (G m₁m₂)/ (r/2)²
= (4 G m₁m₂)/(r²) = 4F
The gravitational force becomes four times.

Q2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Ans: The acceleration due to gravity of a freely falling body is independent of the mass of the falling body. Thus, both heavy and light objects fall with the same acceleration.

Q3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m.)
Ans: Here, the mass of the earth, M = 6 x 10²⁴ kg
Mass of the object, m = 1 kg, radius of the earth, R = 6.4 x 10⁶ m, G = 6.7 x 10^-11 Nm²kg^-2
The magnitude of the gravitational force between the earth and the object is given by:
The force of gravitation between them,


Q4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller force or the same as the force with which the moon attracts the earth? Why?
Ans: 

• Earth attracts the moon with an equal and opposite force as the moon attracts the Earth. 
• This is in accordance with Newton’s law of gravitation according to which gravitational force is the mutual force of attraction which is also equal as per to the third law of motion.

Q5. If the moon attracts the Earth, why does the earth not move towards the moon? 
Ans:

• The moon revolves around the earth because the gravitational force of the earth on the moon provides the necessary centripetal force for circular motion.
• The moon also attracts the earth, due to which the earth also revolves around the moon in a circular orbit, but due to the very large mass of the earth, the motion of the earth is not observed.
• It may be kept in mind that there is no linear motion of Earth towards the moon or the moon towards Earth. The gravitational force between them keeps on changing the direction of motion of the moon or earth in a circular orbit.

Q6. What happens to the force between two objects, if 
(a) the mass of one object is doubled? 
(b) the distance between the objects is doubled and tripled? 
(c) the masses of both objects are doubled? 
Ans: Let masses = m₁ and m₂, the distance between masses = r.
According to the universal law of gravitation F ∝ m₁m₂ and F ∝ 1/r₂.
(a) If one mass, say m₁ is doubled, and F ∝ m₁, then F gets doubled.
(b) If distance is doubled and F ∝ 1/r², force becomes one-fourth, i.e. 1/(2)².
If the distance is tripled and F ∝ 1/r², force becomes one-ninth, i.e. 1/(3)².
(c) If both masses are doubled and F ∝ m₁m₂, force becomes 4 times, i.e. (2m₁)(2m₂) = 4m₁m₂

Q7. What is the importance of universal law of gravitation? 
Ans: The universal law of gravitation has successfully explained the phenomena that were earlier considered to be separate:
(a) It explains the force which holds us on earth.
(b) It explains the motion of planets around the sun.
(c) It explains the motion of the moon around the Earth.
(d) It explains the occurrence of tides in the ocean.

Q8. What is the acceleration of free fall?
Ans: When a body falls freely and accelerates at every point of its motion due to gravitational force alone. This is called the acceleration of free fall. This acceleration is known as the acceleration due to gravity on the earth’s surface. It’s denoted by ‘g’, and its value is 9.8m/s2, and it’s constant for all objects close to the earth’s surface (irrespective of their masses).

Q9. What do we call the gravitational force between the earth and an object? 
Ans: Gravity is the name of the gravitational force between the earth and the object.

Q10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.] 
Ans: Earth is not a perfect sphere. It is flattened at the poles. Thus, the value of ‘g’ is greater at the poles than at the equator (as g ∝ 1/r²).  This means the weight of gold bought at the poles becomes lesser at the equator, and the friend does not agree with the weight.

Q:11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans: 

• A sheet crumpled into a ball has a small surface area as compared to that of an unfolded sheet. Therefore, the unfolded sheet will experience more friction due to air as compared to the crumpled ball in spite of the same force of gravity acting upon it.
• The larger friction of air slows down the unfolded sheet so it will fall slower than the sheet crumpled into a ball.

Q12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth? 
Ans: Here, mass of object, m = 10 kg, g on the earth’s surface = 9.8 ms^-2 
Weight of the object on the earth’s surface = mg
= 10 kg x 9.8 ms^-2
= 98.0 kg ms^-2 = 98 N
The weight of an object on the moon is 1/6th that of the object on Earth, so the weight of the object on the moon:
= 1/6 x 98 N= 16.34 N.

Page No. 112

Q13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate 
(a) the maximum height to which it rises.
(b) the total time it takes to return to the surface of the earth. 
Ans: 
(a) Initial velocity, u = 49 ms^-1, Final velocity, v = 0
Maximum height, H = ?
Acceleration due to gravity, g = + 9.8 ms^-2 (downward)

As object is moving upward, so a = -gFrom relation v² = u² + 2as , we have a = -g , s = H
v² = u² – 2gH


⇒ H = 122.5 m
(b) When the ball returns to the surface of the earth, displacement, s = 0
∴ Relation, s = ut + at² gives s = ut -gt²
0 = (49 ms^-1) t -x 9.8 t²
⇒ t (49 -4.9 t) = 0
t = 0 or 49 – 4.9 t = 0 ⇒ t =
As t ≠ 0, so t = 10s
The maximum height is 122.5 m and the total time is 10 s.

Q14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. 
Ans: Initial velocity, u = 0, height, h = 19.6 m and g = 9.8 ms^-2
Final velocity,

= 19.6 ms^-1.

Q15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What are the net displacement and the total distance covered by the stone?
Ans: Initial velocity, u = 40 ms^-1
g = -10 ms^-2 (upward motion)
Final velocity, v = 0
During upward motion, g = -10 ms^-2

Net displacement on returning back = zero
Total distance = 80 m + 80 m = 160 m

Q16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m. 
Ans: Mass of earth, M_e = 6 x 10²⁴ kg
Mass of sun M_s = 2 x 10³⁰ kg and
Distance, r = 1.5 x 10¹¹ m
Gravitational constant, G = 6.67 x 10^-11 Nm²/kg²
Force, 

= 3.56 x 10²² N

Q17. A stone is allowed to fall from the top of a tower 100 m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet. 
Ans:  u₁ = 0
Let the stones meet at a height h above the ground. Stone 1 covers a distance (100 – h); g = 10 m/s².
(100 – h) = 0 + 
⇒ 100 – h = 5 t² …(i)
u₂ = 25 m/s
Stone 2 covers a distance h
g = – 10m/s²
h = u₂t – 
⇒ h = 25 t – 5 t² …(ii)
From (i) and (ii), we get 
100 – 25 t + 5 t² = 5 t² 
100 – 25 t = 0
or t = 4 s (Stones meet 4 s after they are thrown)
h = 25(4) – 5(4)² = 100 – 80 = 20 m above the ground.

Q18. A ball thrown up vertically returns to the thrower after 6s. Find
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches.
(c) its position after 4s.
Ans: 

Time of ascend = Time of descending

⇒  Time taken to reach the top, t = 3 s.
⇒ Final velocity, v = 0, g = -9.8 ms^-2.
(a) Initial velocity, u = v – gt = 0 – (-9.8) (3)
= 29.4 ms^-1
(b) Maximum height reached, h = (v² – u²)/g = (29.4)²/(2 x 9.8)
= 44.1 m
(c) After 4s, the ball has started falling and has fallen by some distance h’ for 1s.
Here, initial velocity u’ = 0, t = 1s
g = + 9.8 ms^-1
h’ = u’t +  = 0  + 
The ball is at a height, (44.1 – 4.9) = 39.2 m above the ground.

Q19. In what direction does the buoyant force on an object immersed in a liquid act?
Ans: An object immersed in a liquid experiences buoyant force in the upward direction.

Q20. Why does a block of plastic released under water come up to the surface of water?
Ans:

• Two forces act on an object immersed in water. One is the gravitational force, which pulls the object downwards, and the other is the buoyant force, which pushes the object upwards. 
• If the upward buoyant force is greater than the downward gravitational force, then the object comes up to the surface of the water as soon as it is released within the water. 
• Due to this reason, a block of plastic released under water comes up to the surface of the water.

Q21. The volume of 50g of a substance is 20 cm³. If the density of water is 1g cm^-3, will the substance float or sink?
Ans: 

• If the density of an object is more than the density of a liquid, then it sinks into the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.
• Here, the density of the substance = 
• The density of the substance is more than the density of water (1 g cm⁻³). Hence, the substance will sink in water.

Q22. The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm^-3? What will be the mass of the water displaced by this packet? 
Ans:

• Density of the 500 g sealed packet
• The density of the substance is more than the density of water (). Hence, it will sink in water.
• The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 cm³ of water has a mass of 350 g.

Page No. 91

Q1. Which of the following has more inertia?
(a) A rubber ball and a stone of the same size?
(b) A bicycle and a train?
(c) A five rupees coin and one-rupee coin?
Ans: Inertia is the property of an object to resist a change in its state of motion.
More mass = More inertia
(a)  A stone has more mass than a rubber ball of the same size. Therefore, the stone has more inertia.
(b) A train has much more mass than a bicycle. Therefore, the train has more inertia.
(c) A five-rupee coin is heavier than a one-rupee coin. Therefore, the five-rupee coin has more inertia.

Q2. In the following example, try to identify the number of times the velocity of the ball changes.
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.
Ans: The velocity of the ball changes four times in the following ways:
(a) Player 1 changes the velocity by kicking it.
(b) Player 2 changes the velocity by kicking it to the goal.
(c) The goalkeeper changes the velocity by, collecting the ball.
(d) The goalkeeper changes the velocity by kicking the ball.

Q3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Ans: The leaves of the tree have inertia of rest. When the branch is shaken, it moves, but the leaves tend to be in a state of rest due to their inertia. Thus, they get detached from the tree.

Q4. Why do you fall in the forward direction when a moving bus apply brakes to stop and fall back when it accelerates from rest?
Ans: 

• In a running bus, our speed is equal to the speed of the bus. As a moving bus brakes to a stop, the lower part of our body being in contact with the bus comes to rest, but the upper part due to inertia of motion remains in the state of motion.
  The bus stops suddenly, Passenger jerks forward
• Hence, we fall in the forward direction. When the bus accelerates from rest, the feet come into motion while the upper part of the body remains at rest due to the inertia of rest; hence, we fall backwards.

Page No. 97

Exercises

Q1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Ans: 

• Yes, it is possible for the object to be travelling with a non-zero velocity if it experiences a net zero external unbalanced force.
• This is due to the inertia of motion. If the body is initially moving with some velocity on a smooth surface, then it will continue to move with the same velocity, though the net external force acting on the body is zero.
  Example: When we stop pedalling a moving bicycle, the bicycle begins to slow down and finally comes to rest. This is again because of the friction forces acting opposite to the direction of motion. The force of friction opposes the motion of the bicycle. If there were no unbalanced force of friction and no air resistance, a moving bicycle would go on moving forever.

Q2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Ans: When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to the inertia of rest, therefore, dust comes out of it.

Q3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ans: The luggage kept on the roof possesses inertia. As the bus moves, brakes to stop or takes a turn, the luggage might fall off. Thus, it should be tied with a rope.

Q4. A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because 
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Ans: When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball. Therefore, the correct answer is (c).
If the surface of the level ground is lubricated (with oil or some other lubricant), the friction that arises between the ball and the ground will reduce, which will enable the ball to roll for a longer distance.

Q5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes
(Hint: 1 metric tonne = 1000 kg.)
Ans: 
Initial velocity of truck (u) = 0
Time (t) = 20 s
Distance covered (S) = 400 m
Acceleration (a) = ?
Mass of truck (m) 7 tonne = 7000 kg
Force on truck We know ; (F) = ?
We know:
S = ut  1/2 at²
400 = 0 × 20 1/2 × a × (20)²
400 = 200a
a = 2ms^-2
Force on truck (F) = ma
= 7000 x 2
= 14000 N

Q6. A stone of 1 kg is thrown with a velocity of 20 ms^–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Ans:  
Mass, m = 1kg; Initial velocity, u = 20 ms^-1; 
Final velocity, v = 0;
distance travelled, s = 50 m
Acceleration, 

Force exerted = ma
= 1 kg x (- 4) ms^-2 = – 4N
Friction = 4 N against the direction of motion.

Q7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: 
(a) the net accelerating force and (b) the acceleration of the train
Ans: (a) Given, the force exerted by the train (F) = 40,000 N
The force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)
Therefore, the net accelerating force = the sum of all forces

= 40,000 N + (-5000 N) 

= 35,000 N
(b) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg
The total mass of the train is 18000 kg.
As per the second law of motion, F = ma (or: a = F/m)
Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)
= 35,000/18,000 = 1.94 ms^-2
The acceleration of the train is 1.94 ms^-2

Q8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms^–2?
Ans: Here, the mass of the automobile, m = 1,500 kg
Acceleration, a = 1.7 ms^-2 
Force between the vehicle and road (F) can be calculated as 
F = ma = 1500 kg x ( -1.7 ms^-2)
= -2,550 kg ms^-2
= -2,550 N

Negative sign indicates that the force is in a direction opposite to the motion of the vehicle.

Page No. 98

Q9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)² 
(b) mv² 
(c) 1/2mv² 
(d) mv
Ans: (d) mass x velocity
The momentum of an object is defined as the product of its mass m and velocity v.
Momentum = mass x velocity
Hence, the correct answer is mv, i.e., option (d).

Q10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Ans: Since the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. This implies that the magnitude of the opposing frictional force is equal to the force exerted on the cabinet, which is 200 N. Therefore, the total friction force is -200 N.

Q11. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal force cancel each other Comment on this logic and explain why the truck does not move.
Ans: The logic given by the student is wrong because the action and reaction forces do not act on the same body, but act on different bodies.
The reason for not moving the truck is as follows:

• The truck is massive, so it has a very large inertia. Moreover, it is parked on the ground; there is a frictional force between the truck and the ground. 
• The force exerted by us on the truck is insufficient to overcome the force of friction; so these are balanced forces on the truck; (force exerted by us plus force of friction in opposite direction); that is, the net force is zero; and hence, the truck does not move.
  

Q12. A hockey ball of mass 200 g travelling at 10 ms^–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^–1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans:
Mass of ball (m) = 200 g = 0.2 kg
Initial velocity of ball (u₁) = 10 ms^–1
Final velocity of ball (u₂) = 5 ms^–1
(Negative sign denotes that the ball is moving in the opposite direction)
Initial momentum of ball = mu₁ = 0.2 × 10 = 2 Ns
Final momentum of ball = mu₂ = 0.2 × (- 5) = -1 Ns
Change in momentum = Final momentum – Initial momentum
=  -1 – 2 Ns  = -3 Ns

Negative sign denotes that change in momentum is in the direction opposite to the direction of initial momentum of the ball.

Q13. A bullet of mass 10g travelling horizontally with a velocity of 150 ms^–1 strikes a stationary wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
Ans:
Mass of bullet (m) = 10 g = 0.01 kg
The initial velocity of a bullet (u) = 150 ms^–1
The final velocity of a bullet (v) = 0
Time (t) = 0.03 secs
Acceleration on the bullet (a) = ?
Force acting on wooden block (F) = ?
Distance penetrated by the bullet (s) = ?
We know:
v = u + at
0 = 150 + (a × 0.3)
a × 0.03 = -150
a = -5000 ms^-2 (Negative sign indicates that the velocity of the bullet is decreasing.)

We know:

S = ut + 0.5at²
s = 150 × 0.03 + (-5000) × (0.03)² = 4.5 + 2.25 = 6.75m
We know:
F = ma
Force acting on bullet (F) = ma  = 0.01 × (-5000) = -50N

Negative sign denotes that wooden block exerts force in the direction, opposite to the direction of motion of the bullet.

Q14. An object of mass 1 kg travelling in a straight line with a velocity of 10 ms^–1 collides with and sticks to a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Ans: For object:
m₁ = 1 kg
u₁ = 10 ms¹
For wooden block:
m₂ = 5 kg
u₂ = 0
Momentum just before collision:
= m₁u₁ + m₂u₂
= 1 × 10 + 5 × 0
= 10 kg ms^–1
Mass after collision = (m₁ + m₂)
= 1 + 5 = 6 kg
Let velocity after collision = v
Momentum after collision = 6 × v
Using the law of conservation of momentum:
Momentum after collision = Momentum before collision
6 × v = 10
v = 1.67 ms^–1

Q15. An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms^–1 to 8 ms^–1 in 6s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Ans: 
Mass of object (m) = 100 kg
Initial velocity (u) = 5 ms^–1
Final velocity (v) = 8 ms^–1
Time (t) = 6 s
Initial momentum (P_initial) = mu = 100 × 5 = 500 Ns
Final momentum (P_final) = mv = 100 × 8 = 800 Ns
Force exerted on the object (F) = [(mv – mu) / t] = (800 – 500)/6 = 300/6 = 50N

Q16. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions. 
Ans: As per the law of conservation of momentum, the total momentum before the collision between the insect and the car is equal to the total momentum after the collision. Therefore, the change in the momentum of the insect is much greater than the change in momentum of the car (since force is proportional to mass).
Akhtar’s assumption is partially right. Since the mass of the car is very high, the force exerted on the insect during the collision is also very high.
Kiran’s statement is false. The change in momentum of the insect and the motorcar is equal to the conservation of momentum. The velocity of an insect changes accordingly due to its mass as it is very small compared to the motorcar. Similarly, the velocity of a motorcar is very insignificant because its mass is very large compared to the insect.
Rahul’s statement is completely right. As per the third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. However, Rahul’s suggestion that the change in the momentum is the same contradicts the law of conservation of momentum.

Q17. How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm and does not rebound? Take its downward acceleration to be 10ms^–2.
Ans:  
Height, h = 80 cm = 0.8 m
Mass, m = 10 kg ; Initial velocity, u = 0
Acceleration, a = 10 ms^-2
Final velocity, v = 

Momentum transferred = mv = 10 kg x 4 ms^-1
= 10 kg x 4 ms^-1
= 40 kgms^-1

Page No. 99

Q1. The following is the distance-time table of an object in motion:

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing or zero?
(b) What do you infer about the forces acting on the object?
Ans: 
(a) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.

(b) Force, F = ma ⇒ F ∝ a. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well

Q2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms^–2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Ans: 
Mass of car, m = 1200 kg
Let each person apply force F on the motor car. Thus, the force ‘2F applied by 2 persons balances the frictional force due to the ground force applied by three persons = 3F
Effective force = 3F – friction = 3F – 2F = F
Thus force F produces an acceleration of 0.2 ms^-2.
F = ma = 1200 x 0.2 = 240 N
Thus, force applied by each person = 240 N

Q3. A hammer of mass 500 g, moving at 50 ms^–1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Ans: 
Mass of hammer, M = 500 g = 0.5 kg
The initial velocity of the hammer, u = 50 ms^-1 
The final velocity of the hammer, v = 0 
Time, t = 0.01 s
= – 5000 ms^-2
The force applied by the nail, F = ma
= (0.5) x (-5000 ms^-2)
= – 2500 N (opposite to the motion of the hammer)

Q4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required. 
Ans:
Mass of motor car, m = 1200 kg 
Initial velocity of car, u = 90 km/h

Final velocity of car, v = 18 km/h

Time, t = 4s
Acceleration, 
Momentum change = mv – mu or  m(v – u)
= 1200 (5 – 25) = – 24000 kg ms^-1
Magnitude of force, F = ma
= 1200 x (- 5) = – 6000 N
Acceleration, momentum change and force are opposing the motion of the motorcar, as indicated by a negative sign.

Page No. 74

Q1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.

Ans: Yes, an object can have zero displacements even if it has moved through a distance.

Example: If an athlete runs around a circular path of radius ‘r’ and comes back to the initial point, then distance covered = 2πr, displacement = zero.

Q2. A farmer moves along the boundary of a square field of side 10 m in the 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Ans: Given,

• Side of the square field = 10 m
• Perimeter of the square = 4 × 10 = 40 m
• Time to complete one round = 40 s
• Total time = 2 minutes 20 seconds = 2 × 60 + 20 = 140 s

Speed of the farmer = Perimeter / Time = 40 / 40 = 1 m/s

Distance covered in 140 s = 1 × 140 = 140 m

Number of complete rounds = Total distance / Perimeter = 140 / 40 = 3.5

After 3 complete rounds (120 m, 120 s), the farmer is back at the starting point (e.g., point A at (0,0)). In the remaining 0.5 round (20 m, 20 s), the farmer moves halfway around the square. Starting at A (0,0), the path is:

• A to B (10,0): 10 m
• B to C (10,10): 10 m

After 0.5 round, the farmer is at C (10,10). The displacement is the straight-line distance from A (0,0) to C (10,10):

Displacement = √((10-0)² + (10-0)²) = √(100 + 100) = √200 = 10√2 ≈ 14.14 m

Magnitude of displacement = 14.14 m

Q3. Which of the following is true for displacement?

(a) It cannot be zero.

(b) Its magnitude is greater than the distance travelled by the object.

Ans:

(a) False; Displacement can be zero if an object returns to its initial position after moving. For example, an athlete running around a circular track and returning to the starting point has zero displacement.

(b) False; The magnitude of displacement is always less than or equal to the distance travelled. Displacement equals distance only when the motion is in a straight line without reversing direction. For example, in circular motion, displacement is zero when returning to the start, while distance is non-zero.

Page No. 76

Q1. Distinguish between speed and velocity.

Ans:

Q2. Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?

Ans: Average speed measures the total distance covered over a specific time period, while average velocity refers to the total displacement during that same time. The magnitudes of average speed and average velocity will be equal when the total distance traveled matches the displacement.

Q3. What does the odometer of an automobile measure? 

Ans: An odometer, also known as an odograph, is a device that calculates the distance an automobile has travelled by measuring the circumference of the wheel as it rotates.

Q4. What does the path of an object look like when it is in uniform motion?

Ans: The path of the object will be a straight line at the instant of measurement when it is in uniform motion.

Q5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, which is 3 × 10⁸ ms^–1.

Ans: Speed of signal = 3 × 10⁸ ms^–1

Time in which signal reaches ground = 5 min = 5 × 60 = 300 s

Distance of spaceship from the ground level = speed × time = 3 × 10⁸ × 300 = 9 × 10¹⁰ mPage No. 77

Q1. When will you say a  body is in

(i) Uniform acceleration (ii) Non-uniform acceleration?

Ans:

(i) If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the body is said to be in uniform acceleration. 

Example: The motion of a freely falling body.

(ii) If an object travels in a straight line and its velocity changes by unequal amounts in equal intervals of time, then the body is said to be in non-uniform acceleration. 

Example: If a car is travelling along a straight road and passes through a crowd, suffers an unequal change in velocity, in equal intervals of time.

Q2. A bus decreases its speed from 80 km h⁻¹ to 60 km h⁻¹ in 5 s. Find the acceleration of the bus.

Ans:

• Initial velocity, u=80u=80km/h
• Final velocity, v=60v=60km/h
• Time, t=5t=5s

Solution:

1. Convert velocities from km/h to m/s:

Use the first equation of motion: v=u+atv=u+at

• 16.67=22.22+a⋅5
• a⋅5=16.67−22.
• a⋅5=−5.55
• 

The acceleration of the bus is −1.11m/s²  (negative sign indicates deceleration or retardation).

Q3. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h^-1 in 10 minutes. Find its acceleration.

Ans:  Given, 

Initial velocity, u = 0 km /h

 Final velocity, v = 40 km/h   =40 × ( 5/ 18) = 11.11 m/s

Time, t = 10 min = 10 × 60 = 600 sec

Acceleration, a = ?

Consider the formula, v = u + at

  ⇒ 11.11 = 0 + a × 600

⇒ 11.11 = 600 a

⇒ a = 11.11/600 = 0.0185 m/s²

Page No. 81

Q1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Ans:

• When the motion is uniform, the distance-time graph is a straight line with a slope.
• When the motion is non-uniform, the distance-time graph is not a straight line. It can be any curve.

Q2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Ans: If the distance-time graph is a straight line parallel to the time axis, the body is at rest.

Q3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Ans: If the speed-time graph is a straight line parallel to the time axis, the object is moving uniformly.

Q4. What is the quantity which is measured by the area occupied below the velocity-time graph? 

Ans: The area beneath the velocity-time graph corresponds to the area of the rectangle OABC, calculated as OA multiplied by OC. Here, OA represents the object’s velocity, and OC indicates time. Thus, the shaded area can be expressed as:

The area under the velocity-time graph = velocity × time.

By substituting the value of velocity as displacement divided by time into this equation, we find that the area under the velocity-time graph represents the total displacement of the object.

Page No. 82

Q1. A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes. Find

(a) the speed acquired, (b) the distance travelled.

Ans: Given, 

 Initial velocity, u = 0 ms^-1

 Acceleration, a = 0.1 ms^-2

Time, t = 2 min = 120 s

(a) Speed, v = u + at = 0 + 0.1 x 120 = 12 ms^-1

(b) Distance, s = 720 m

The speed acquired is 12 ms^-1 and the total distance travelled is 720 m.

Q2. A train is travelling at a speed of 90 km h^-1. Brakes are applied so as to produce a uniform acceleration of 0.5 ms^–2. Find how far the train will go before it is brought to rest.

Ans: Given the initial speed of the train, u= 90 km/h = 25 m/s

Final speed of the train, v = 0 m/s (finally the train comes to rest)

Acceleration = – 0.5 m s^-2

According to the third equation of motion:

v² = u² + 2as (where s is the distance covered by the train)

⇒ (0)² = (25)² + 2 (- 0.5) s

⇒ 

The train will cover a distance of 625 m at an acceleration of -0.5ms^-2 before it comes to rest.

Q3. A trolley, while going down an inclined plane, has an acceleration of 2 cm/s^–2. What will be its velocity 3 s after the start?

Ans: Initial Velocity of trolley, u = 0 cms^-1; Acceleration, a = 2 cms^-2; Time, t = 3 s

We know that final velocity, v = u + at = 0 + 2 x 3 cms^-1

Therefore, The velocity of the train after 3 seconds = 6 cms^-1

Page No. 83

Q4. A racing car has a uniform acceleration of 4 ms^-2. What distance will it cover in 10 s after the start?

Ans: Initial Velocity of the car, u = 0 ms^-1; Acceleration, a = 4 m s^-2; Time, t = 10 s

We know Distance, s = ut + (1/2) at²

Therefore, Distance covered by car in 10 second = 0 × 10 + (1/2) × 4 × 10² = (1/2) × 400 = 200 m

Q5. A stone is thrown in a vertically upward direction with a velocity of 5 m/s^-1. If the acceleration of the stone during its motion is 10 m/s^–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Ans: Given,

 The initial velocity of stone, u = 5 ms^-1

Downward or negative acceleration, a = 10 ms^-2

We know that: 2as = v²– u²

⇒ 0 = (5)² + 2 x(-10) x s

⇒ 0 = 25 – 20s

⇒ s = 25/20 = 1.25 m

The height attained by stone, s = 1.25 m

We know that: v = u + at

⇒ 0 = 5 + (–10) × t

⇒ 0 = 5 − 10t

⇒ t = 5/10 = 0.5 s

Thus, the stone will attain a height of 1.25 m and the time taken to attain the height is 0.5 s.

Page No. 85

Q1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Ans: Here, the diameter of the circular track = 200 m.

The radius of the circular track, r = 100 m.

Let the athlete start moving from A, which is treated as a reference point.

Given:

• Diameter of the circular track = 200 m
• Radius, r = 100 m
• Time for one round = 40 s
• Total time = 2 minutes 20 seconds = 2 × 60 + 20 = 140 s

Circumference of the track = 2πr = 2 × (22/7) × 100 = 4400/7 ≈ 628.57 m

Speed = Circumference / Time = 628.57 / 40 ≈ 15.71 m/s

Distance covered in 140 s = 15.71 × 140 ≈ 2200 m

Number of rounds = Distance / Circumference = 2200 / 628.57 ≈ 3.5

After 3 complete rounds, the athlete is back at the starting point (e.g., point A). After 0.5 round more, the athlete is at the opposite point on the circular track (e.g., point B, 180° from A). Since the track has a diameter of 200 m, the displacement is the straight-line distance from A to B, which is the diameter:

Displacement = 200 m

Distance covered = 2200 m, Displacement = 200 m

Q2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in Jogging (a) from A to B and (b) from A to C? 

Ans: Let Joseph jog from A to B and back to C as shown.

Given : 

• Distance from A to B = 300 m
• Time from A to B = 2 minutes 50 seconds = 2×60+50=1702×60+50=170s
• Distance from B to C = 100 m
• Time from B to C = 1 minute = 6060s
• Total distance from A to C = 300 m + 100 m = 400 m
• Total time from A to C = 170 s + 60 s = 230 s
• Displacement from A to B = 300 m (in the direction of motion)
• Displacement from A to C = 300 m – 100 m = 200 m (since he jogs back 100 m)

Solution:

(a) From A to B:

(b) From A to C:

Q3. Abdul while driving to school computes the average speed for his trip to be 20 km h^–1. On his return trip along the same route, there is less traffic and the average speed is 40 km h^–1. What is the average speed for Abdul’s trip? Ans: Let the distance between Abdul’s home and school be d km.Speed while going to school = 20 km/hSpeed while returning home = 40 km/hTotal distance covered = d + d = 2d kmTime taken while going = d ÷ 20 hoursTime taken while returning = d ÷ 40 hoursTotal time taken = d ÷ 20 + d ÷ 40= (2d + d) ÷ 40= 3d ÷ 40 hoursNow,Average speed = Total distance ÷ Total time= 2d ÷ (3d ÷ 40)= (2d × 40) ÷ 3d= 80 ÷ 3 km/hFinal Answer:Average speed = 80⁄3 km/h or approximately 26.67 km/hQ4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^–2 for 8.0 s. How far does the boat travel during this time?Ans: Given,  initial velocity of the boat = 0 m/s, Acceleration = 3 m/s² Time period = 8 seconds From the second equation of motion, s =ut+12at² Thus, the total distance travelled by the boat in 8 seconds = 0 + 1/2 (3) (8)² = 96 metersTherefore, the motorboat covers a distance of 96 meters in 8 seconds. Q5. A driver of a car travelling at 52 km h^–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3 km h^–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? Ans:The total displacement of each car can be determined by calculating the area under the speed-time graph.For the first car, the displacement is given by the area of triangle AOB:=12×OB×OADisplacement of the first car= 1/2 ×OB×OAHere, OB=5 seconds and OA=52 km/h, which converts to 14.44m/s. Therefore, the area of triangle AOB is:12×(5)×(14.44)=36 metersFor the second car, the displacement is represented by the area of triangle COD:=12×OD×OCHere, OD=10 seconds and OC=3km/h, which converts to 0.83 m/s. Thus, the area of triangle COD is12×(10)×(0.83)=4.15 metersIn conclusion, the first car is displaced by 36 meters, while the second car is displaced by 4.15 meters. Therefore, the first car, travelling at 52 km/h, moved farther after applying the brakes.Q6. Figure below shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions :(a) Which of the three is travelling the fastest?(b) Are all three ever at the same point on the road?(c) How far has C travelled when B passes A?(d) How far has B travelled by the time it passes C?Ans: (a) since the slope of line B is the greatest, B is travelling at the fastest speed.(b) since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.(c) since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.Since the initial point of the object, C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 kmWhen A passes B, the distance between the origin and C is 8kmTherefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km(d) the distance that object B has covered at the point where it passes C is equal to 9 graph units.Therefore, the total distance travelled by B when it crosses C = 9*(4/7) = 5.14 kmPage No. 86Q7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^–2, with what velocity will it strike the ground? After what time will it strike the ground?Ans: Initial velocity of the ball, u = 0 (as it is dropped)Height, h = 20 m, Acceleration, a = 10 ms^-2,s = 20m a = 10m/s² t = ? s = ut + 1/2at² 20 = 0 × t + 1/2 × 10 × t² 20 = 0 + 5t² t² = 4 t = 2s v = u + at = 0 + 10 × 2= 20m/s OR v² = u² + 2asv² = 0 + 2 × 10 × 20 v² = 400 v = 20m/s The ball will strike the ground after 2 sec with the velocity of 20m/s.v = u + at20 = 0 + 10t∴ t = 20/10or time, t = 2 sTherefore, the ball reaches the ground after 2 seconds.Q8. The speed-time graph for a car is shown in figure below.(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.(b) Which part of the graph represents uniform motion of the car?Ans: (a)In the velocity-time graph,Distance = Area of the v-t graphHere we approximate the area using the area of the triangle,Distance travelled = Area of triangle AOBThe shaded area, which is equal to 1/2 × 4 × 6 = 12 m represents the distance travelled by the car in the first 4 s.(b)The part of the graph in red colour between times 6 s to 10 s represents the uniform motion of the car.Q9. State which of the following situations are possible and give an example for each of these.(a) An object with a constant acceleration but with zero velocity.(b) An object moving in a certain direction with acceleration in the perpendicular direction.(c) an object moving with acceleration but with uniform speed.Ans: (a) When an object is thrown upwards, it comes to a momentary rest at the highest point. Thus velocity is zero, but the acceleration due to the gravitational pull of the earth still acts upon it.(b) In a uniform circular motion, the speed remains constant, but there is varying velocity as it changes its direction, so there always be acceleration which is given by centripetal force.(c) When an object is thrown in the forward direction, then during its motion in the horizontal direction, the acceleration due to the gravity of the earth acts in the vertically downward direction.Q10. An artificial satellite is moving in a circular orbit with a radius of 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.Ans: Satellite completes one round in 24 hoursThe radius of the orbit r=42250 kmThe circumference C of the orbit is given by:C=2πrSubstituting the value of r:C= 2×3.1416×42250 km C≈ 265,571.6 km Speed v = C/TV= 265,571.6 km/24 hoursV ≈ 11, 065.48 km/h

Page No. 61

Q1. What is a tissue?
Ans: A group of cells that are similar in structure and/or work together to achieve a particular function forms a tissue.

Q2. What is the utility of tissues in multi-cellular organisms?
Ans:

• Tissues are made up of a group of cells carrying a specialised function. Each specialised function is taken up by a different tissue. Since these cells of a tissue carry out only a particular function, they do it very efficiently. 
• The use of tissues in multicellular organisms is to provide structural and mechanical strength. 
• Example: In human beings, muscle cells contract or relax to cause movement, nerve cells carry messages, and blood flows to transport gases, food, hormones, waste materials and so on. Likewise, in plants, vascular tissues (xylem, phloem) conduct water and food from one part of the plant to other parts. 
• So, multicellular organisms show a division of labour through tissues.
  

Page No. 65

Q1. Name types of simple tissues.
Ans: Simple permanent tissues are of three types:
(i) Parenchyma
(ii) Collenchyma
(iii) Sclerenchyma
Parenchyma tissue is of further two types:
(i) Aerenchyma
(ii) Chlorenchyma

Q2. Where is apical meristem found?
Ans: Apical meristem is present at the growing tips of stems and roots.

Q3. Which tissue makes up the husk of coconut?
Ans: Sclerenchyma tissue makes up the husk of coconut. These tissues cause the plant to become stiff and hard. The cells of this tissue are dead and their cell walls are thickened because of the presence of lignin.

Q4. What are the constituents of phloem?
Ans: The constituents of phloem are:

• Sieve tubes
• Companion cells
• Phloem parenchyma
• Phloem fibres

Page No. 69

Q1. Name the tissue responsible for movement in our body.
Ans: Two tissues jointly are responsible for the movement of our body, namely:

• Muscular tissue
• Nervous tissue
  

Q2. What does a neuron look like?
Ans: A neuron consists of a cell body with a nucleus and cytoplasm, from which long thin hair-like parts called dendrites arise. Each neuron has a single long part called the axon.

Q3. Give three features of cardiac muscles.
Ans:
Three features of cardiac muscles are:
(i) Cardiac muscles are involuntary.
(ii) Cardiac muscle cells are cylindrical, branched and uninucleate.
(iii) Cardiac muscles show rhythmic contraction and relaxation.

Q4. What are the functions of areolar tissue?
Ans: Areolar tissue acts as a supportive and packing tissue between organs lying in the body cavity, and also helps in the repair of tissues.

Page No. 70

Q1. Define the term “tissue”.
Ans: A group of cells that are similar in structure and work together to achieve a particular function is called tissue.

Q2. How many types of elements together make up the xylem tissue? Name them.
Ans: Xylem is composed of the following elements:

• Tracheids
• Vessels
• Xylem parenchyma
• Xylem fibres

Q3. How are simple tissues different from complex tissues in plants?
Ans:

Q4. Differentiate between parenchyma, collenchyma and sclerenchyma, on the basis of their cell wall.
Ans: The differences between the cell walls of parenchyma, collenchyma and sclerenchyma are:

Q5. What are the functions of the stomata?
Ans: The functions of stomata are:
(i) Stomata allow gaseous exchange between the plant and the atmosphere. 
(ii) These are sites of transpiration in plants.

Q6. Diagrammatically show the difference between the three types of muscle fibres.
Ans: The three types of muscle fibres are: 
(i) Striated muscles(ii) Smooth muscles (unstriated muscle fibre)(iii) Cardiac muscles

Q7. What is the specific function of the cardiac muscle?
Ans: The cardiac muscles are branched and cylindrical. They are uninucleated and are involuntary in nature. Throughout one’s lifetime, the cardiac muscles bring about the rhythmic contraction and relaxation of the heart.

Q8. Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.
Ans:

Q9. Draw a labelled diagram of a neuron.
Ans:
Q10. Name the following:
(a) Tissue that forms the inner lining of our mouth.
(b) Tissue that connects muscle to bone in humans.
(c) Tissue that transports food in plants.
(d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix.
(f) Tissue present in the brain.
Ans: 
(a) Simple squamous epithelium 
(b) Tendon 
(c) Phloem 
(d) Adipose tissue 
(e) Blood 
(f) Nervous tissue

Q11. Identify the type of tissue in the following: 
Skin, Bark of Tree, Bone, Lining of Kidney Tubule, Vascular Bundle.
Ans:
Skin: Epithelial tissue (Squamous epithelium) 
Bark of Tree: Cork (protective tissue) 
Bone: Skeletal tissue (connective tissue)
Lining of Kidney Tubules: Cuboidal epithelial tissue 
Vascular Bundle: Complex permanent tissue—xylem and phloem

Page No. 71

Q12. Name the regions in which parenchyma tissue is present.
Ans: Parenchyma tissues are found in:

• The pith of stems and roots.
• When parenchyma contains chlorophyll it is called as chlorenchyma, it is found in green leaves.
• Parenchyma found in aquatic plants have large air cavities that enable them to float and are hence called aerenchyma.

Q13. What is the role of the epidermis in plants?
Ans: Role of the epidermis in plants:
(i) It acts as a protective tissue, covering the plant body. 
(ii) It protects the plant from excessive heat or cold and from the attack of parasitic fungi and bacteria. 
(iii) It allows the exchange of gases and transpiration through stomata. 
(iv) The cuticle of the epidermis checks the excessive evaporation of water.

Q14. How does the cork act as a protective tissue?
Ans: The cork cells are dead and do not have any intercellular spaces. The cell walls of the cork cells are coated with suberin (a waxy substance). Suberin makes these cells impermeable to water and gases, or in simpler words, it makes them waterproof and blocks gases from passing through. Thus, it protects underlying tissues from desiccation (loss of water from the plant body), infection and physical damage. 

Q15. Complete the table:

Ans:

Page No. 51

Q1. Who discovered cells and how?
Ans:

• The cell was first discovered by Robert Hooke in 1665. He examined thin slices of cork under a self-made microscope and saw a multitude of tiny hollow spaces that he remarked looked like the walled compartments of a honeycomb. 
• He termed these spaces as ‘cell’ meaning ‘small room’ in Latin.
  Microscope and Cork cells

Q2. Why is the cell called the structural and functional unit of life?
Ans:

• All living organisms are made up of cells. This shows that the cell is the structural unit of life. 
• Each living cell can perform certain basic functions that are characteristic of all living forms. 
  Example: 
  (i) Phagocytic cells eat or kill unwanted or foreign particles inside the body (e.g., WBCs). 
  (ii) Some cells secrete enzymes and hormones, e.g., pancreatic cells, small intestinal cells, and liver cells.

Page No. 53

Q1. How do substances like CO₂ and water move in and out of the cell? Discuss.
Ans: CO₂ moves by diffusion – This cellular waste accumulates in high concentrations in the cell, whereas the concentration of CO₂ in the external surroundings is comparatively lower. This difference in the concentration level inside and outside of the cell causes the CO₂ to diffuse from a higher (within the cell) region to a lower concentration.
H₂O diffuses by osmosis through the cell membrane. It moves from a region of higher concentration to a lower concentrated region through a selectively permeable membrane until equilibrium is reached.

Q2. Why is the plasma membrane called a selectively permeable membrane?
Ans: The plasma membrane allows or permits the entry and exit of some materials in and out of the cell and prevents the movement of some other materials through it. Hence, it is called a selectively permeable membrane.Page No. 55

Q1. Fill in the gaps in the following table illustrating the differences between prokaryotic and eukaryotic cells.

Ans:Page No. 57

Q1. Can you name the two organelles we have studied that contain their genetic material?
Ans: Two organelles that contain their genetic material are Mitochondria and Plastids. Mitochondria help in respiration in the cell, while plastids are responsible for the process of photosynthesis in leaves. 

Q2. If the organization of a cell is destroyed due to some physical or chemical influence, what will happen?
Ans:

• Cell organelles are responsible for the organization and proper functioning of a cell, as each of them performs some specific functions. 
• Naturally, if any of these organelles are destroyed, the cell will not be able to perform many basic functions like photosynthesis, respiration, nutrition, etc., and may also result in the stopping of all life activities in the cell. Due to the cell damage, the lysosome bursts, and their enzymes digest such cells.

Q3. Why are lysosomes known as suicide bags?
Ans:

• Lysosomes are the cell organelles involved in the digestion of any foreign material that enters the cell, as they contain digestive enzymes. 
• In case any cell is dead or damaged, the lysosome bursts to release the digestive enzymes to digest its cell. 
• Thus, these are known as ‘suicide bags’.

Q4. Where are proteins synthesized inside the cell?
Ans:

• Proteins are the building blocks of the body that are made up of various amino acids. They are synthesized inside the ribosomes, are the small structures present in the cytoplasm, or might be attached to the surface of the endoplasmic reticulum. The endoplasmic reticulum is called the rough endoplasmic reticulum due to the presence of ribosomes on its surface. 
• In simpler words, proteins are the body’s building blocks, made of amino acids. They are created inside ribosomes, which are small structures found in the cytoplasm or attached to the endoplasmic reticulum. When ribosomes are attached to the endoplasmic reticulum, it’s called the rough endoplasmic reticulum because the ribosomes give it a rough surface. 

Page No. 59

Q1. Make a comparison and write down ways in which plant cells are different from animal cells.
Ans:

Q2. How is a prokaryotic cell different from a eukaryotic cell?
Ans:

Q3. What would happen if the plasma membrane ruptures or breaks down?
Ans:

• The plasma membrane is the selectively permeable membrane that surrounds the cell and allows the entry and exit of selected materials of the cell. 
• If it ruptures, the contents of the cell will come in direct contact with the surrounding medium, and not only unwanted material will be able to enter freely into the cell, but useful material will also find its way out of the cell easily. 
• This will seriously disrupt the various metabolic activities of the cell and will result in its imminent death.

Q4. What would happen to the life of a cell if there was no Golgi apparatus?
Ans:

• If there were no Golgi apparatus, the material synthesized by the endoplasmic reticulum would not be carried to the various parts inside and outside of the cell.
  
• Also, as the Golgi apparatus performs the function of storage and modification of the material synthesized in the cell, these materials would not be stored and modified further.
• Moreover, there will be no production of lysosomes, which will cause the accumulation of waste material, viz., worn out and dead cell organelles within the cell, which will ultimately lead to cell death.

Q5. Which organelle is known as the powerhouse of the cell? Why
Ans:

• Mitochondria are known as the powerhouse of the cell because these are the sites of cellular respiration.
  Mitochondria
• The energy that is released by the mitochondria is in the form of ATP molecules and is required for various chemical activities needed for. 
• The energy stored in ATP is used by the body to make new chemical compounds and to perform mechanical work. 
• Thus, mitochondria are known as the powerhouse of cells.

Q6. Where do the lipids and proteins constituting the cell membrane get synthesized?
Ans:

• Lipids and proteins are the essential parts of the plasma membrane, which are synthesized through the endoplasmic reticulum. 
• The endoplasmic reticulum is found to be of two types based on the substances they synthesize. 
• The smooth endoplasmic reticulum is responsible for the synthesis of lipids, while the rough endoplasmic reticulum is responsible for the synthesis of proteins. 

Q7. How does an Amoeba obtain its food?
Ans:

• Amoeba takes in food using temporary finger-like extensions of the cell surface, which fuse over the food particle, forming a food vacuole as shown in the figure. 
• Inside the food vacuole, complex substances are broken down into simpler ones, which then diffuse into the cytoplasm. 
• The remaining undigested material is moved to the surface of the cell and thrown out.
  Nutrition in Amoeba

Q8. What is osmosis?
Ans: It is the passage of solvent from a region of high concentration to a region of low concentration through a semipermeable membrane.

Q9. Carry out the following osmosis experiment
Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now,
(a) Keep cup A empty.
(b) Put one teaspoon of sugar in cup B
(c) Put one teaspoon salt in cup C
(d) Put one teaspoon of sugar in the boiled potato cup D.
Keep these for two hours. Then observe the four potato cups and answer the following
(i) Explain why water gathers in the hollowed portion of B and C.
(ii) Why is potato A necessary for this experiment?
(iii) Explain why water does not gather in the hollowed-out portions of A and D.
Ans: 
(i) 

• Water gathers in the hollowed portion of cups B and C because of the process of endosmosis (moving in of the solvent). The potato wall acts as a semi-permeable membrane. 
• As cups B and C are filled with sugar and salt, respectively, and their outer part is in contact with the water, the concentration of water outside the cups is higher than inside the cups. 
• So, water moves from its higher concentration towards the lower concentration, i.e., inside the cup. 

(ii) Potato A is necessary for this experiment because:

• Potato A acts as a control of the experiment. It is very necessary to compare the results of the experiment. 
• It shows that if the concentration of water is the same on both sides, there will be no movement of water. 

(iii) Water does not gather in the hollowed-out portions of A and D because:

• Water does not gather in the hollowed-out portions of A as it does not contain a hypertonic solution, so there is no concentration difference and hence no movement of solvent. 
• Water does not gather in cup D as the cells of the boiled potato are dead, and the potato wall is no longer semi-permeable. Hence, no osmosis occurs.

Q10. Which type of cell division is required for the growth and repair of the body and which type is involved in the formation of gametes?
Ans:

• For growth and repair, mitotic division (mitosis) is involved as this type of division keeps the chromosome number constant.
• For gamete formation, meiosis is involved as a reduction of chromosome number is necessary for this case.

Page No. 39

Q1. What are canal rays?
Ans: Canal rays are a type of positively charged radiation discovered by E. Goldstein in 1886. They are produced in a gas discharge and are significant for the following reasons:

• They helped in the identification of protons, a key subatomic particle.
• Canal rays consist of positive ions that move towards the cathode in a discharge tube.
• Their discovery contributed to the understanding of atomic structure.

Goldstein’s experimental setup
Q2. If an atom contains one electron and one proton, will it carry any charge or not?
Ans: No, it will not carry any charge because the number of protons (positively charged) is equal to the number of electrons (negatively charged).

Page No. 41
Q1. On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Ans: According to Thomson’s model, an atom is made up of a sphere of positive charge with electrons embedded within it. This arrangement is designed to create a stable electrostatic balance.

• The atom is neutral because it contains an equal number of electrons and positively charged particles.
• The positive charge of the sphere balances the negative charge of the electrons.
• Thus, the overall charge of the atom is zero, making it electrically neutral.

Thomson’s Model of an atom

Q2. On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nucleus of an atom?
Ans: According to Rutherford’s model of the atom:

• Protons are present in the nucleus, giving it a positive charge.
• The nucleus contains most of the atom’s mass.
• Neutrons are also found in the nucleus, contributing to the overall mass.

Rutherford’s Model of an atom
Q3. Draw a sketch of Bohr’s model of an atom with three shells.
Ans: 

A line diagram showing different energy levels of energies E₁, E₂ and E₃ for the electrons

Q4. What do you think would be the observation if the α-particle scattering experiment is carried out using a foil of a metal other than gold?
Ans: If the metal taken is heavier than gold, then the deviation shown by α-particles will be more, and if the metal taken is lighter than gold, then the deviation shown will be less due to less repulsion between positively α-particles and nucleus.

Q5. Name the three sub-atomic particles of an atom.
Ans: The three sub-atomic particles of an atom are:
(i) Protons
(ii) Electrons
(iii) Neutrons

Q6. Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Ans: Number of neutrons = Atomic mass – Number of protons 
For a helium atom:

• Atomic mass = 4 u
• Number of protons = 2
• Therefore, number of neutrons = 4 – 2 = 2

Page No. 42

Q1. Write the distribution of electrons in carbon and sodium atoms.
Ans: 

 (a) Electronic distribution in a carbon atom:

• Atomic number: 6
• Number of electrons: 6
• Distribution: K = 2, L = 4

(b) Distribution of electrons in a sodium atom:

• Atomic number: 11
• Number of electrons: 11
• Distribution: K = 2, L = 8, M = 1

Q2. If K and L shell of an atom are full, then what would be the total number of electrons in the atom?
Ans. The maximum number of electrons in an atom with full K and L shells is:

• K shell: 2 electrons
• L shell: 8 electrons

Therefore, the total number of electrons in the atom is:

• Total: 10 electrons

Page No. 44

Q1. How will you find the valency of chlorine, sulphur and magnesium?
Ans: To find the valency of chlorine, sulphur, and magnesium:

Chlorine (Cl):

• Atomic number: 17
• Electrons: 2, 8, 7
• Can gain 1 electron to achieve stability (like argon).
• Valency: 1

Sulphur (S):

• Atomic number: 16
• Electrons: 2, 8, 6
• Can gain 2 electrons to achieve stability (like argon).
• Valency: 2

Magnesium (Mg):

• Atomic number: 12
• Electrons: 2, 8, 2
• Can lose 2 electrons to achieve stability (like neon).
• Valency: 2

Q2. If number of electrons in an atom is 8 and number of protons is also 8, then 
(i) what is the atomic number of the atom and 
(ii) what is the charge on the atom?
Ans:
(i) The atomic number is equal to the number of protons in an atom. In this case:
Atomic number = Number of protons = 8
(ii)  The charge of an atom is determined by the balance between protons and electrons.

• Since the number of electrons (8) is equal to the number of protons (8), the atom has no overall charge.
• Thus, the charge on the atom is zero.

Q3. With the help of Table 4.1, find out the mass number of oxygen and sulphur atom.

Ans: 
(a) To find the mass number of Oxygen:
Number of protons = 8
Number of neutrons = 8
Atomic number = 8
Atomic mass number = Number of protons + number of neutrons = 8 + 8 = 16
Therefore, mass number of oxygen = 16
(b) To find the mass number of Sulphur:
Number of protons = 16
Number of neutrons = 16
Atomic number = 16
Atomic mass number = Number of protons + number of neutrons = 16 + 16 = 32

Page No. 45

Q1. For the symbol H, D and T tabulate three sub-atomic particles found in each of them.
Ans: 

Hydrogen (H): The most common isotope, consisting of a single proton and no neutrons.

Deuterium (D): A heavier isotope of hydrogen, with one neutron in addition to the proton.

Tritium (T): A radioactive isotope of hydrogen, containing two neutrons and one proton.

Q2. Write the electronic configuration of any one pair of isotopes and isobars.
Ans: 

Isotopes : Isotopes have the same electronic configuration because they have the same number of electrons and protons, differing only in the number of neutrons.
Example : ¹²C₆ and ¹⁴C₆ are isotopes, have the same electronic configuration as (2, 4)

Electronic Configuration(2,4) of Carbon atom

Isobars: Isobars have the same mass number but differ in atomic number, leading to different electronic configurations.

Example :  ²²Ne₁₀ and ²²Na₁₁ are isobars. They have different atomic number but mass number is same. 

Page No. 46, 47 & 48

Q1. Compare the properties of electrons, protons and neutrons.

Ans: ElectronProtonNeutron(a) It is negatively charged.(a) It is positively charged.(a) It is neutral.(b) Its absolute mass is equal to 9.1 x 10^-31 kg. (b) Its absolute mass is equal to 1.673 x 10^-27 kg. (b) Its mass is slightly more than that of protons. Its absolute mass is 1.675 x 10^-27 kg.

Q2. What are the limitations of J.J. Thomson’s model of the atom?
Ans: Limitations of JJ Thomson’s model of an atom:
(i) The results of experiments carried out by other scientists could not be explained by this model
(ii) The positive charge is spread all over could not be justified.

Q3. What are the limitations of Rutherford’s model of the atom?
Ans: Limitations of Rutherford’s model of an atom
(i) According to classical electromagnetic theory, a moving charged particle, such as an electron, under the influence of attractive force, loses energy continuously in the form of radiations. As a result of this, the electron should lose energy and, therefore, should move in even smaller orbits, ultimately falling into the nucleus. But the collapse does not occur. There is no explanation for this behaviour.
(ii) According to Rutherford, the electrons revolve around the nucleus in fixed orbits. However, Rutherford did not specify the number of orbits and the number of electrons in each orbit.

Q4. Describe Bohr’s model of the atom.
Ans: Bohr made a bold new suggestion that particles at the atomic level would behave differently from macroscopic (bigger) objects.
According to Bohr’s theory:

• Electrons travel in specific paths called orbits or energy levels.
• While in these orbits, electrons do not emit energy.
• Electrons can move to a higher energy level when they absorb energy.
• When electrons drop back to a lower energy level, they release energy.

This model helped explain the stability of atoms and their emission spectra.

Bohr’s Model of an atom

Q5. Compare all the proposed models of an atom given in this chapter.
Ans: Thomson modelRutherford modelBohr model(a) The positive charge is spread all over and electrons are embedded in it like seeds in the watermelon. (a) There is a positively charged centre called the nucleus where the mass of the atom is concentrated.(a) Electrons revolve in certain discrete orbits called energy levels.(b ) The positive and negative charges are equal, therefore the atom is neutral.(b) Electrons revolve around the nucleus in well-defined orbits.(b) While revolving in these orbits, electrons do not radiate energy.(c) The nucleus is very small compared to the overall size of the atom.
(c) The size of the nucleus is very small as compared to the size of an atom.(c) These orbits are represented by K, L, M, N or 1, 2, 3, 4.

Q6. Summarize the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Ans: The distribution of electrons in the shells of an atom is known as electronic configuration. To write the electronic configuration for an element, follow these steps:

• Determine the total number of electrons using the element’s atomic number.
• Distribute these electrons across the shells in order of increasing energy, following the 2n² rule.
• Fill the shell with the lowest energy first, then proceed to the next higher energy shells.
• List the number of electrons in each shell, starting from the lowest, separated by commas. For example, sodium (atomic number 11) has:

Electronic configuration of Na: K = 2, L = 8, M = 1 

The maximum number of electrons in different shells is:

• K-shell (1st shell) = 2
• L-shell (2nd shell) = 8
• M-shell (3rd shell) = 18
• N-shell (4th shell) = 32

Key points to remember:

• The outermost shell can hold a maximum of 8 electrons.
• Electrons fill shells in a step-wise manner; inner shells must be filled before outer shells.

Q7. Define valency by taking examples of silicon and oxygen.
Ans: The valency of an atom is the number of hydrogen atoms which combine with one atom of an element.
For example, 

• Silicon (atomic number 14) has the following electronic distribution:
  K = 2, L = 8, M = 4.
  In the outermost shell, there are 4 electrons; hence, one atom of silicon will combine with 4 atoms of hydrogen. So, the valency of silicon is 4.
• In the case of oxygen (atomic number 8), the electronic distribution in various shells is given below:
  K = 2, L = 6
  There are six valence electrons in the atom of oxygen. So, one atom of oxygen will combine with 2 atoms of hydrogen to make a noble gas configuration. Hence, the valency of oxygen is 2.

Q8. Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.
Ans: 
(i) Atomic number: The atomic number of an atom is equal to the number of protons in the nucleus of an atom.
Atomic number = Number of protons.
(ii) Mass number: The total mass of the atom is due to the sum of the masses of protons and neutrons present in the nucleus. The mass number of an atom is equal to the sum of the number of protons and neutrons in the nucleus.
Mass number = Number of protons + Number of neutrons.
(iii) Isotopes: Isotopes are the atoms of the same element having the same atomic number but different mass numbers. For example, Hydrogen has three isotopes: protium (1), deuterium (2), and tritium (3).
(iv) Isobars: Isobars are the atoms of the different elements having the same mass number and different atomic numbers. For example, Calcium (atomic number 20) and argon (atomic number 18) both have a mass number of 40.

Uses of isotopes:
(i) Cobalt-60 is a radioactive isotope of cobalt. It is used in radiotherapy for cancer.
(ii) ¹⁴C is a radioactive isotope of carbon. It is used in carbon dating.

Q9. Na⁺  has completely filled K and L shells. Explain.
Ans: Sodium has the atomic number 11. It has 11 electrons in its orbitals, wherein the number of protons is equal to the number of electrons. Hence, its electronic configuration is K-2, L-8, and M-1. The one electron in the M shell is lost, and it obtains a positive charge since it has one more proton than electrons and obtains a positive charge, Na⁺. Na⁺ is formed when one electron is lost by a sodium atom, as given below:

The new electronic configuration is K-1; L-8 which is the filled state. Hence it is very difficult to eliminate the electron from a filled state as it is very stable.

Q10. If bromine atom is available in the form of, say, two isotopes  ⁷⁹Br₃₅ (49.7%) and ⁸¹Br₃₅ (50.3%) calculate the average atomic mass of bromine atom.
Ans: Average atomic mass of Br
= ( (79 x 49.7) + (81 x 50.3) ) / 100
= (3926.3 + 4074.3) / 100
= 8000.6 / 100
= 80.006 u

Q11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes  in the sample?
Ans: It is given that the average atomic mass of the sample of element X is 16.2 u.
Let the percentage of isotope 18 / 8 X be y%. Thus, the percentage of isotope 16 / 8 X will be (100 – y) %.
Therefore,
18 x (y/100) + 16 x ((100-y)/100) = 16.2
18y/100 + ( 16(100 – y) / 100 ) = 16.2
(18y+1600-16y )/100 = 16.2
18y + 1600 – 16y = 1620
2y + 1600 = 1620
2y = 1620 – 1600
y = 10
Therefore, the percentage of isotope 18 / 8 X is 10%.
And, the percentage of isotope 16 / 8 X is (100 – 10) % = 90%.

Q12. If Z = 3, what would be the valency of the element? Also, name the element.
Ans. Z = 3
Electronic configuration is 2, 1.
Valence electron is equal to 1.
Valency is also equal to 1.
The name of the element is lithium.

Q13. Composition of the nuclei of two atomic species X and Y are given as under

Give the mass numbers of X and Y. What is the relation between the two species?
Ans: Mass number of X = 6 + 6 = 12 
Mass number of Y = 6 + 8 = 14
Atomic number of X = 6 
Atomic number of Y = 6
X and Y are isotopes since both the elements have same atomic number but different mass number.

Q14. For the following statements, write T for ‘True’ and F for ‘False’.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1 / 2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Ans. (a) False (b) False (c) True (d) False

Q15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron
Ans: (a) Atomic nucleus
An atomic nucleus is the dense central core of an atom, composed of protons and neutrons held together by the strong nuclear force.

Q16. Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers
Ans: (c) different number of neutrons

Q17. Number of valence electrons in Cl ^–ion are:
(a) 16
(b) 8
(c) 17
(d) 18
Ans: (b) 8
Cl^– has 8 valence electrons. Cl^– has 17 + 1 = 18 electrons. Its electronic configuration is 2, 8, 8.

Q18. Which one of the following is a correct electronic configuration of sodium?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1
Ans: (d)  2 ,8 ,1 
2 ,8 ,1 is the correct electronic configuration of sodium.

Q19. Complete the following table.

Ans: 
Atomic number(Z) =Number of protons
Mass number = Number of neutrons + atomic number
=> Mass number(A) = Number of neutrons + number of neutrons

Page No. 27

Q1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate ethanoic acid → sodium ethanoate carbon dioxide water
Ans: 

Mass of reactants = 5.3 g + 6 g = 11.3 g
Mass of products = 2.2 g + 0.9 g + 8.2 g = 11.3 g
Mass of reactants = Mass of products
Therefore, the law of conservation of mass is proven.

Q2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3g of hydrogen gas?
Ans: Since hydrogen and oxygen combine in the ratio of 1:8 by mass, 3g of hydrogen gas will react completely with 24 g of oxygen gas.

Q3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Ans: Dalton’s postulate that “atoms can neither be created nor destroyed,” is a result of the law of conservation of mass.

Q4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Ans: Atoms combine in a fixed ratio to form compounds, which can explain the law of definite proportions.Page No. 30

Q1. Define atomic mass unit.
Ans: It is defined as equal to 1/12th of the mass of 1 atom of C-12. It is called unified mass denoted by ‘u’ these days.

Q2. Why is it not possible to see an atom with naked eyes?
Ans: The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom of an element does not exist independently.Page No. 34

Q1. Write down the formulae of 
(a) sodium oxide
(b) aluminium chloride
(c) sodium sulphide
(d) magnesium hydroxide
Ans:
(a) Formula of Sodium Oxide

(b) Formula of Aluminium Chloride

(c) Formula of Sodium Sulphide

(d) Formula of Magnesium Hydroxide

Q2. Write down the names of compounds represented by the following formulae:
(a) Al₂(SO₄)₃
(b) CaCl₂
(c) K₂SO₄
(d) KNO₃
(e) CaCO₃
Ans: 
(a) Aluminium sulphate 
(b) Calcium chloride 
(c) Potassium sulphate 
(d) Potassium nitrate 
(e) Calcium carbonate

Q3. What is meant by the term chemical formula?
Ans: The chemical formula of a compound is a symbolic representation of its composition.
Chemical Formula of Water

Q4. How many atoms are present in 
(a) H₂S molecule and
(b) PO₄^3- ion?
Ans: 
(i) H₂S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in totality.
(ii) PO₄^3- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in totality.Page No. 35

Q1. Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
Ans: 
Molecular mass of H₂ = 2 × Atomic mass of H
= 2 × 1
= 2 u
Molecular mass of O₂ = 2 × Atomic mass of O
= 2 × 16
= 32 u
Molecular mass of Cl₂ = 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u
Molecular mass of CO₂ = Atomic mass of C₂ × Atomic mass of O
= 12+ (2+16) = (12 + 32)u 
= 44 u
Molecular mass of CH₄= Atomic mass of C4 × Atomic mass of H
= 12+ (4 x 1)u = (12 + 4)u 
= 16 u
Molecular mass of C₂H₆ = 2× Atomic mass of C₆× Atomic mass of H
= (2 x 12 + 6 x 1)u = (24 + 6)u 
= 30 u
Molecular mass of C₂H₄ = 2 x Atomic mass of C₄ × Atomic mass of H
= (2 x 12 + 4 x 1)u = (24 + 4)u 
= 28 u
Molecular mass of NH₃ = Atomic mass of N₃ × Atomic mass of H
= (14 + 3 x 1)u = (14 + 3)u 
= 17 u
Molecular mass of CH₃OH = Atomic mass of C₃ × Atomic mass of H Atomic mass of O Atomic mass of H
= (12 + 3 x 1 + 16 + 1)u = (12 + 3 + 17)u 
= 32 u

Q2. Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Ans:
(i) Formula unit mass of ZnO 
= 65 + 16 = 81 u 
(ii) Formula unit mass of Na₂O
= 2 x 23 + 16 = 46 + 16 = 62 u
(iii) Formula unit mass of K₂CO₃ 
= 2 x 39 + 12 + 3 x 16
= 78 + 12 + 48 = 138 uPage No. 36

Q1. A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. 
Calculate the percentage composition of the compound by weight.
Ans: Percentage of boron = (mass of boron / mass of the compound) x 100
= (0.096g / 0.24g) x 100  
= 40%
Percentage of oxygen = 100 – percentage of boron
= 100 – 40 
= 60%

Q2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Ans:  When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
Given that
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
Find out
We need to find out the mass of carbon dioxide that will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen.
Solution
First, let us write the reaction taking place here.
C + O₂ → CO₂
As per the given condition, when 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
3g + 8g →11 g ( from the above reaction)
The total mass of reactants = mass of carbon + mass of oxygen
= 3g+8g
= 11g
The total mass of reactants = Total mass of products
Therefore, the law of conservation of mass is proved.
Then, it also depicts that carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.
Thus, it further proves the law of constant proportions.
3 g of carbon must also combine with 8 g of oxygen only.
This means that (50−8) = 42g of oxygen will remain unreacted.
The remaining 42 g of oxygen will be left un-reactive. In this case, too, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.

Q3. What are polyatomic ions? Give examples.
Ans: Polyatomic ions are ions that contain more than one atom, but they behave as a single unit.
Example: CO₃^2-, H₂PO₄^– 

Q4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Ans: The following are the chemical formula of the above-mentioned list:
(a) Magnesium chloride – MgCl₂
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO₃)₂
(d) Aluminium chloride – AlCl₃
(e) Calcium carbonate – CaCO₃

Q5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Ans:  The following are the names of the elements present in the following compounds:
(a) Quick lime – Calcium and oxygen (CaO)
(b) Hydrogen bromide – Hydrogen and bromine (HBr)
(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO₃)
(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K₂SO₄)

Q6. Calculate the molar mass of the following substances.
(a) Ethyne, C₂H₂ 
(b) Sulphur molecule, S₈ 
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus =31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃
Ans:  Listed below is the molar mass of the following substances:
(a) Molar mass of Ethyne C₂H₂= 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g
(b) Molar mass of Sulphur molecule S₈ = 8 x Mass of S = 8  x 32 = 256g
(c) Molar mass of Phosphorus molecule, P₄ = 4 x Mass of P = 4 x 31 = 124g
(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g
(e) Molar mass of Nitric acid, HNO₃ =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+
3×16 = 63g

Page No. 15

Q1. What is meant by a substance?
Ans: A pure substance is one that is made up of only one kind of particle, either atoms or molecules. It has definite composition and distinct properties.  Examples of Pure SubstancesA pure substance consists of only one type of particle, either atoms or molecules. It has a definite composition and distinct properties. Examples include:

• Oxygen
• Carbon

In contrast, a mixture contains two or more pure substances. For instance:

• Sea water is a mixture of salt and water.
• Soil contains various organic and inorganic materials.

Key points about mixtures:

• Mixtures can be separated into their components through physical processes.
• Each component retains its own properties.

Types of mixtures include:

• Homogeneous mixtures have a uniform composition (e.g., sugar in water).
• Heterogeneous mixtures have a non-uniform composition (e.g., sand and salt).

In summary, a pure substance has consistent properties, while a mixture contains multiple substances that can vary in composition.

Q2. List the points of differences between homogeneous and heterogeneous mixtures.

Ans:

Page No. 18

Q1. Differentiate between homogeneous and heterogeneous mixtures with examples.
Ans: The following are the differences between heterogeneous and homogenous mixtures.

Q2. How are sol, solution and suspension different from each other?
Ans:

Q3. To make a saturated solution, 36g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.

Ans: Mass of solute (NaCl): 36 g

Mass of solvent (H₂O): 100 g

Mass of solution: 136 g (NaCl + H₂O)

Concentration: Calculated as follows:

• Concentration = (Mass of solute / Mass of solution) × 100
• Concentration = (36 g / 136 g) × 100
• Concentration = 26.47%

Thus, the concentration of the solution is 26.47%.Page No. 19

Q1. Classify the following as chemical or physical changes: 

• cutting of trees, 
• melting of butter in a pan, 
• rusting of almirah, 
• boiling of water to form steam, 
• passing of electric current, through water and the water breaking down into hydrogen and oxygen gases, 
• dissolving common salt in water, 
• making a fruit salad with raw fruits, and
• burning of paper and wood.

Ans: The following is the classification into physical and chemical change:

Q2. Try segregating the things around you as pure substances or mixtures.
Ans: Listed below are the classifications based on pure substances and mixtures:
 Page No. 22

Exercises

Q1. Which separation techniques will you apply for the separation of the following? 
(a) Sodium chloride from its solution in water. 
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride. 
(c) Small pieces of metal in the engine oil of a car. 
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water. 
(g) Tea leaves from tea.
(h) Iron pins from sand. 
(i) Wheat grains from husk. 
(j) Fine mud particles suspended in water.
Ans: (a) In water, sodium chloride in its solution can be separated through the process of evaporation (as well as crystallization). 
(b) The sublimation technique is appropriate as ammonium chloride supports sublimation. 
(c) Tiny metal pieces in the engine oil of a car can be filtered manually.
(d) Chromatography can be used to separate different pigments from an extract of flower petals.
(e) The technique of churning can be applied to separate butter from curd. It is based on the concept of difference in density. 
(f) To separate oil from water, which are two immiscible liquids which vary in their densities, using a funnel can be an effective method. 
(g) Tea leaves can be manually separated from tea using simple filtration methods. 
(h) Iron pins can be separated from sand either manually or with the use of magnets as the pins exhibit strong magnetic quality, which can be a key characteristic taken into consideration. 
(i) The differentiating property between husk and wheat is that there is a difference in their mass. If treated with a small amount of wind energy, a remarkable variation in the moving distance is noticed. Hence, to separate them, the sedimentation/winnowing procedure can be applied. 
(j) Due to the property of water, sand or fine mud particles tend to sink in the bottom as it is denser, provided they are undisturbed. Through the process of sedimentation/decantation, water can be separated from fine mud particles, as the technique is established on obtaining clear water by tilting it out.Page No. 23

Q2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Ans:  Steps for Making Tea

• Heat a cup of milk, which acts as the solvent.
• Add tea powder or leaves, the solute, to the boiling milk.
• Observe that the tea powder remains insoluble while boiling.
• Add sugar to the boiling solution and stir.
• Sugar, being a solute, is soluble in the milk.
• Continue stirring until the sugar completely dissolves, achieving saturation.
• Once the raw smell of tea leaves disappears, remove the solution from heat.
• Filter the mixture to separate the tea powder, which becomes the residue.
• The liquid that passes through is the filtrate, containing the dissolved sugar and milk.

Q3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K? 
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain. 
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature? 
(d) What is the effect of change of temperature on the solubility of a salt?
Ans: (a) Given:
Mass of potassium nitrate required to produce a saturated solution in 100 g of water at 313 K = 62g
To find:
Mass of potassium nitrate required to produce a saturated solution in 50 g of water =?
Required amount = 62 x 50/100 = 31
Hence, 31 g of potassium nitrate is required.
(b) The solubility of potassium chloride in water is decreased when a saturated solution of potassium chloride loses heat at 353 K. Consequently, Pragya would observe crystals of potassium chloride, which would have surpassed its solubility at low temperatures.
(c) As per the given data, that is
Solubility of potassium nitrate at 293K = 32 g
Solubility of sodium chloride at 293K = 36 g
Solubility of potassium chloride at 293K = 35 g
Solubility of ammonium chloride at 293K = 37g
We can observe from this data that ammonium chloride has the highest solubility at 293K.
(d) Effect of change of temperature on the solubility of salts:
The table clearly depicts that the solubility of the salt is dependent upon the temperature and increases with an increase in temperature. With this, we can infer that when a salt arrives at its saturation point at a specific temperature, there is a propensity to dissolve more salt through an increase in the temperature of the solution.

Q4. Explain the following, giving examples. 
(a) Saturated solution 
(b) Pure substance 
(c) Colloid 
(d) Suspension
Ans: (a) Saturated solution: It is the state in a solution at a specific temperature when a solvent is no longer soluble without an increase in temperature. Example: Excess carbon leaves off as bubbles from a carbonated water solution saturated with carbon.
(b) Pure substance: A substance is said to be pure when it comprises only one kind of molecule, atom or compound without adulteration with any other substance or any divergence in the structural arrangement. Examples: Sulphur, diamonds etc.
(c) Colloid: A Colloid is an intermediate between solution and suspension. It has particles of various sizes that range between 2 to 1000 nanometers. Colloids can be distinguished from solutions using the Tyndall effect. Tyndall effect is defined as the scattering of light (light beam) through a colloidal solution. Examples: Milk and gelatin.
(d) Suspension: It is a heterogeneous mixture that comprises solute particles that are insoluble but are suspended in the medium. These particles that are suspended are not microscopic but visible to bare eyes and are large enough (usually larger than a micrometer) to undergo sedimentation.

Q5. Classify each of the following as a homogeneous or heterogeneous mixture. 
soda water, wood, air, soil, vinegar, filtered tea.
Ans: The following is the classification of the given substances into homogenous and heterogenous mixtures.

Q6. How would you confirm that a colorless liquid given to you is pure water?
Ans: We can confirm if a colorless liquid is pure by setting it to boil. If it boils at 100°C, it is said to be pure. But if there is a decrease or increase in the boiling point, we infer that water has added impurities and, hence not pure.Page No. 24

Q7. Which of the following materials fall in the category of a “pure substance”? 
(a) Ice 
(b) Milk
(c) Iron 
(d) Hydrochloric acid 
(e) Calcium oxide 
(f) Mercury 
(g) Brick 
(h) Wood 
(i) Air
Ans: The following substances from the above-mentioned list are pure substances:

• Iron
• Ice
• Hydrochloric acid
• Calcium oxide
• Mercury

Q8. Identify the solutions among the following mixtures.
(a) Soil
(b) Seawater
(c) Air
(d) Coal
(e) Soda water
Ans: The following are the solutions from the above-mentioned list of mixtures:

• Sea water
• Air
• Soda water

Q9. Which of the following will show the “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution
Ans:  Milk and starch solution demonstrate the Tyndall effect because they are colloidal solutions. In these solutions, light is scattered, making its path visible.

• The Tyndall effect occurs when light passes through a colloid.
• Colloidal solutions contain particles that are small enough to scatter light.
• Examples of colloids include milk and starch solutions.

Q10. Classify the following into elements, compounds, and mixtures. 
(a) Sodium 
(b) Soil 
(c) Sugar solution 
(d) Silver 
(e) Calcium carbonate 
(f) Tin 
(g) Silicon 
(h) Coal 
(i) Air 
(j) Soap 
(k) Methane 
(i) Carbon dioxide 
(m) Blood
Ans:
 

Q11. Which of the following are chemical changes? 
(a) Growth of a plant 
(b) Rusting of iron 
(c) Mixing of iron filings and sand 
(d) Cooking of food 
(e) Digestion of food 
(f) Freezing of water 
(g) Burning of a candle
Ans: Among the options listed, the following are considered chemical changes:

• Rusting of iron
• Cooking of food
• Digestion of food
• Burning of a candle

The growth of a plant is a complex process involving both chemical and physical changes, while mixing iron filings and sand and freezing water are not chemical changes.

Page No. 3

Q1. Which of the following are matter?
Chair, air, love, smell, hate, almonds, thought, cold, lemon water, the smell of perfume.
Ans: The following substances are matter: Chair, Air, Almonds, Lemon water, and the smell of perfume (The smell is caused by volatile substances which are matter, as they occupy space and have mass).

Matter around us

Q2. Give reasons for the following observation.
The smell of hot sizzling food reaches you several meters away, but to get the smell from cold food, you have to go close.
Ans: When the air is heated, the particles in it gain more kinetic energy and move faster. This is why the smell of hot food travels farther, allowing a person to sense it even from several meters away.

Q3. A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?
Ans: The particles of every matter have a force of attraction between them. This force keeps the particles together in a matter. In the case of water, the force of attraction between particles is less in comparison to solids. Thus, water molecules flow easily, giving way to a diver.

Showing less intermolecular force between liquid molecules.

Q4. What are the characteristics of the particles of matter?
Ans:  The characteristics of particles of matter are:
(i) Presence of intermolecular spaces between particles.
(ii) Particles are in constant motion.
(iii) They attract each other.Page No. 6

Q1. The mass per unit volume of a substance is called density. 
(density=mass/volume). Arrange the following in the order of increasing density – air, exhaust from the chimneys, honey, water, chalk, cotton and iron.
Ans: The following substances are arranged in increasing density: Air < Exhaust from chimney <  Cotton < Water <  Honey < Chalk < Iron.

Q2. (a) Tabulate the differences in the characteristics of states of matter.
(b) Comment upon the following: rigidity, compressibility, fluidity, filling a gas container, shape, kinetic energy and density.
Ans:  (a) The differences in the characteristics of the three states of matter solid, liquid and gas are:-

(b)
(i) Rigidity: It is the property of matter that continues to remain in its shape when treated with an external force.
(ii) Compressibility: Particles have the ability to reduce their intermolecular space when an external force is applied, which increases their density. This characteristic is called compressibility.
(iii) Fluidity: It is the ability of a substance to flow or move about freely.
(iv) Filling the gas container: The particles in a container take their shape as they randomly vibrate in all possible directions.
(v) Shape: It is the definite structure of an object within an external boundary.
(vi) Kinetic energy: Motion allows particles to possess energy, which is referred to as kinetic energy. The increasing order of kinetic energy possessed by various states of matter is Solids < Liquids < Gases.
Mathematically, it can be expressed as K.E = 1/2 mv², where ‘m’ is the mass and ‘v’ is the velocity of the particle.
(vii) Density: It is the mass of a unit volume of a substance. It is expressed as d = M/V, where ‘d’ is the density, ‘M’ is the mass and ‘V’ is the volume of the substance

Q3. Give reasons 
(a) A gas fills completely the vessel in which it is kept. 
(b) A gas exerts pressure on the walls of the container.
(c) A wooden table should be called a solid. 
(d) We can easily move our hand in the air, but to do the same through a solid block of wood, we need a karate expert.
Ans: 
(a) There is a low force of attraction between gas particles. The particles in the filled vessel are free to move about.
(b) Gaseous particles have the weakest attraction force and move randomly in all directions. When a gas particle hits the walls of its container, it applies a force, creating pressure on the walls.
(c) The hardwood table has a clear shape and volume. The wood particles are tightly packed and do not change to fit the shape of a container. This gives the table its solid properties.
(d) The air particles are spread far apart with a lot of space between them, which is why we can move our hands freely through the air. However, in a solid block, the particles are held tightly together by a strong force of attraction, leaving little or no space between them. That’s why breaking a solid block would require the strength of a karate expert.

Q4. Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why?
Ans: 

• The mass per unit volume of a substance is called density (density = mass/volume). 
• As the volume of a substance increases, its density decreases. In general, solids have higher density than liquids. Water is also a liquid, so it should also have less density than that of a solid, which is ice.
• Though ice is solid, it has a cage-like structure; hence, there are a large number of empty spaces between its particles. These spaces are larger than the spaces between the particles of water. Thus, for a given mass of water, the volume of ice is greater than that of water.
• Hence, the density of ice is less than that of water. A substance with a lower density than water can float on water. Therefore, ice floats on water.

Page No. 9

Q1. Convert the following temperature to Celsius scale:
(a) 300 K 
(b) 573 K
Ans: To convert a temperature from the Kelvin scale to the Celsius scale, you simply subtract 273 from the given Kelvin temperature.

(a)  (300 – 273)°C = 27°C
(b) (573 – 273)°C = 300°C

Q2. What is the physical state of water at:
(a) 250°C 
(b) 100°C
Ans: 
(a) At 250°C, the physical state of water is gas as the temperature is beyond its boiling point.
(b) At 100°C, it is in the transition state (both liquid and gaseous states) as the water is at its boiling point. Hence, it would be present in both liquid and gaseous state.

Q3. For any substance, why does the temperature remain constant during the change of state?
Ans: It is due to the latent heat as the heat supplied to increase the temperature of the substance is used up to transform the state of matter of the substance; hence, the temperature stays constant. Latent heat is the heat energy needed to change a substance from one form to another without changing its temperature. For example, when ice melts into water, it absorbs heat (latent heat of fusion) without increasing in temperature. Similarly, when water evaporates into steam, it also absorbs heat (latent heat of vaporization) without a temperature change.

Q4. Suggest a method to liquefy atmospheric gases.
Ans: To transform a gas into a liquid, it is necessary to bring its constituent particles or molecules closer. This can be achieved with atmospheric gases by either increasing the pressure or lowering the temperature. Page No. 10

Q1. Why does a desert cooler cool better on a hot dry day?
Ans: A desert cooler works better on hot, dry days because the high temperature and low humidity increase the rate of evaporation. This greater evaporation results in more effective cooling.

Q2. How does the water kept in an earthen pot (matka) become cool during summer?
Ans: An earthen pot has tiny pores that allow water to seep through and evaporate from its surface. This evaporation requires energy, which is drawn from the water inside the pot, making the water cooler.

Q3. Why does our palm feel cold when we put some acetone or petrol or perfume on it?
Ans: Acetone, petrol, and perfume are highly volatile substances. When applied to our palm, they evaporate quickly, absorbing heat from the skin and making our palm feel cold.

Q4. Why are we able to sip hot tea or milk faster from a saucer rather than a cup?
Ans: A saucer has a larger surface area than a cup, which allows for faster evaporation. This rapid evaporation cools the tea or milk more quickly, enabling us to sip it faster.

Q5. What type of clothes should we wear in summer?
Ans: In summer, it is advisable to wear light-colored cotton clothes. Light colors reflect heat, and cotton is breathable, allowing sweat to evaporate and providing a cooling effect on the skin.

Exercises 

Q1. Convert the following temperature to Celsius scale.
(a) 293K 
(b) 470K
Ans: To convert a temperature from the Kelvin scale to the Celsius scale, you simply subtract 273 from the given Kelvin temperature because 0°C=273K.
(a) 293K= (293 – 273)°C = 20°C
(b) 470K= (470 – 273)°C = 197°C

Q2. Convert the following temperatures to the kelvin scale.
(a) 25°C 
(b) 373°C
Ans: 
0°C = 273K
(a) 25°C = (25+273)K = 298K
(b) 373°C = (373+273)K = 646K

Q3. Give reason for the following observations:
(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.
Ans: 
(a) At room temperature, naphthalene balls undergo sublimation, which means they change directly from a solid to a gaseous state without undergoing the intermediate state, i.e., the liquid state.

Naphthalene Balls

(b) It is because perfumes contain a volatile organic solvent that can easily diffuse through air and, hence, carry the fragrance to people sitting several metres away.

Q4. Arrange the following substances in increasing order of forces of attraction between the particles— water, sugar, oxygen.
Ans: Oxygen (gas) < Water (liquid) < Sugar (solid)

Q5. What is the physical state of water at: 
(a) 25°C, (b) 0°C, (c) 100°C?
Ans: The physical state of water at different temperatures is as follows:
(a) At 25°C, water is in a liquid state (typical room temperature).
(b) At 0°C, water is at its freezing point, so both solid (ice) and liquid (water) phases can be observed.
(c) At 100°C, water is at its boiling point, resulting in the presence of both liquid water and gaseous water (water vapor).

Q6. Give two reasons to justify.
(a) Water at room temperature is a liquid.
(b) An iron almirah is a solid at room temperature.
Ans:

(a) Water remains in a liquid state at room temperature for two main reasons:

1. Its melting or freezing point is lower than room temperature, while its boiling point is higher (100°C).
2. Water has no fixed shape and flows to take the shape of its container, which indicates that it occupies a fixed volume but does not have a defined shape.

(b) An iron almirah is a solid at room temperature for the following reasons:

1. Both the melting and boiling points of iron are above room temperature, meaning it remains solid under these conditions.
2. An iron almirah is rigid and maintains a definite shape, and metals typically have a high density, further confirming that it is solid at room temperature.

Q7. Why is ice at 273 K more effective in cooling than water at the same temperature?
Ans: At 273 K, ice will absorb heat energy or latent heat from the medium during melting to transform into water. As a result, ice has a greater cooling impact than water at the same temperature since water does not absorb the excess heat from the medium.

Q8. What produces more severe burns, boiling water or steam?
Ans: Steam produces severe burns. It is because it is an exothermic reaction that releases a high amount of heat, which it consumes during vaporization.

Q9. Name A, B, C, D, E and F in the following diagram showing change in its state.

Ans: Interconversion of three states of matter: Using temperature or pressure, any state of matter can be turned into another.
(A) Solid to Liquid → Melting (or) fusion (or) liquefaction
(B) Liquid to Gas → Evaporation (or) vaporization
(C) Gas to liquid → Condensation
(D) Liquid to Solid → Solidification
(E) Solid to Gas → Sublimation
(F) Gas to Solid → Deposition