Solution

AUGUST 2025 REVISION EXAMINATION

Class 11 - Physics
SECTION A (20 X 1 =20)
1.
(d) F = mg
Explanation:
according to newton's second law of motion
F = ma
here acceleration is due to gravity and is equal to g.
So that for a freely falling body
F = mg

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2.

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(d) F1 + F2 + F3 = 0
Explanation:
F1 + F2 + F3 = 0
Tu
3.
(c) 2ma

g+a

Explanation:
am

e
ntr
air
Ce
iS

When the balloon descends down with acceleration a,

mg - U = ma
When the balloon moves up with acceleration a,
Sr

U - (m - m0)g = (m - m0)a
On adding the two equations,
⇒ mg - mg + m0g = ma + ma - m0a

2ma
⇒ m0 =
g+a

4.
(d) impulse
Explanation:
Impulse = Change in momentum

5.
(b) car will cover less distance before stopping
Explanation:
Being lighter than a truck, the car has less kinetic energy. On applying brakes with the same force, the car will cover less
distance before coming to rest.

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 6. (a) (Rg)1/2
Explanation:
−
−−
At the top of the swing, v c = √Rg

7.
(b) F = ma - Fp
Explanation:
A pseudo-force Fp appears on the body in the opposite direction of the accelerating non-inertial frame.

8.
mv
(d) 2
eastward and is due to the friction on the tyres exerted by the road
Explanation:
Here, mass of car = m
As it starts from rest, u = 0 final velocity along east v = v^i , t = 2sec
From v = u + at

n
^ v ^
v i = 0 + a⃗ × 2, a⃗ = i
2

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⃗
F = ma⃗ =
mv

2
^
i , force of car is mv

2
east ward
This is due to friction on the tyres exerted by the road.

9. (a) zero
Explanation:
Tu
Apparent weight = m(g - a)
= 50(9.8 - 9.8) = 0

10. (a) ∑ F ⃗ = 0
am

Explanation:
e
The vector sum of all the forces acting on a body is zero.
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11. (a) Mv
air

M−m

Explanation:
Mv = (M - m)v'
Ce

′ Mv
v =
iS

M−m

12.
(c) T1 > T2
Explanation:
Sr

2
mv
T = + mg cos θ
r

∴ T ∝ cos θ

T1 ∘ √3/2 √3

T2
=
cos 30

cos 60
∘
= =
1
or T1 > T2
1/2

g
13. (a) 3

Explanation:
M2 − M1 10−5 g
a= M2 + M1
g =
10+5
× g =
3

14. (a) the products must move in opposite directions

Explanation:
Let, M = mass of nucleus at rest. m1 and m2 are masses of two smaller nuclei. v1 and v2 are the velocities of respective masses
Now, According to the law of conservation of momentum,
Initial momentum before disintegration = final momentum after disintegration
m1v1 + m2v2 = 0
m2
v2 = − ⋅ v1
m1

As masses, m1 and m2 cannot be negative, v1, and v2 having opposite signs and so the two smaller nuclei move in opposite
directions.

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15. (a) Option (ii)
Explanation:
Velocity of light is maximum in vacuum.
16.
(b) False
Explanation:
False. Force may be along v ⃗ , opposite to v ⃗ , normal to v ⃗ or make some angle with v ⃗ .

17. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
Both A and R are true and R is the correct explanation of A.
18.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:

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Both A and R are true but R is not the correct explanation of A.

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19. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
Both A and R are true and R is the correct explanation of A.
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20.
(d) A is false but R is true.
Explanation:
A is false but R is true.
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e SECTION B (5 X 2 =10)
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21. Due to its small speed, the stone remains in contact with the windowpane for a longer duration. It transfers its motion to the pane
and breaks it into pieces. But the particles of windowpane near the hole are unable to share the fast motion of the bullet and so
air

remain undisturbed.
22. When a person climbs up a pole, he presses the pole downward with his feet and the pole, in turn, pushes the person upwards with
Ce

an equal force. If the pole is greasy, its surface becomes slippery and the person is not able to press it. As there is no action, there
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will be no reaction. Hence it becomes difficult for the person to climb up.
23. Force by the first boy is A = 30 N
Force by the second boy is B = 60 N
the resultant force is R
Sr

−−−−−−−−−−−−−−−−
2 2
R = √A + B + 2AB cos θ

the angle of the resultant force with force applied by the first boy is 90
B sin θ
tan 90 =
A+B cos θ
B sin θ
∞ =
A+B cos θ

A + B cos θ = 0
A = -B cosθ
−A −1
cos θ = =
B 2

θ = 120o
−−−−−−−−−−−−−−−−−−−−−−−−− −
2 2
R = √30 + 60 + 2 × 30 × 60 × cos 120

R = 51.96 N
This is the effective pull.
24. Let m, m1, and m2 be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.
Initial momentum of the system (parent nucleus) = 0
Let v1 and v2 be the respective velocities of the daughter nuclei having masses m1 and m2.
Total linear momentum of the system after disintegration = m1v1 + m2v2
According to the law of conservation of momentum:
Total initial momentum = Total final momentum

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 0 = m1v1 + m2 + v2
−m2 v2
v1 =
m1

Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.
OR
dp
The net force, Fnet = dt

Also, the rate of change of momentum = slope of the graph.

From the graph, slope AB = slope C D
And slope(BC ) = slope(DE) = 0
Therefore, the force acting on the particle is equal in regions AB and C D and in regions BC and DE (which is zero).
25. The Newton's third law of motion explains that to move forward and equal and opposite reaction force is acted on the ground. If
the horse pushes the ground with some force, the ground, in turn, exerts equal and opposite force on the feet of the horse. This
reaction force is the reason for the motion of the horse. Empty space does not have a surface for friction and thus no reaction
force.
Therefore, a horse cannot pull a cart and run in empty space.

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SECTION C (6 X 3 = 18)
26. mass of the monkey, m = 40kg,

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Tensile strength of the rope, T = 600N (max tension rope can hold without breaking)
Here, the rope will break if reaction (R) exceeds the tension (T) applied, i.e. R > T
a. a = 6m/s2
For upward accelerated motion the net acceleration is (g + a) instead of g, hence R = m (g + a) = 40 (10 + 6) = 640 N.
Tu
Therefore the rope will break, as R > T
b. a = 4m/s2
For downward accelerated motion the net acceleration is (g - a) instead of g, hence R = m (g – a) = 40 (10 – 6) = 240 N.
Therefore the rope will not break as R < T
am

c. v= 5m/s (constant) a = 0
e
R = mg = 40 × 10 = 400 N. Therefore the rope will not break as R < T
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d. For freefall, net acceleration on the body is zero, a = g; R = m (g – a) = m (g – g). Therefore R = zero (Rope will not break)
air

27. i. To just prevent the block from sliding down the inclined plane:
In this case, the applied force (F) must balance the force trying to pull the block down, which is the component of weight along
Ce

the incline minus the frictional force.

iS

∘
W∥ = mg sin θ = (0.5 × 9.8 × sin 30 ) = 2.45 N

∘
W⊥ = mg cos θ = (0.5 × 9.8 × cos 30 ) = 4.24 N

f = μN = 0.2 × 4.24 = 0.848 N

To prevent the block from sliding down:

Sr

F = W∥ − f = 2.45 − 0.848 = 1.602 N

ii. To just move the block up the inclined plane:

Here, the applied force must overcome both the component of weight acting down the incline and the frictional force, which now
acts in the opposite direction (down the incline).
F = W∥ + f = 2.45 + 0.848 = 3.298 N

iii. To move the block up the incline with an acceleration of 0.2 m/s : 2

The net force (F ) required must provide enough force to overcome the weight component, the frictional force, and also accelerate
the block. Using Newton's second law:
F − (W∥ + f ) = ma

F = ma + W∥ + f = (0.5 × 0.2) + 2.45 + 0.848 = 0.1 + 2.45 + 0.848 = 3.398 N

OR
i. When stone is dropped just after from the window of a stationary train,
Force on stone, F = Force due to gravity = mg = 0.1 kg × 10 m/s2 = 1 N in the vertically downward direction.
ii. As the train is running with constant velocity, acceleration is zero in horizontal direction i.e. direction of motion of train.
Hence no force in horizontal direction. So, only force due to gravity in vertically downward direction exists. In this case too,
force is same as in (a) i.e., F = 1 N downwards.

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 iii. Net Force, F = Force due to gravity = mg = 0.1 k g × 10 m/s2 = 1 N
∴ F = 1 N vertically downwards

iv. Here, a = 1 ms-2 in horizontal direction towards the motion of train and the stone being at rest relative to the train,
Hence, net force F = ma = 0.1 × 1 = 0.1 N.
28. Mass of the body, m = 3 kg
Initial speed of the body, u = 2 m/s
Final speed of the body, v = 3.5 m/s
Time, t = 25 s
Acceleration is given by
v−u
a=
t
3.5−2 1.5 −2
= = = 0.06m/s
25 25

Using Newton's second law of motion, force is given as:

F = ma
∴ F = 3 × 0.06 = 0.18N

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The force acts in the direction of the motion since the direction of the body remains unchanged.
29. Coefficient of static friction is defined as the ratio of static frictional force to the normal reaction.

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Angle of friction is the angle between the resultant of limiting friction and normal reaction with the normal reaction. Here in the
figure below, the angle of friction is denoted by α .

Tu
tan α =
Fs
...(i), Fs = force of static friction, R = normal reaction
am

R e
Coefficient of static friction
The limiting value of static frictional force is directly proportional to the normal reaction i.e.
ntr
F αR ⇒ F = μ R , μ = constant of proportionality here known as coefficient of static friction
s s s s
air

Or
Fs
μs = ...(ii)
Ce

From (i) and (ii)

iS

μ = tan α .
s

This is the required relation between coefficient of static friction and angle of friction.
30. In a closed cage, the inside air is bound with the cage.
i. As the acceleration is zero, there is no change in the weight of the cage.
Sr

ii. In this case, the reaction R is given by

R - Mg = Ma or R = M (g + a)
Thus the cage will appear heavier than before.
iii. In this case, the reaction R is given by
Mg - R = Ma or R = M (g - a)
Thus the cage will appear lighter than before.

31.

Let m be the mass per unit length of the string. Each element of the string falls freely. Initial velocity at the end is zero. When the

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 string has fallen through distance y, its velocity v at that instant is given by
v2 - 02 = 2 gy
−−−
or v = √2gy
After this instant, the length of the string that falls in small time dt is
dy = vdt
Mass of length dy of the string
= mdy = mvdt
Momentum transferred to the table in small time dt is
dp = (mvdt). v mv2dt
∴ Force exerted on the table,
dp
F1 =
dt
= mv
2
= m (2gy) = (2my)g
Since a length y of the string already lies on the table which also exerts force on the table given by
F2 = weight of length y = (my) g
Total force exerted on the table is

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F = F1 + F2 = (2 my)g + (my)g = 3(my)g

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= 3 × Weight of the part of the string on the table.
SECTION D (4 X 5 =20)
32. Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time. So, with
impulse, you can calculate the change in momentum, or you can use impulse to calculate the average impact force of a collision.
The formula for impulse is: Impulse = Force × time.
Tu
am

e
ntr
air

Consider a small displacement element Δs under force F, which is represented by a strip KLMN. As the displacement Δs is
extremely small, hence force F for entire strip KLMN may be taken as constant.
Ce

Therefore, Work done during elementary expansion will be equal to

iS

ΔW = FΔx = area of strip KLMN ...(i)

Total work done by the variable force can be calculated by dividing the whole path into such elementary parts and in each case the
work done will be equal to area of small shaded strips like KLMN. Therefore, total work done for a given displacement is given
Sr

by:
W = ΣΔW = Σ area of various strips = total area under F-s graph ...(ii)
Thus, the work done by a variable force is given by the area under F-s curve.
33. Newton's second law of motion
It states that the rate of change of linear momentum of a body is directly proportional to the external force applied to the body and
this change takes place always in the direction of the applied force.
Let m is the mass of a body that is moving along a straight line with an initial velocity u. It is uniformly accelerated to velocity, v
in time, t with a constant force, F.
The initial momentum of the body, p1 = mv.
Final momentum of the body, p2 = mv
The change in momentum = p2 - p1 = m(v - u)
m(v−u)
The rate of change in momentum = t

Now, the applied force is given by,

F ∝ The rate of change in momentum
m(v−u)
F∝ t
km(v−u)
F= t

F = kma

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 Where k is constant of proportionality i.e, k = 1
F = ma
Thus, the magnitude of the force can be calculated by multiplying the mass of the body and the acceleration produced in it. Hence,
the second law of motion gives us a measure of property.
One newton is the force needed to accelerate one kilogram of mass at the rate of one metre per second squared. 1 N = 1 kg m/s2.
34. Banking of road means raising the outer edge of the road with respect to inner edge so that the road makes an angle with the
horizontal.
When a vehicle moves along a curve, the force of friction provides the necessary centripetal force. But this force has a limit =
μmg. When the speed of the vehicle increases beyond this maximum value, the banking of roads is necessary so that the
component of normal reaction of the road towards the centre provides the necessary centripetal force.
From the figure, for vertical equilibrium:
Total upwards force = total downwards force
N cosθ = mg + f sinθ
mg = N cos θ - f sin θ ...(i)
As the forces N sin θ, f cos θ provide the necessary centripetal force.

n
2
mv

r
= N sin θ + f cos θ ...(ii)

itio
Thus dividing (ii) by (i),
2
mv

r N sin θ+fcosθ
=
mg N cos θ−f sin θ

As f = μ N s

Thus,
v
2

=
sin θ+ μ
s
cos θ
Tu
rg cos θ− μ sin θ
s

tan θ+μs
=
1− μs tan θ

tan θ+tan λ
= = tan(θ + λ)
am

1−tan λ tan θ
−−−−−−−−− −
where tan λ = μ
v = √rg tan(θ + λ) s
e
ntr
air
Ce
iS

OR
Sr

Horizontal force is given by, F = 60 kgf = 60 × g N = 60 × 10 N = 600 N

Mass of body is given by A, m1 = 10 kg
Mass of body B is given by, m2 = 20 kg
Total mass of the system is given by, m = m1 + m2 = 30 kg
By Using Newton's second law of motion, the acceleration (a) produced in the system can be calculated as:
F = ma
F 600 2
a= = = 20m/s
m 30

a. Both the bodies A and B will move with this acceleration as shown in the image. Now when force F is applied on body A, say
the tension acting in the string between A and B is T.

The equation of motion can be written as:

F - T = m1a
∴ T = F - m1a
= 600 − 10 × 20 = 400 N ,Hence tension= 400N.
b. In this case two bodies A and B will also move with the same acceleration, a = 20 m/s2 but in the opposite direction as shown
in the figure. Now when the force F is applied on body B, say the tension in the string between A and B is T' in this case.

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 The equation of motion can be written as:
F - T' = m2a
T' = F - m2a
′
∴ T = 600 − 20 × 20 = 200N

35. Definition of force

i. A force is a push or pulls on an object caused by its interaction with another object.
ii. If two objects interact, a force is exerted on each of the objects.
Absolute and gravitational units

i. Newton (N)is the absolute unit of force. 1 N = 1 kg m/s2.

ii. The gravitational unit is the kilogram-force (Kgf). 1Kgf is the amount of force exerted by gravity on a body of mass 1 kg.
Relation between the absolute and gravitational unit of force:
We know that the value of g is approximately 9.8m/s2

n
Hence, 1 kgf = 9.8 N

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SECTION E CASE BASED QUESTIONS (3 X 4 = 12)
2

36. i. Frictional force f = mv

R
−−
fr
∴ v = √
m
..(i) Tu
Maximum friction, f max = μN = μmg ...(ii)
Putting in eqn. (i)
−−−−
μmgr
vmax = √
m
−−−
= √μgr ...(iii)
ii. In order to oppose skidding, frictional force (f) comes into play. This force of friction between the road and the tyres provides
am

the necessary centripetal force mv

, where m is the mass of the vehicle.
r
e
This frictional force f creates another problem. It produces an outward torque h × f about the centre of gravity (G) of the
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vehicle, where h is the height of centre of gravity. This creates a tendency of overturning. Since the line of action of normal
reaction N and weight mg pass through the centre of gravity G, they do not produce a counter torque to oppose this tendency
air

of overturning.
iii. The frictional force f creates an outward torque h × f about the centre of gravity (G) of the cycle, where h is the height of
Ce

centre of gravity. This creates a tendency of overturning. To avoid this situation, the rider has to lean towards the centre of
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curvature of the road. Now, the line of action of normal reaction N does not pass through the centre of gravity.
So, N gives rise to a counter torque that opposes the torque due to friction i.e., h × f = b × n, where b is the horizontal
distance of C.G. and point of ground contact A.
Sr

iv. As
∴ tan θmax = μ = 0.57
37. i. Newton's cradle demonstrates the third law of motion. When one of the balls is lifted and released, it strikes the remaining
stationary balls and sends force through all of them to push the ball on the other end away.
ii. In Newton's cradle, the balls are hung from the crossbars by light wire, with the balls at the point of an inverted triangle. This
ensures that the balls can only swing in one plane, parallel to the crossbars.
iii. True
iv. System which is free from any external force.
The law of conservation only works in a system, which is free from any external force.
Such a system is called closed system. The Newton's cradle is not a closed system throughout its operation.
v. During the small time between the first ball's impact and the 5th ball swinging out, that momentum is conserved.
Law of conservation only works in a closed system. But the Newton's cradle is not a closed system throughout its operation.
When 5th ball swings out away from the rest of the balls, it is affected by the force of gravity, which brings the ball down.
But, the horizontal line of balls at rest, functions as a closed system, free from any influence of any force other than gravity.
It's here, during the small time between the first ball's impact and the 5th ball swinging out, that momentum is conserved.
38. i. The frictional force acts between the pulley and rope.
ii. As, tension, T = mg
⇒ T ∝ m

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 So, the side having 8 kg mass will have more tension.
iii. As, μR = f = 20 N
μ = = = (∵ R = mg)
20 20 1

R 14×10 7

iv.

Due to friction, tension at of all points of the thread is nor alike

T1 - T2 = f

f = 8g - 6g = 2g = 20 N (∵ g = 10 ms-2)

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⇒

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Tu
am

e
ntr
air
Ce
iS
Sr

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