Newton S Laws Of Motion(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

NEWTON S LAWS OF MOTION MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be

W_{1}

.If the bird flies about inside the cage, the reading of the spring balance is

W_{2}

. Which of the following is true

W1=W2
W1>W2
W1
Nothing definite can be predicted

Discuss Question

Answer: Option A. -> W1=W2
:
A
Again, the bird is not flying because of the spring force, or normal forse from the cage, right! So,

W_{1} = W_{2}

Question 22. A 60 kg man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from 60 kg to 50 kg for a while and then comes back to the original mark. What should we conclude

The lift was in constant motion upwards
The lift was in constant motion downwards
The lift while in constant motion upwards, is stopped suddenly
The lift while in constant motion downwards, is suddenly stopped

Discuss Question

Answer: Option C. -> The lift while in constant motion upwards, is stopped suddenly
:
C
For retarding motion of a lift R = m(g + a) for downward motion
R = m(g – a) for upward motion
Since the weight of the body decrease for a while and then comes back to original value it means the lift was moving upward and stops suddenly.
Note: Generally we use R = m(g + a) for downward motion
R = m(g – a) for upward motion
Here a = acceleration, but for the given problem a = retardation

Question 23. The acceleration of block B in the figure will be

m2g(4m1+m2)
2m2g(4m1+m2)
2m1g(m1+4m2)
2m1g(m1+m2)

Discuss Question

Answer: Option A. -> m2g(4m1+m2)
:
A
When the block

m_{2}

moves downward with acceleration a, the acceleration of mass

m_{1}

will be 2a because it covers double distance in the same time in comparison to

m_{2}

.
Let T is the tension in the string.
By drawing the free body diagram of A and B

T = m_{1} 2 a - - - - - - - - - - (i)

M_{2} g ­ - 2 T = m_{2} a - - - - - - - - - - - (i i)

By solving (i) and (ii)
By solving (i) and (ii)

a = \frac{m_{2} g}{(4 m_{1} + m_{2})}

Question 24. In the above problem, if we consider the two masses to be a system, the tension is an external force.

True
False
10 cm
80 cm

Discuss Question

Answer: Option B. -> False
:
B
Tension is the internal force in this case. Gravitational force is the only external force.

Question 25. A spring balance A shows a reading of 2 kg, when an aluminium block is suspended from it. Another balance B shows a reading of 5 kg, when a beaker full of liquid is placed in its pan. The two balances are arranged such that the Al – block is completely immersed inside the liquid as shown in the figure. Then

The reading of the balance A will be more than 2 kg
The reading of the balance B will be less than 5 kg
The reading of the balance A will be less than 2 kg. and that of B will be more than 5 kg
The reading of balance A will be 2 kg. and that of B will be 5 kg.

Discuss Question

Answer: Option C. -> The reading of the balance A will be less than 2 kg. and that of B will be more than 5 kg
:
C
Due to buoyant force on the aluminium block the reading of spring balance A will be less than 2 kg but it increase the reading of balance B.

Question 26. At what minimum acceleration should a monkey slide down a rope whose breaking strength is two-thirds of the monkey's weight?

Discuss Question

Answer: Option B. -> g3
:
B
When a monkey is sliding down a rope, the net force acting on it is given by
ma = mg - T, where T is the breaking strength of the rope

\Rightarrow

ma = mg -

\frac{2}{3}

\Rightarrow

a =

\frac{g}{3}

Question 27. A weightless ladder, 20 ft long rests against a frictionless wall at an angle of 60⁰ with the horizontal. A 150 pound main is 4 ft from the top of the ladder. A horizontal force is needed to prevent it from slipping choose the correct magnitude from the following

175 lb
100 lb
70 lb
150 lb

Discuss Question

Answer: Option C. -> 70 lb
:
C
Since the system is in equilibrium therefore

\sum F_{x} = 0 a n d \sum F_{y} = 0 ∴ F = R_{2} a n d W = R_{1}

Now by taking the moment of forces about point B.

F . (B C) + W . (E C) = R_{1}

(AC) [from the figure EC = 4 cos 60

F . (20 s i n 60) + W (4 c o s 60) = R_{1} (20 c o s 60)

10 \sqrt{3} F + 2 W = 10 R_{1} [A s R_{1} = W] ∴ F = \frac{8 W}{10 \sqrt{3}} = \frac{8 \times 150}{10 \sqrt{3}} = 701 b

A Weightless Ladder, 20 Ft Long Rests Against A Frictionless...

Question 28. A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight placed on a spring balance, It records a weight of 25 N. The bird (mass = 0.5kg) flies upward in the cage with an acceleration of

2 m / s^{2}

. The spring balance will now record a weight of

24 N
25 N
26 N
27 N

Discuss Question

Answer: Option B. -> 25 N
:
B
Since the cage is closed and we can treat bird cage and air as a closed (Isolated ) system. In this condition the force applied by the bird on the cage is an internal force due to this reading of spring balance will not change.

Question 29. A body weighs 8 gm, when placed in one pan and 18gm, when placed in the other pan of a false balance. The beam is horizontal (when both the pans are empty), the true weight of the body is

13 gm
12 gm
15.5 gm
15 gm

Discuss Question

Answer: Option B. -> 12 gm
:
B
For given condition true weight

= \sqrt{W_{1} W_{2}} = \sqrt{8 \times 18} = 12 g m

Question 30. Two blocks of mass 4 kg and 6 kg are placed in contact with each other on a frictionless horizontal surface. If we apply a push of 5 N on the heavier mass, the force on the lighter mass will be