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Physics NEET Questions — Free MCQ Practice

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Two identical balls A and B having velocities of 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be

A.

-0.5 m/s and 0.3 m/s

B.

0.5 m/s and -0.3 m/s

C.

-0.3 m/s and 0.5 m/s

D.

0.3 m/s and 0.5 m/s

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Explanation

(b) Key Idea :

In elastic collision,kinetic energy of the system remains unchanged and momentum is also conserved.

It is given that mass of balls are same and collision is perfectly elastic (e=1) so their velocities will be interchanged.

Thus, $v_{A}^{'} = v_{B} = - 0.3 m / s, v_{B}^{'} = v_{A} = 0.5 m / s$

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Two rotating bodies A and B of masses m and 2m with moments of inertia $I_{A}$ and $I_{B} (I_{B} > I_{A})$ have equal kinetic energy of rotation. If $L_{A}$ and $L_{B}$ be their angular momenta respectively, then

A.

$L_{A} = \frac{L_{B}}{2}$

B.

$L_{A} = 2 L_{B}$

C.

$L_{B} > L_{A}$

D.

$L_{A} > L_{B}$

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Explanation

(c) As we know that, the kinetic energy of a rotating body,

$K E = \frac{1}{2} I ω^{2} = \frac{1}{2} \frac{I^{2} ω^{2}}{I} = \frac{L^{2}}{2 l}$

Also, angular momentum, $L = l ω$

Thus, $K_{A} = K_{B}$

$\Rightarrow \frac{1}{2} \frac{L_{A}^{2}}{I_{A}} = \frac{1}{2} \frac{L_{B}^{2}}{I_{B}} \Rightarrow {(\frac{L_{A}}{L_{B}})}^{2} = \frac{I_{A}}{I_{B}} \Rightarrow \frac{L_{A}}{L_{B}} = \sqrt{\frac{I_{A}}{I_{B}}}$

$L \propto \sqrt{I} ∴ L_{A} < L_{B [∵ I_{B} > I_{A}]}$

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A solid sphere of mass m and radius R is rotating about its diameter. A soild cyclinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation $(E_{s p h e r e} / E_{c y l i n d e r})$ will be

A.

2:3

B.

1:5

C.

1:4

D.

3:1

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Explanation

(b) Key Idea KE of a rotating rigid body, KE= $\frac{1}{2} I ω^{2}$

$∴ K E$ of sphere, $K_{s} = \frac{1}{2} I ω_{1}^{2}$

= $\frac{1}{2} \times \frac{2}{5} m R^{2} ω_{2}^{2} = \frac{1}{5} m R^{2} ω_{1}^{2}$

KE of cylinder, $K_{c} = \frac{1}{2} \frac{1}{2} m R^{2} ω_{2}^{2} = \frac{1}{4} m R^{2} ω_{2}^{2}$

$∴ \frac{K_{s}}{K_{c}} = \frac{\frac{m R^{2} ω_{1}^{2}}{5}}{\frac{m R^{2} ω_{1}^{2}}{4}} = \frac{4}{5} \frac{ω_{1}^{2}}{ω_{2}^{2}}$

$= \frac{4}{5} \frac{ω_{1}^{2}}{{(2 ω_{1})}^{2}} = \frac{1}{5} (g i v e n, ω_{2} = 2 ω_{1})$

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From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ?

A.

$13 M R^{2} / 32$

B.

$11 M R^{2} / 32$

C.

$9 M R^{2} / 32$

D.

$15 M R^{2} / 32$

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Explanation

The moment of inertia of a solid disc about its perpendicular axis through the center is MR^2/2. The moment of inertia of a circular hole of radius R/2 is (MÏ€R^4)/(64Ï€(R/2)^2) = MR^2/16. Therefore, the remaining part has a moment of inertia of MR^2/2 - MR^2/16 = 13MR^2/32.

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Two particles A and B. move with constant velocities v₁ and v₂. At the initial moment, their position vectors are $r_{1}$ and $r_{2}$ respectively. The condition for particles A and B for their collision is-

A.

$\frac{r_{1} - r_{2}}{|r_{1} - r_{2}|} = \frac{v_{2} - v_{1}}{|v_{2} - v_{1}|}$

B.

$r_{1} \cdot v_{1} = r_{2} \cdot v_{2}$

C.

_{$r_{1} \times v_{1} = r_{2} \times v_{2}$}

D.

r₁-r₂=v₁-v₂

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A force F= $a \hat{i}$ $+ 3 \hat{j}$ $+ 6 \hat{k}$ is acting at a point r= $2 \hat{i}$ - $6 \hat{j}$ -12 $\hat{k}$ . The value of α for which angular momentum about origin is conserved is

A.

-1

B.

2

C.

zero

D.

1

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Explanation

Key Concept : When the resultant external torque acting on a system is zero, the total angular momentum of a system remains constant. This is the principle of the conservation of angular momentum.

Given, force F=α $\hat{i}$ $+ 3 \hat{j}$ $+ 6 \hat{k}$ is acting at a point r= $2 \hat{i}$ - $6 \hat{j}$ -12 $\hat{k}$
As, angular momentum about origin is conserved i.e. τ=constant

=> Torque,τ=0

=> r x F=0

| $\hat{i}$ $\hat{j}$ $\hat{k}$ |
|2-6 -12| =0
|α 3 6|

=> (-36+36) $\hat{i}$ -(12+12α) $\hat{j}$ +(6+6α) $\hat{k}$ =0

=> 0 $\hat{i}$ -12(1+α) $\hat{j}$ +6(1+α) $\hat{k}$ =0

=> 6(1+α)=0=> α=-1

So, value of α for angular momentum about origin is conserved, α=-1

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An automobile moves on a road with a speed of 54 km h^-1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m². If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to the wheel is

A.

6.66 kgm²s^-2

B.

8.58 kgm²s^-2

C.

10.86 kgm²s^-2

D.

2.86 kgm²s^-2

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Explanation

As velocity of an automobile,

v=54km/h

=54 x 5/18

=15m/s

Angular velocity of a vehicle, v=ω_or

=> ω_{o=v/R=15/0.45=100/3rad/s

$S o, a n g u l a r a c c e l e r a t i o n o f a n a u t o m o b i l e, α = Δ ω / t = ω_{f} - ω_{o} / t = 0 - 100 / 3 / 15 = - 100 / 45 r a d / s^{2} T h u s, a v e r a g e t o r q u e t r a n s m i t t e d b y i t s b r a k e s t o w h e e l τ = I α 3 x 100 / 45 = 6.66 k g m^{2} s^{- 2}$}

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A body of mass (4m) is lying in xy-plane at rest. It suddenly explodes into three pieces. Two pieces each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is

A.

mv²

B.

3/2mv²

C.

2mv²

D.

4mv²

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A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis.A massless string is wound round the cylinder with one end attached to it and other hanging freely.Tension in the string required to produce an angular acceleration of 2 rev/ s² is

A.

25N

B.

50N

C.

78.5N

D.

157N

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Explanation

The moment of inertia of a solid cylinder about its axis is (1/2)MR^2. The torque required for an angular acceleration Î± is IÎ±. Substituting the given values, we get Ï„ = (1/2) Ã— 50 Ã— (0.5)^2 Ã— 2Ï€ Ã— 2 = 157 N-m. The tension in the string provides this torque.

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The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle θ without slipping and slipping down the incline without rolling is

A.

5:7

B.

2:3

C.

2:5

D.

7:5

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Physics MCQs for NEET — Practice Questions with Answers

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