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NEET NEET Questions — Free MCQ Practice

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If the kinetic energy of a body increases by 0.1%, the percent increase of its momentum will be

A.

0.05%

B.

0.1%

C.

1.0%

D.

10%

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Explanation

$p = \sqrt{2 mE} ∴ p \propto \sqrt{E}$

Percentage increase in p $\approx$ $\frac{1}{2}$ (percentage increase in E)

= $\frac{1}{2}$ (0.1%) = 0.05%

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A particle P moving with speed v undergoes a head-on elastic collision with another particle Q of identical mass but at rest. After the collision-

A.

both P and Q move forward with speed $\frac{v}{2}$ .

B.

both P and Q move forward with speed $\frac{v}{\sqrt{2}}$ .

C.

P comes to rest and Q moves forward with speed v.

D.

P and Q move in opposite directions with speed $\frac{v}{\sqrt{2}}$ .

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Explanation

In a head on collision, velocities after collision are given as

$v_{1} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} u_{1} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}} a n d v_{2} = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} u_{2} + \frac{2 m_{1} u_{1}}{m_{1} + m_{2}} P u t t i n g m_{1} = m_{2} = m a n d u_{2} = 0 v_{1} = 0; v_{2} = u_{1} h e r e u_{1} = v_{(f o r P)} H e n c e c h o i c e 3 i s c o r r e c t$

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The moments of inertia of two freely rotating bodies A and B are $I_{A} a n d I_{B}$ respectively. $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{A} a n d K_{B}$ are their kinetic energies, then

A.

$K = K_{B}$

B.

$K_{A} > K_{B}$

C.

$K_{A} < K_{B}$

D.

$K_{A} = 2 K_{B}$

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Explanation

Kinetic energy $E = \frac{L^{2}}{2 I}$

If angular momenta are equal then $E α \frac{1}{I}$

Kinetic enery E =K [Given in the problem]

If $I > I_{B} t h e n K_{A} < K_{B}$

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A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

A.

Both together only when the angle of inclination of the plane is $45^{0}$

B.

Both together

C.

Hollow cylinder

D.

Solid cylinder

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Explanation

$W h e n a b o d y r o l l s d o w n t h e p l a n e : T o r q u e e q u a t i o n : f . R = I . α = I . \frac{a}{R} f = I . \frac{a}{R^{2}} F o r c e e q u a t i o n : m g \sin θ - f = m a m g \sin θ - I . \frac{a}{R^{2}} = m a m g \sin θ = I . \frac{a}{R^{2}} + m a a = \frac{m g \sin θ}{\frac{I}{R^{2}} + m} H e r e; I = m k^{2} a = \frac{m g \sin θ}{\frac{m k^{2}}{R^{2}} + m} = \frac{g \sin θ}{1 + \frac{k^{2}}{R^{2}}}$

Acceleration of a rolling body down an inclined plane is :

$a = \frac{g \sin θ}{1 + \frac{k^{2}}{r^{2}}} a_{(solid cylinder)} = \frac{gsin θ}{1 + \frac{1}{2}} = \frac{2}{3} gsin θ = 0.67 gsin θ \begin{aligned} a_{(holow cylinder)} & = \frac{gsinθ}{1 + \frac{k^{2}}{r^{2}}} = \frac{gsin θ}{1 + \frac{r^{2}}{r^{2}}} = \frac{gsin θ}{2} = 0 .5 gsin θ \end{aligned} S ince S = \frac{1}{2} {at}^{2}; t = \sqrt{\frac{2 S}{a}}; t \propto \frac{1}{\sqrt{a}} \begin{array}{l} \end{array}$

The time for the solid cylinder is less. It reaches the bottom earlier. Choice D is correct.

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The moment of inertia of a rod about an axis through its centre and perpendicular to it is $\frac{1}{12} M L^{2}$ (where M is the mass and L is the length of the rod). The rod is bent in the middle so that the two halves make an angle of 60 $°$ . The moment of inertia of the bent rod about the same axis would be

A.

$\frac{1}{48} M L^{2}$

B.

$\frac{1}{12} M L^{2}$

C.

$\frac{1}{24} M L^{2}$

D.

$\frac{M L^{2}}{8 \sqrt{3}}$

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A man is sitting on a rotating table with his arms stretched outwards. When he suddenly folds his arms inside, then

A.

his angular velocity will decrease

B.

his angular velocity remains constant

C.

his moment of inertia decreases

D.

angular momentum increases

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Explanation

Decrease of distance of masses from the axis results in decrease in M.I. as $I = \sum_{} m r^{2}$

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A body of mass M and radius R is rolling horizontally without slipping with speed v. It then rolls up a hill to a maximum height h. If $h = \frac{5 v^{2}}{6 g}$ , what is the M.I of the body?

A.

$\frac{M R^{2}}{2}$

B.

$\frac{2}{3} M R^{2}$

C.

$\frac{3}{4} M R^{2}$

D.

$\frac{2}{5} M R^{2}$

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Explanation

$M g h = \frac{1}{2} M v^{2} + \frac{1}{2} l ω^{2} = \frac{1}{2} M v^{2} + \frac{1}{2} \times I ω^{2} = \frac{1}{2} M v^{2} + \frac{1}{2} \frac{I v^{2}}{R^{2}}; a s h = \frac{5 v^{2}}{6 g} m g \times \frac{5 v^{2}}{6 g} = \frac{1}{2} M v^{2} + \frac{1}{2} \frac{I v^{2}}{R^{2}} (\frac{5 M v^{2}}{6} - \frac{M v^{2}}{2}) \frac{2 R^{2}}{v^{2}} = I M (\frac{5}{6} - \frac{3}{6}) \times 2 R^{2} = I, ∴ \frac{2}{3} M R^{2} = I w h i c h g i v e s I = \frac{2}{3} M R^{2} (c h o i c e 2)$

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One projectile moving with velocity v in space, gets burst into 2 parts of masses in the ratio 1:3. The smaller part becomes stationary. What is the velocity of the other part?

A.

4v

B.

v

C.

$\frac{4 v}{3}$

D.

$\frac{3 v}{4}$

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Explanation

$m v = (\frac{3}{4} m) v' ∴ v' = \frac{4 v}{3}$

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One circular ring and one circular disc both having the same mass and radius. The ratio of their moments of inertia about the axes passing through their centres and perpendicular to planes will be

A.

1:1

B.

2:1

C.

1:2

D.

4:1

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Explanation

$I_{r i n g} = m R^{2}, I_{d i s c} = \frac{1}{2} m R^{2}$

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A wheel of radius R rolls on the ground with a uniform velocity v. The velocity of topmost point relative to the bottommost point is

A.

v

B.

2v

C.

v/2

D.

zero

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Explanation

In rolling motion, the topmost point has an instantaneous velocity equal to the velocity of the center of the wheel plus the peripheral velocity. Since the center's velocity is v and the peripheral velocity is also v, the velocity of the topmost point relative to the bottommost point is 2v.

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