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Physics NEET Questions — Free MCQ Practice

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For the same total mass, which of the following will have the largest moment of inertia about an axis passing through the centre of gravity and perpendicular to the plane of the body?

A.

A disc of radius a

B.

a ring of radius a

C.

a square lamina of side 2 a

D.

four rods forming square of side 2 a.

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Explanation

$l_{disc} = M \frac{a^{2}}{2}; l_{ring} = {Ma}^{2} l_{square lamina} = M (\frac{L^{2} + B^{2}}{12}) = M \frac{4 (a^{2} + a^{2})}{12} = \frac{2 {Ma}^{2}}{3} l_{rods} = [\frac{M}{4} x \frac{4 a^{2}}{12} + \frac{M}{4} x {(\frac{2 a}{2})}^{2}] x 4 = \frac{4 {Ma}^{2}}{3}$

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The radius of gyration of a uniform rod of length L about an axis passing through its centre of mass is

A.

$\frac{L}{2 \sqrt{3}}$

B.

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

C.

$\frac{L}{\sqrt{3}}$

D.

$\frac{L}{\sqrt{2}}$

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Explanation

$K = \sqrt{\frac{I}{M}} = \sqrt{\frac{(m L^{2} / 12)}{m}} = \frac{L}{2 \sqrt{3}}$

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Two uniform, thin identical rods each of mass M and length l are joined together to form a cross. What will be the moment of inertial of the cross about an axis passing through the point at which the two rods are joined and perpendicular to the plane of the cross

A.

$\frac{M l^{2}}{12}$

B.

$\frac{M l^{2}}{6}$

C.

$\frac{M l^{2}}{4}$

D.

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

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Explanation

$I = 4 [\frac{M}{2} \frac{{(l / 2)}^{2}}{3}] = \frac{M I^{2}}{6}$

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If the linear momentum is increased by 50%, the kinetic energy will increase by

A.

50%

B.

100%

C.

125%

D.

25%

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Explanation

$\begin{array}{l} Let p_{1} = p, p_{2} = p_{1} + 50 % of p_{1} = p_{1} + \frac{p_{1}}{2} = \frac{3 p_{1}}{2} \\ \begin{aligned} E & \propto p^{2} \end{aligned} \end{array} \Rightarrow \begin{aligned} \frac{E_{2}}{E_{1}} & = {(\frac{p_{2}}{p_{1}})}^{2} \end{aligned} = {(\frac{3 p_{1} / 2}{p_{1}})}^{2} = \frac{9}{4} \Rightarrow \begin{aligned} E_{2} & = 2 .25 E_{2} = E_{1} + 1 .25 E_{1} \end{aligned} ∴ \begin{aligned} E_{2} & = E_{1} + 125 % of E_{1} \end{aligned}$

i.e., Kinetic energy will increase by 125%.

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A solid sphere, a hollow sphere, and a disc, all having the same mass and radius, are placed at the top of a smooth incline and released. Least time will be taken in reaching the bottom by

A.

the solid sphere

B.

the hollow sphere

C.

the disc

D.

all will take the same time

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Explanation

Acceleration of all bodies on smooth surface = g sin $θ$ (equal)

$∴$ All bodies will take equal time.

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A bullet of mass m and velocity v is fired into a block of mass M and sticks to it. The final velocity of the system equals

A.

$\frac{M}{m + M} . v$

B.

$\frac{m}{m + M} . v$

C.

$\frac{m + M}{m} . v$

D.

None of these

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Explanation

$mv = (m + M) v' ∴ v' (\frac{m}{m + M}) v$

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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.

Physics MCQs for NEET — Practice Questions with Answers

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