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The rotational KE of a body is E and its moment of inertia is I. The angular momentum is

$2 \sqrt{(EI)}$

$\sqrt{(2 EI)}$

$\frac{E}{I}$

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Explanation

$\begin{aligned} E & = \frac{1}{2} {Iω}^{2} \\ ω & = \sqrt{\frac{2 E}{I}} \\ Now, L & = Iω = \sqrt{2 EI} \end{aligned}$

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If rotational kinetic energy is 50 % of translational kinetic energy, then the body is

Ring

Cylinder

Hollow sphere

Solid sphere

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Explanation

For a cylinder, the rotational kinetic energy is half of its translational kinetic energy. This is because the moment of inertia of a cylinder about its axis is (1/2) MR^2, where M is the mass and R is the radius.

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Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration a. The centre of mass has an acceleration

zero

1/2 a

2 a

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Explanation

The center of mass of a system of particles moves as if the entire mass of the system were concentrated at that point and all external forces were applied there. For two identical particles, one at rest and the other with acceleration 'a', the center of mass acceleration is (a/2).

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A solid sphere rolls without slipping down a $30 °$ inclined plane. If g = 10 $m / s^{2}$ , the acceleration of the rolling sphere is

5 ${ms}^{- 2}$

$\frac{7}{25} {ms}^{- 2}$

$\frac{25}{7} {ms}^{- 2}$

$\frac{15}{7} {ms}^{- 2}$

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Explanation

$\begin{aligned} a & = \frac{gsin θ}{1 + \frac{k^{2}}{r^{2}}} = \frac{gsin 30^{\circ}}{1 + \frac{2}{5}} = \frac{10 \times \frac{1}{2}}{\frac{7}{5}} \\ = \frac{5}{7} \times 5 = \frac{25}{7} {ms}^{- 2} \end{aligned}$

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If the equation for the displacement of a particle moving on a circular path is given by $θ = 2 t^{3} + 0.5$ , where $θ$ is in radian and t is in second, then the angular velocity of the particle after 2s is

8 rad/s

12 rad/s

24 rad /s

36 rad/s

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Explanation

$ω = \frac{d θ}{d t} = 6 t^{2}, A t t = 2 s ω = 6 {(2)}^{2} = 24 r a d / s$

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A pan containing a layer of uniform thickness of ice is placed on a circular turntable with its centre coinciding with the centre of the turn table. The turntable is now rotated at a constant angular velocity about a vertical axis passing through its centre and then driving is withdrawn. There is no friction between the table. As the ice melts

the angular velocity of the system decreases

the angular velocity of the system increases

the angular velocity of the system remains unchanged

the moment of inertia of the system decreases

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Explanation

When Ice melts, water tends to move away from the axis, M.I increases, by conservation of angular momentum, angular velocity decreases.

Choice A is correct.

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The motion of planets in the solar system is an example of the conservation of

mass

Linear momentum

Angular momentum

Energy

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Explanation

Where L = Angular momentum, m = mass of the planet

$\frac{d A}{d t}$ = Arial velocity of the line which joins the sun and the planet, which is constant and angular momentum is constant.

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A circular disc is to be made by using iron and aluminium, so that it acquires maximum moment of inertia about its geometrical axis. It is possible with

iron and aluminium layers in alternate order

aluminium at interior and iron surrounding it

iron at interior and aluminium surrounding it

either '1' or '3'

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Explanation

Moment of Inertia of a body about an axis is $m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + . . . . . . . .$

M.I increase if heavier particles are more away from the axis. Since density of iron is more as compared to that of aluminium. Therefore, MI of disc will be maximum if iron is more away from geometrical axis. Therefore, M.I of disc is maximum if aluminium is at interior and iron surrounds it.

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A particle of mass 15 kg has an initial velocity $v_{i} = \hat{i} - 2 \hat{j}$ m/s. It collides with another body and the impact time is 0.1 s, resulting in a velocity $v_{f} = 6 \hat{i} + 4 \hat{j} + 5 \hat{k}$ m/s after impact. The average force of impact on the particle is

$15 |5 \hat{i} + 6 \hat{j} + 5 \hat{k}|$

$15 |5 \hat{i} + 6 \hat{j} - 5 \hat{k}|$

$150 |5 \hat{i} - 6 \hat{j} + 5 \hat{k}|$

$150 |5 \hat{i} + 6 \hat{j} + 5 \hat{k}|$

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Explanation

$\begin{aligned} F = \frac{Δp}{Δt} & = \frac{m (v_{t} - v_{i})}{4 t} \\ = \frac{15 [(6 \hat{i} + 4 \hat{j} + 5 \hat{k}) - (\hat{i} - 2 \hat{j})]}{0 .1} \\ = 150 (5 \hat{i} + 6 \hat{j} + 5 \hat{k}) \end{aligned}$

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Two Circular discs A and B are of equal masses and thicknesses but made of metal with densities $d_{A} and d_{B}$ ( $d_{A} > d_{B}$ ). If their moments of inertia about an axis passing through their centers and perpendicular to circular faces be $I_{A} and I_{B}$ , then

$I_{A} = I_{B}$

$I_{A} > I_{B}$

$I_{A} < I_{B}$

$I_{A} \geq I_{B}$

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Explanation

Let M be the mass of each disc and let $R_{A} a n d R_{B}$ be the radii of disc A and B respectively, them

$M = {πR}_{A}^{2} {td}_{A} = {πR}_{B}^{2} {td}_{B} as d_{A} > d_{B}; R_{A}^{2} < R_{B}^{2} M . I .' s are given as I_{A} = \frac{1}{2} M \times R_{A}^{2}; I_{B} = \frac{1}{2} M \times R_{B}^{2} \frac{I_{A}}{h_{B}} = \frac{R_{A}^{2}}{R_{B}^{2}} < 1; ∴ I_{A} < I_{B}$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.