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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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A body is rolling without slipping on a horizontal surface and its rotational kinetic energy is equal to the translational kinetic energy. The body is

Disc

Sphere

Cylinder

Ring

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Explanation

$(d) K_{R} = K_{T} \Rightarrow \frac{1}{2} {mv}^{2} (\frac{K^{2}}{R^{2}}) = \frac{1}{2} {mv}^{2} \Rightarrow∴ \frac{K^{2}}{R^{2}} = 1$

i.e. the body is a ring.

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The centre of a wheel rolling on a plane surface moves with a speed $v_{0}$ . A particle on the rim of the wheel at the same level as the centre will be moving at speed

zero

$v_{0}$

$\sqrt{2} v_{0}$

$2 v_{0}$

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Moment of inertia of a uniform cylinder of mass M, radius R and length l about an axis passing through its center and normal to its axis would be

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

$\frac{{MR}^{2}}{2}$

$M [\frac{l^{2}}{12} + \frac{R^{2}}{4}]$

$M [\frac{l^{2}}{12} + \frac{R^{2}}{2}]$

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A particle moves along a circle of radius $\frac{20}{π} m$ with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begin, the tangential acceleration is

640 $π m / s^{2}$

160 $π m / s^{2}$

40 $π m / s^{2}$

40 $m / s^{2}$

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Explanation

$ω = \frac{v}{r} = \frac{80}{20 / π} = 4 π, ω_{0} = 0, θ = 2 π (n) = 4 π (As n = 2) ω^{2} = {ω^{2}}_{0} + 2 αθ \Rightarrow α = \frac{ω^{2} - {ω^{2}}_{0}}{2 θ} = \frac{16 π^{2}}{2 \times 4 π} = 2 π Tangential acceleration a_{t} = rα = \frac{20}{π} \times 2 π = 40 m / s^{2}$

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If the radius of the earth is suddenly contracted to half of its present value, then the duration of the day will be of

6 hours

12 hours

18 hours

24 hours

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Explanation

$Angular momentum of earth is constant . Iω = constant \frac{2}{5} {MR}^{2} \times \frac{2 π}{T} = Constant T α R^{2}, if radius becomes half then time period of rotation becomes 1 / 4 of the prvious value . T_{new} = 24 / 4 = 6 hrs .$

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A wheel has angular acceleration of 3.0 rad/ $s e c^{2}$ and an initial angular speed of 2.00 rad/sec. In a time of 2 sec it has rotated through an angle (in radian) of

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Explanation

$θ = ω_{0} t + \frac{1}{2} α t^{2} = 2 \times 2 + \frac{1}{2} \times 3 \times 2^{2} = 10 r a d i a n (c h o i c e 1)$

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

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