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Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement (s) is (are) correct ?

Both cylinder P and Q reach the ground at the same time

Cylinder P has larger linear acceleration than cylinder Q

Both cylinder P and Q reaches the ground with same translational kinetic energy

Cylinder Q reaches the ground with larger angular speed

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Explanation

(a)

$I_{P} > I_{Q} a_{P} = \frac{g sinθ}{I_{P} + {mR}^{2}} a_{Q} = \frac{g sinθ}{I_{Q} + {mR}^{2}} a_{P} < a_{Q} \Rightarrow V = u + at \Rightarrow t \propto \frac{1}{a} t_{P} > t_{Q} V^{2} = u^{2} + 2 as \Rightarrow u \propto a \Rightarrow V_{P} < V_{Q} Translational K . E . = \frac{1}{2} {mV}^{2} \Rightarrow TR {KE}_{P} < TR {KE}_{Q} V = ωR \Rightarrow ω \propto V \Rightarrow ω_{P} < ω_{Q} .$

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A soldier is firing 20 bullets per second from his gun having a muzzle speed of 150 m/s. The mass of each bullet is 50 g. If they strike the wall and rebound with the same speed, then the force on the wall is

75 N

150 N

300 N

600 N

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Explanation

$\vec{F} = \frac{d \vec{P}}{dt} = 2 mvn (Rate of change of momentum) = (Force) \begin{array}{l} = 2 \times 20 \times \frac{50}{1000} \times 150 \\ = 300 N \end{array}$

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The minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination 45 $°$ is

2/7

1/3

1/2

2/5

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Explanation

Minimum friction coefficient rolling without slipping on the inclined plane is:

$\begin{aligned} μ \geq \frac{\tan θ}{1 + \frac{R^{2}}{k^{2}}} \\ μ \geq \frac{\tan 45}{1 + \frac{5}{2}} = 2 / 7 \end{aligned}$

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A wheel is rolling along a horizontal ground with a speed o 10 m/s. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to

$2 \sqrt{10} m / s$

10 m/s

$\sqrt{10} m / s$

$10 \sqrt{2} m / s$

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Explanation

$(4) T h e p o i n t w i l l h a v e t w o v e l o c i t i e s, o n e d u e t o t r a n s l a t i o n a l m o t i o n, v_{c m} a n d a n o t h e r d u e t o r o t a t i o n a l m o t i o n, R ω . F o r r o l l i n g, v_{c m} = R ω = 10 m / s B o t h v e l o c i t i e s a r e p e r p e n d i c u l a r t o e a c h o t h e r f o r t h i s p o i n t . v_{n e t} = \sqrt{10^{2} + 10^{2}} = 10 \sqrt{2} m / s$

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A sphere of mass m moving with constant velocity hits another sphere of the same mass at rest. If e is the coefficient of restitution. The ratio of their velocities after the collision is

1 + e

$\frac{1 + e}{2}$

$\frac{1 + 2 e}{1 - 2 e}$

$\frac{1 - e}{1 + e}$

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Explanation

$mu = m v_{1} + {mv}_{2} \Rightarrow v_{2} + v_{1} = u \to 1 e = \frac{v_{2} - v_{1}}{u - 0} \Rightarrow v_{2} - v_{1} = eu \to 2 by solving 1 and 2, v_{1} = \frac{(1 - e)}{2} u v_{2} = \frac{(1 + e)}{2} u$

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A mass M is moving with constant velocity parallel to the X-axis. Its angular momentum w.r.t. origin is:

Zero

Remains constant

goes on increasing

goes on decreasing

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A particle is rotating under the central force, in which one of the following quantities is constant:

torque

angular momentum

linear momentum

none of these

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Explanation

Under central force, torque is zero. So angular momentum is constant.

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What minimum friction coefficient is required for hollow sphere rolling without slipping on the inclined plane with an angle of inclination 37 $°$ ?

$\frac{3}{10}$

$\frac{3}{14}$

$\frac{3}{16}$

None of these

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Explanation

$(1) Force equation : mgsinθ - μmgcosθ = ma . . . . . (i) Torque equation : (μmgcosθ) R = Iα = \frac{2}{3} {mR}^{2} (\frac{a}{R}) μmgcosθ = \frac{2}{3} ma (ii) Solving both equations : a = \frac{3}{5} gsinθ And, μ \geq \frac{2}{5} tanθ μ \geq \frac{3}{10} S h o r t c u t : μ_{m i n} = \frac{g \tan θ}{1 + \frac{R^{2}}{k^{2}}} w h e r e k i s r a d i u s o f g y r a t i o n$

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The disc is placed on the rough horizontal surface given only angular velocity $ω_{0}$ . The velocity of the centre of mass when it starts pure rolling is:

$\frac{ω_{0} R}{2}$

$\frac{ω_{0} R}{3}$

$\frac{ω_{0} R}{4}$

$\frac{ω_{0} R}{6}$

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Explanation

$T o r q u e e q u a t i o n, f . R = I α \Rightarrow α = \frac{f . R}{I} = \frac{2 f . R}{m R^{2}} = \frac{2 f}{m R} v e l o c i t y a f t e r t i m e t, v = \frac{f}{m} t \Rightarrow f = \frac{m v}{t} A n g u l a r v e l o c i t y a f t e r t i m e t, ω = ω_{0} - α t \frac{v}{R} = ω_{0} - \frac{2 f}{m R} t = ω_{0} - \frac{2 m v}{m R . t} t = ω_{0} - \frac{2 v}{R} v = \frac{ω_{0} R}{3}$

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A force 2N is along line y = 3x + 4. The torque about the origin is:

$\frac{5}{\sqrt{10}} Nm$

$\frac{7}{\sqrt{10}} Nm$

$\frac{6}{\sqrt{10}} Nm$

$\frac{8}{\sqrt{10}} Nm$

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Explanation

$\begin{aligned} r = \frac{{ax}_{1} + {by}_{1} + c}{\sqrt{a^{2} + b^{2}}} = \frac{4}{\sqrt{10}} \\ τ = r \times F = 2 \times \frac{4}{\sqrt{10}} = \frac{8}{\sqrt{10}} Nm \end{aligned}$

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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.