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A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is v, the power delivered to the particle is

$\frac{{Mv}^{2}}{T}$

$\frac{1}{2} \frac{{Mv}^{2}}{T^{2}}$

$\frac{{Mv}^{2}}{T^{2}}$

$\frac{1}{2} \frac{{Mv}^{2}}{T}$

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Explanation

$The kinetic energy of particle = \frac{1}{2} {Mv}^{2} Power = \frac{Energy}{Time} P = \frac{1}{2} \frac{{Mv}^{2}}{T}$

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A body of mass 1 kg is thrown upwards with a velocity $20 {ms}^{- 1} .$ It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? $(g = 10 {ms}^{- 2})$

20 J

30 J

40 J

10 J

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Explanation

Key Idea The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.

Initially, body posses only kinetic energy and after attaining a height the kinetic energy is zero.

Therefore, loss of energy=KE-PE

$= \frac{1}{2} {mv}^{2} - mgh = \frac{1}{2} \times 1 \times 400 - 1 \times 18 \times 10 = 200 - 180 = 20 J$

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A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

Mg/k

2Mg/k

4 Mg/k

Mg/2k

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Explanation

Use the law of conservation of energy. Let x be the extension in the spring.

Applying conservation of energy-

$m g x - \frac{1}{2} k x^{2} = 0 - 0$

$\Rightarrow x = \frac{2 m g}{k}$

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K is the force constant of a spring. The work done in increasing its extension from $l_{1}$ to $l_{2}$ will be

$K (l_{2} - l_{1})$

$\frac{K}{2} (l_{2} + l_{1})$

$K (l_{2}^{2} - l_{1}^{2})$

$\frac{K}{2} (l_{2}^{2} - l_{1}^{2})$

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Explanation

(d)

$A t e x t e n s i o n l_{1}, t h e s t o r e d e n e r g y = \frac{1}{2} K l_{1}^{2} A t e x t e n s i o n l_{2}, t h e s t o r e d e n e r g y = \frac{1}{2} K l_{2}^{2} W o r k d o n e i n i n c r e a \sin g i t s e x t e n s i o n f r o m l_{1} t o l_{2} = \frac{1}{2} K (l_{2}^{2} - l_{1}^{2})$

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A ball of mass m moving with a speed u undergoes a head-on elastic collision with a ball of mass nm initially at rest. The fraction of initial energy transferred to the heavier ball is

$\frac{n}{1 + n}$

$\frac{n}{{(1 + n)}^{2}}$

$\frac{2 n}{{(1 + n)}^{2}}$

$\frac{4 n}{{(1 + n)}^{2}}$

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Explanation

$K_{i} = \frac{1}{2} m u^{2}$

By conservation of momentum mu = $m v_{1} + n m v_{2}$

$u = v_{1} + n v_{2} (1) v_{2} - v_{1} = u_{1} - u_{2} (2) ∴ v_{2} - v_{1} = u (1) + (2) 2 u = (n + 1) v_{2} v_{2} = \frac{2 u}{n + 1}$

Kinetic energy transferred $K_{t} = \frac{1}{2} (n m) v_{2}^{2}, K_{t} = \frac{2 n m u^{2}}{{(n + 1)}^{2}}, \frac{K_{t}}{K_{i}} = \frac{4 n}{{(n + 1)}^{2}}$

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If a body of mass m moving with velocity u collides head-on elastically with another identical body at rest. After collision , velocity of the second body will be

zero

data insufficient

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Explanation

Between elastic collision of two identical masses, velocities are interchanged.

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A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

$\frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$

$\frac{1}{3} v \hat{i} + \frac{2}{3} v \hat{j}$

$\frac{2}{3} v \hat{i} + \frac{1}{3} v \hat{j}$

$\frac{3}{2} v \hat{i} + \frac{1}{4} v \hat{j}$

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A particle is moving in a circular orbit with constant speed. Select wrong alternate

Its linear momentum is conserved

Its angular momentum is conserved

It is moving with variable velocity

It is moving with variable acceleration

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Explanation

Direction of linear velocity always keeps on changing. Hence , linear momentum is varying

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One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively I_A and I_B such that

$I_{A} = I_{B}$

$I_{A} > I_{B}$

$I_{A} < I_{B}$

$\frac{I_{A}}{I_{B}} = \frac{d_{A}}{d_{B}}$

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Explanation

$M . O . I . o f s o l i d s p h e r e : I_{A} = \frac{2}{5} M R^{2} M . O . I . o f h o l l o w s p h e r e : I_{B} = \frac{2}{3} M R^{2}$

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A couple produces: [NTSE 1995; CBSE PMT 1997; DCE 2004]

Purely linear motion

Purely rotational motion

Linear and rotational motion

No motion

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Explanation

A couple consists of two equal and opposite forces which cause pure rotational motion.

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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.