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NEET Practice Questions (MCQs) with Answers & Solutions

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A stone is projected with an angle $ \theta $ and velocity $V_0$ from point P. It strikes the ground at point Q. If the both P and Q are on same horizontal line, then find average velocity.

A. $ V_0 cos \theta $ B. $ V_0 sin \theta $ C. $ V_0 cos { \theta \over 2 } $ D. $ V_0 sin { \theta \over 2 } $

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Explanation

For a projectile motion where the initial and final points are on the same horizontal line, the average velocity is the horizontal component of the initial velocity, which is \( V_0 \cos \theta \).

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Angle of projection of a projectile is changed, keeping initial velocity constant. Find the rate of change of maximum height. Range of the projectile is R.

A. R/4 B. R/3 C. R/2 D. R

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A body travelling in a circle at constant speed.

A. has a constant velocity B. is not accelerated C. has an outward radial acceleration D. has an inward radial acceleration

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Explanation

When a body travels in a circle at constant speed, it is undergoing circular motion. In circular motion, even though the speed is constant, the direction of the velocity is constantly changing. This change in direction means there is an acceleration, called centripetal acceleration, which acts towards the center of the circle. Therefore, the body has an inward radial acceleration.

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If f is the frequency of a body moving in a circular path with constant speed. a is its centrifugal acceleration, so.

A. $ a \alpha f $ B. $ a \alpha f^2 $ C. $ a \alpha f^3 $ D. $ a \alpha 1/f $

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Explanation

Centrifugal acceleration $ a_r = { v^2 \over r } = \left( { 2 \pi r \over T } \right) ^2 { 1 \over r } = 4 \pi^2 r f^2 $

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Following question is Assertion - Reason type question. Choose Assertion : At the highest point of projectile motion the velocity is not zero. Reason : Only the verticle component of velocity is zero. Where as horizontal component still exists

A. If both Assertion - Reason are true, reason is correct explanation of Assertion. B. If both Assertion - Reason are true but reason is not correct explanation of Assertion. C. If Assertion is true but Reason is false. D. If Assertion is true but Reason is false.

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Explanation

In projectile motion, at the highest point, the vertical component of the velocity is zero but the horizontal component of the velocity still exists. Thus, the total velocity is not zero. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

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A particle is moving in a circle of radius R with constant speed. The time period of the particle is T Now after time t = T/6 .Average velocity of the particle is

A. $ { 3R \over T } $ B. $ { 6R \over T } $ C. $ { 2R \over T } $ D. $ { 4R \over T } $

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A particle is moving in a circle of radius R with constant speed. The time period of the particle is T Now after time t = T/6 Range of a projectile is R and maximum height is H. Find the area covered by the path of the projectile and horizontal line.

A. $ { 2 \over 3 } RH$ B. $ { 5 \over 3 } RH$ C. $ { 3 \over 5 } RH$ D. $ { 6 \over 5 } RH$

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Explanation

$ A = \int _0^R y dx $ $ take \, y = x tan \theta - { gx^2 \over 2V_0 x cos \theta } $

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Volume, pressure and temperature of an ideal gas are V, P and T respectively.
If mass of molecule is m, then its density is [ $k_B = Boltzmann`s constant$]

A. $ { P \over k_B T } $ B. $ { Pm \over k_B T } $ C. $ { P \over k_B TV } $ D. $ m k_B T $

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Explanation

$ PV = RT \Rightarrow PV = { M \over M_o } RT $ $ \therefore {PMo \over RT } ={ M \over V} = \rho $
$ \rho = { PMo \over RT} = {P \times m \times N_A \over RT} = { Pm \over \left( { R \over N_A} \right)} T = { Pm \over k_B T } $

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The temperature of an ideal gas at atmospheric pressure is 300 K and volume $1m^3$. If temperature and volume become double, then pressure will be

A. $ 4 \times 10^5 Nm^{-2} $ ` B. $ 2 \\times 10^5 Nm^{-2} $ C. $ 1 \times 10^5 Nm^{-2 } $ D. $ 0.5 \times 10^5 Nm^{-2 } $

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Explanation

$ PV = \mu RT \Rightarrow P \alpha { T \over V } $ If T and V both doubled, then pressure remains same

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At 100 K and 0.1 atmospheric pressure, the volume of helium gas is 10 litres. If volume and pressure are doubled, its temperature will change to

A. 127 K B. 400 K C. 25 K D. 200 K

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Explanation

$ PV = \mu RT \Rightarrow PV \alpha T $ If P and V doubled, then T becomes four times,

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