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Physics MCQs for NEET — Practice Questions with Answers

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The periodic time of a simple pendulum is 3.3 s. Now if the point of support of the pendulum starts moving along the vertically upward direction with a velocity $v = kt ( where k = 2.1 m/s^2 )$, then the new periodic time is……s. ${ Take g = 10 m/s^2 }$

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Explanation

Initial periodic time $T_1 = 2 \pi \sqrt { l \over g} …..(1) $ When pendulum moves along vertical direction, effective acceleration $ g_{eff} = g+a $ where ‘a’ inaccleration of pendulum. $now , a = { d \nu \over dt} = { d (kt) \over dt } = k = 2.1 ms^{-2} $ $ \therefore New periodic time T_2 = 2 \pi \sqrt { l \over g_{eff} } .....(2) $ $ \therefore { T_2 \over T_1 } = \sqrt { g \over g_{eff} }^1$

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A block is placed on a horizontal table. The table executes S.H.M. along the horizontal plane with a period T. The coefficient of static friction between the table and block is $ \mu $ . The maximum amplitude of oscillation should be...... so that the block does not slide off the table.

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Explanation

Block will not slide if $ \mu mg \geq ma \Rightarrow \mu g \geq a $ To prevent the block from sliding the maximum acceleration of table must be $a_max = \mu g$ Now maximum accleration $ a_{max} = \omega^2 A $ $ \mu^2 A_{max} = \mu g $ $ \therefore A_{max} = { \mu g \over \omega^2 }= { \mu g T^2 \over 4 \pi^2 } $

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A horizontal plank is executing SHM along the vertical direction with angular frequency ù. A coin is placed on top of this plank. If the amplitude of oscillation is increased gradually, for what maximum amplitude will the coin be on the verge of loosing contact with the plank?

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Explanation

At the upper most end, $ when mg = R + m \omega^2A$ coin will loose contact. Taking R=0 $ m \omega^2 A = mg $ $ A = { g / \omega ^ 2} $

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 For the following questions, statement as well as the reason(s) are given.

questionshas four options. Select the correct option.

Statement – 1 : If a spring having spring constant k is divided into equal parts, then the spring constant of each part will be 2k.

Statement – 2 : When the length of the elastic spring is increased ( stretched ) byx, then the amount of work required to be done is $ 1/2 kx^2 $

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Explanation

Force required to increase the length by x in F = kx….. (1) After spring is divided into 2 equal parts, F = k ' x ' where x ' = x/2 = k' x/2 …..(2) From (1) and (2) ; k' = 2 k

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Assertion – Reason type questions : For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. Statement – 1 : The periodic time of a S.H.O. depends on its amplitude and force constant. Statement – 2 : The elasticity and inertia decides the frequency of S.H.O.

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Explanation

Frequency of SHM depends on elasticity & inertia.

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Assertion – Reason type questions : For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. Statement – 1 : For small amplitude, the motion of a simple pendulum is a S.H.M. with periodic time $ T = 2 \pi \sqrt { l \over g} $. . For large amplitudes, periodic time is greater than $ 2 \pi \sqrt { l \over g } $ Statement – 2 : For large amplitude, the speed of the bob is more when it passes through the mid-point ( equilibrium point ).

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Explanation

$ Restoring force F = - mg sin \theta OR $ $ F = - mg _e where g_e = g sin \theta $ $ F = - mg _e where g_e = g sin \theta $ $ If \theta is small , sin \theta \approx \theta $ $ \therefore Effective value of g is g_e \theta $ $ For large oscillation, g sin \theta \lt g \theta ( sin \theta \lt \theta ) $ $ \therefore T \gt 2 \pi \sqrt { l^1 \over g } $

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Assertion – Reason type questions : For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option Statement – 1 : Periodic time of a simple pendulum is independent of the mass of the bob. Statement – 2 : The restoring force does not depend on the mass of the bob.

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Explanation

$ Restoring force F= - mg sin \theta $ which depends on “m”

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Assertion – Reason type questions : For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. Statement – 1: The periodic time of a simple pendulum increases on the surface of moon. Statement – 2 : Moon is very small as compared to Earth.

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Explanation

“g” in less on moon $ \therefore form the equation T = 2 \pi \sqrt { l^1 \over g} $

T will increase As compared to earth, moon in small

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 For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option.

Statement – 1: If the length of a simple pendulum is increased by 3%, then the periodic time changes by 1.5%.

Statement – 2 : Periodic time of a simple pendulum is proportional to its length.

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Explanation

$ periodic \; time \,T \alpha \sqrt l $ $ \therefore \triangle T = {1 \over 2 \sqrt l } . \triangle l\; \{ On differentiation \}$ $ \therefore { \triangle T \over T } = 1.5 \% $

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Assertion – Reason type questions : For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. Statement – 1: For a particle executing S.H.M. with an amplitude of 0.01 m and frequency 30 hz, the maximum acceleration is $36p^2 m/s^2 $. Statement – 2 : The maximum acceleration for the above particle is $ \pm ù2A$, where A is amplitude.

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Explanation

$ a_{max} = \omega^2 A = 4 \pi ^2 f^2 A = 4 \pi^2 (30) ^2 \times 0.01 $ Since the oscillator moves between + A& -A, maximum acceleration $ = \pm 36 \pi^2 $

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