NEETshala
Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology
Register free to filter questions
● ● ● ● ●

A trolley is sliding down a frictionless slope having inclination è. If a simple pendulum is suspended on top of this trolley, its periodic time is given by

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ g^2 _{off} = a_x^2 + (g -ay ) ^2 here a_x g sin \theta cos \theta , ay = g sin ^2 \theta $ $ = a_x^2 + g^2 + a_y^2 - 2 ga_y $ $ = a^2 sin ^ 2 \theta cos ^2 \theta + g^2 + g^2 sin^2 \theta - 2 g^2 sin ^2 \theta $ $ = g^2 ( 1 -sin^2 \theta ) $ $= g ^2 cos ^2 \theta $ $ \therefore g_{eff} = g cos \theta $

● ● ● ● ●

A system is executing S.H.M. The potential energy of the systemfor displacement x is $E_1$ and for a displacement of y, the potential energy of the system is $E_2$. The potential energy for a displacement of (x+y) is ………

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ E_1 = {1 \over 2 } m \omega^2 x^2 \Rightarrow \sqrt E_1 = x \sqrt { {1 \over 2 } m \omega^2 } ...(1) $ $ E_2 = {1 \over 2 } m \omega^2 y^2 \Rightarrow \sqrt E_2 = x \sqrt { {1 \over 2 } m \omega^2 } ...(2) $ $ E = {1 \over 2 } m \omega^2 (x + y )^2 \Rightarrow \sqrt E = (x + y ) \sqrt { {1 \over 2 } m \omega^2 } ...(3) $ From (1) ,(2) ,(3) , $ \sqrt E = \sqrt E_1 + \sqrt E_2 $ $or E = E_1 + E_2 + 2 \sqrt { E_1 E_2 } $

● ● ● ● ●

A system is executing S.H.M. with a periodic time of 4/5 s under the influence of force $F_1$. When a force $F_2$ is applied, the periodic time is (2/5) s. Now if $F_1$ and $F_2$ are applied simultaneously along the same direction, the periodic time will be ………

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ \omega_1^2 = { k \over m } = { kx \over mx } = { F_1 \over mx} ...(1) $ $ similarly , \omega_2^2 = { F_2 \over mx} ...(2) $ $ if F_1 and F_2 acts simultaneously ,then angular frequency $ $ w_2 ={ F_1 + F_2 \over mx } .....(3) $ $ From (1) , (2) and (3) ; \omega^2 = \omega_1^2 + \omega_2^2 $ $ now ,use eqn \omega = { 2 \pi \over T} $

● ● ● ● ●

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. ${ Take g = 10 m/s^2 }$

You've reached today's free limit of 20 questions. Log in to keep practising for free.
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$

● ● ● ● ●

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.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
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 } $

● ● ● ● ●

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?

You've reached today's free limit of 20 questions. Log in to keep practising for free.
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} $

● ● ● ● ●

 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 $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
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

● ● ● ● ●

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.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Frequency of SHM depends on elasticity & inertia.

● ● ● ● ●

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 ).

You've reached today's free limit of 20 questions. Log in to keep practising for free.
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 } $

● ● ● ● ●

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.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

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

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

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.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

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