Blog Log in Get Started Free

Physics MCQs for NEET — Practice Questions with Answers

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

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Register free for difficulty & keyword filters

● ● ● ● ●

The mean radius of the earth is R, its angular speed on its own axis is $ω$ and the acceleration due to gravity at earth's surface is g. The cube of the radius of the orbit of a geostationary satellite will be -

$R^{2} g / ω$

$R^{2} ω^{2} / g$

$R g / ω^{2}$

$R^{2} g / ω^{2}$

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

Explanation

(d) Orbital velocity $v_{o} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{g R^{2}}{r}}$ and $v_{o} = r ω$

This gives $r^{3} = \frac{R^{2} g}{ω^{2}}$

● ● ● ● ●

A satellite whose mass is M, is revolving in circular orbit of radius r around the earth. Time of revolution of satellite is

$T \propto \frac{r^{5}}{G M}$

$T \propto \sqrt{\frac{r^{3}}{G M}}$

$T \propto \sqrt{\frac{r}{G M^{2} / 3}}$

$T \propto \sqrt{\frac{r^{3}}{G M^{2} / 4}}$

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

Explanation

(b) $T = \frac{2 πr}{v} . . . (1) a n d \frac{m v^{2}}{r} = \frac{G m_{e} M}{r^{2}} \Rightarrow v^{2} = \frac{G m_{e}}{r} o r, v = \sqrt{\frac{G m_{e}}{r}} . . . (2)$

Putting the value of v from equation (2) in equation (1)

$T = \frac{2 πr}{\sqrt{\frac{G m_{e}}{r}}} = 2 π \sqrt{\frac{r^{3}}{{Gm}_{e}}} \Rightarrow T α \sqrt{\frac{r^{3}}{{GM}_{e}}}$

● ● ● ● ●

The periodic time of a communication satellite is

6 hours

12 hours

18 hours

24 hours

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

Explanation

(d) Communication satellite is a kind of geostationary satellite so its time period is 24 hours.

● ● ● ● ●

Suppose the gravitational force varies inversely as the $n^{th}$ power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to -

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

$R (\frac{n - 1}{2})$

$R^{n}$

$R (\frac{n - 2}{2})$

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

Explanation

(a) ${mω}^{2} R \propto \frac{1}{R^{n}} \Rightarrow m (\frac{4 π^{2}}{T^{2}}) R \propto \frac{1}{R^{n}} \Rightarrow T^{2} \propto R^{n + 1} ∴ T \propto R^{(\frac{n + 1}{2})}$

● ● ● ● ●

The orbital speed of an artificial satellite very close to the surface of the earth is $V_{o}$ . Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is

$4 V_{0}$

$2 V_{0}$

$0.5 V_{0}$

$4 V_{0}$

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

Explanation

If $r = R + h = R + 3 R = 4 R$ then $v = \frac{V_{o}}{2} = 0.5 V_{0}$

● ● ● ● ●

If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would -

Decrease by 2%

Remain unchanged

Increase by 2%

Increase by 1%

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

Explanation

% increase in g = 2(% decrease in R) = 2x1% = 2%

● ● ● ● ●

The distance of a geo-stationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to -

5 R

7 R

10 R

18 R

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

Explanation

(b) 6R from the surface of earth and 7R from the centre.

● ● ● ● ●

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ${ms}^{- 2}$ and radius of earth is 6400 kms)

0 rad/sec

$\frac{1}{800}$ rad/sec

$\frac{1}{80}$ rad/sec

$\frac{1}{8}$ rad/sec

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

Explanation

(b) $g' = g - ω^{2} {Rcos}^{2} λ$

For weightlessness at equator $λ = 0$ and $g' = 0$

$∴ 0 = g - ω^{2} R \Rightarrow ω = \sqrt{\frac{g}{R}} = \frac{1}{800} \frac{rad}{s}$

● ● ● ● ●

A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of is $T_{2} / T_{1}$ is

$\sqrt{2}$

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

Explanation

(d) If acceleration due to gravity is g at the surface of earth then at height R its value becomes $g' = g {(\frac{R}{R + h})}^{2} = \frac{g}{4}$

$T_{1} = 2 π \sqrt{\frac{l}{g}}$ and $T_{2} = 2 π \sqrt{\frac{l}{g / 4}}$

$∴ \frac{T_{2}}{T_{1}} = 2$

● ● ● ● ●

A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be

mgR

$\frac{1}{2}$ mgR

2 mgR

$\frac{1}{4}$ mgR

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

Explanation

(b) $∆ U = \frac{mgh}{1 + \frac{h}{R}} = \frac{1}{2} mgR (∵ h = R)$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Physics 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.