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

Chapter Language Difficulty Show only questions I answered incorrectly Register free to filter questions

● ● ● ● ●

A convex lens of focal length f is placed somewhere in between an object and a screen. The distance between the object and the screen is x. If the numerical value of the magnification product by the lens is m, What is the focal length of the lens ?

A. $ {mx \over (m -1 ) ^2 }$ B. $ {mx \over (m +1 ) ^2 }$ C. $ { (m-1)^2 \over m } x $ D. $ { (m+1)^2 \over m } x $

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

Explanation

x = u + v $ m = { f \over ( f + u ) } = { (f - v ) \over f } $ $ u = { -(m+1) \over m } f $ $ v = (m+1) f $ solved the equation

● ● ● ● ●

A convex lens of focal length f produces a real image x times the size of an object, Then what is the distance of the object fromthe lens ?

A. (x +1 ) f B. (x -1 ) f C. $ \left( { x+1 \over x } \right) f $ D. $ \left( { x-1 \over x } \right) f $

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

Explanation

$ v = f \left( 1 -{ v\over u } \right) = f (1 -x ) $ $ \therefore u = f { (1+x ) \over x } $

● ● ● ● ●

A thin lens has focal length f, and its aperture has diameter d. It forms an image of intensity I. Now, the central part of the aperture upto diameter d/2 is blocked by an opeque paper. The focal length and image intensity will change to.....

A. f and 3I / 4 B. 3f / 4 and I / 2 C. f and I / 4 D. f /2 and I / 2

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

Explanation

Due to blocking of central part, focal lenght does not change. However, theintensity decreases. The amount of light crossing the lens decreases by a factor of $ {\pi ( { d \over 2 }) ^2 \over \pi d^2 } = { 1 \over 4 } , Hence I' = I -{I \over 4 } = { 3I \over 4 } $

● ● ● ● ●

The distance between object and the screen is D. Real images of an object are formed on the screen two positions of a lens seperated by a distance d. What will be the ratio between the sizes of two images ?

A. $ {D^2 \over d^2 } $ B. $ {D \over d } $ C. $ \sqrt {D\over d } $ D. $ {(D-d^2) \over(D+ d)^2 } $

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

Explanation

$ m_1 = { I _1 \over 0 } = { v \over u } = { D+d \over D-d} $ $ m_2 = { I _2 \over 0 } = { v \over u } = { D+d \over D-d} $ $ \therefore {I_1 \over I_2 } = {(D+d)^2 \over (D-d)^2 } or { I_2 \over I_1} = { (D-d)^2 \over (D+d)^2 }$

● ● ● ● ●

A spherical mirror forms an erect image three times the linear size of the object. If the distance between the object and the image is 80 cm, What is the focal length of the mirror ?

A. 30cm B. 40cm C. -15cm D. 15cm

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

Explanation

Knowledge base

● ● ● ● ●

A short linear object of length L lies on the axis of a spherical mirror of focal length of fat a distance u from the mirror. Its image has an axial length L` equal to ................

A. $ L \left[ { f \over ( u -f ) }\right] ^2 $ B. $ L \left[ {(u-f ) \over f }\right] ^2 $ C. $ L \left[ {(u+f ) \over f }\right] ^{1/2} $ D. $ L \left[ {f \over u -f }\right] ^2 $

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

Explanation

for a spherical mirrer $ v ={ uf \over u-f } â€¦(1) $ The value of u for the two ends of the object of length are $ u _1 = \left( u - {L \over 2 } \right) and u_2 = \left(u + {L \over 2 } \right) ....(2) $ solved the eqn (1) and (2)

● ● ● ● ●

A concave mirror of focal length f produces an images n times the size of the object. If the image is real then What is the distance of the object from the mirror ?

A. (n+1 ) f B. $ \left[ { n-1 \over n } \right] f $ C. (n-1)f D. $ \left[ { n+1 \over n } \right] f $

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

Explanation

$ u = ({ n+1 \over n } ) f $

● ● ● ● ●

An object is placed at a distance of f/2 , the from a convex lens the image will be

A. at f, real and inverted B. at , 3f/2 real and inverted C. at one of the foci, virtual and double its size D. at 2f, virtual and erect.

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

Explanation

$ {1 \over v } - {1 \over u } = {1 \over f} Here , u = { -f \over 2 } $

● ● ● ● ●

Angle of prism is A and its one surface is silvered. Light rays falling at an angle at incidence 2Aon first surface return back through the same path after suffering reflection at second silvered surface .What is the refractive index of material ?

A. tanA B. 2sinA C. 2cosA D. cos A/2

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

● ● ● ● ●

The minimum deviation produced by a glass prism of angle 60Â° is 30Â°. If the velocity of light in vaccum is $3 \times 10^8 m/s$. Then what is the velocity of light in glass in m/s ?

A. $ 2.72 \times 10^8 $ B. $ 3.1 \times 10^8 $ C. $ 2.9 \times 10^8 $ D. $ 2.121 \times 10^8 $

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

Explanation

$ n = { sin \left( {A + \delta m \over 2 } \right) \over sin {A \over 2 } }$ $ velocity of light in glass = C_{air } \over n $

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.