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 standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance 1.21 Å between them. The wavelength of the standing wave is :

1.21 Å

2.42 Å

6.05 Å

3.63 Å

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

● ● ● ● ●

In stationary waves, the distance between a node and its nearest antinode is 20 cm. The phase difference between two particles having a separation of 60 cm will be :

Zero

π/2

3π/2

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

Explanation

$\frac{λ}{4} = 20 \Rightarrow λ = 80 c m$ , also $Δ ϕ = \frac{λ}{2 π} . Δ x$

⇒ $Δ ϕ = \frac{60}{80} \times 2 π = \frac{3 π}{2}$

● ● ● ● ●

A standing wave is represented by

$Y = A \sin (100 t) \cos (0 .01 x)$

where Y and A are in millimetre, t is in seconds and x is in metre. The velocity of the wave is :

10⁴ m/s

1 m/s

10^–4m/s

Not derivable from the above data

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

Explanation

By comparing given equation with $y = a \sin (ω t) \cos k x$

⇒ $v = \frac{ω}{k} = \frac{100}{0 .01} = 10^{4} m / s$

● ● ● ● ●

Two waves are approaching each other with a velocity of 20 m/s and frequency n. The distance between two consecutive nodes is :

$\frac{20}{n}$

$\frac{10}{n}$

$\frac{5}{n}$

$\frac{n}{10}$

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

Explanation

Distance between the consecutive node $= \frac{λ}{2},$

but $λ = \frac{v}{n} = \frac{20}{n}$ so $\frac{λ}{2} = \frac{10}{n}$

● ● ● ● ●

The stationary wave produced on a string is represented by the equation $y = 5 \cos (π x / 3) \sin 40 π t$ where x and y are in cm and t is in seconds. The distance between consecutive nodes is :

5 cm

π cm

3 cm

40 cm

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

Explanation

On comparing the given equation with standard equation $\Rightarrow \frac{2 π}{λ} = \frac{π}{3} \Rightarrow λ = 6 c m$ .

Hence, distance between two consecutive nodes ⇒ λ = 3 cm

● ● ● ● ●

The following equations represent progressive transverse waves $Z_{1} = A \cos (ω t - k x)$ , $Z_{2} = A \cos (ω t + k x)$ , $Z_{3} = A \cos (ω t + k y)$ and $Z_{4} = A \cos (2 ω t - 2 k y)$ . A stationary wave will be formed by superposing :

Z₁ and Z₂

Z₁ and Z₄

Z₂ and Z₃

Z₃ and Z₄

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

Explanation

When two waves of equal frequency and traveling in the opposite direction superimpose, then the stationary wave is produced. Hence Z₁ and Z₂ produce stationary waves.

● ● ● ● ●

Two traveling waves $y_{1} = A \sin [k (x - c t)]$ and $y_{2} = A \sin [k (x + c t)]$ are superimposed on the string. The distance between adjacent nodes is :

ct / π

ct / 2π

π / 2k

π / k

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

Explanation

The distance between adjacent nodes $x = \frac{λ}{2}$

Also, $k = \frac{2 π}{λ} .$

Hence, $x = \frac{π}{k} .$

● ● ● ● ●

A string fixed at both ends is vibrating in two segments. The wavelength of the corresponding wave is :

$\frac{l}{4}$

$\frac{l}{2}$

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

Explanation

When a string fixed at both ends vibrates in two segments, it means there is one node at the center. The wavelength of the standing wave is twice the length of the string segment, which is equal to the full length of the string (l).

● ● ● ● ●

A 1 cm long string vibrates with the fundamental frequency of 256 Hz. If the length is reduced to $\frac{1}{4} c m$ keeping the tension unaltered, the new fundamental frequency will be :

256

512

1024

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

Explanation

$n \propto \frac{1}{l}$ ⇒ $\frac{n_{2}}{n_{1}} = \frac{l_{1}}{l_{2}}$

⇒ $n_{2} = \frac{l_{1}}{l_{2}} n_{1} = \frac{1 \times 256}{1 / 4} = 1024 H z$

● ● ● ● ●

Standing waves are produced in a 10 m long stretched string. If the string vibrates in 5 segments and the wave velocity is 20 m/s, the frequency is :

2 Hz

4 Hz

5 Hz

10 Hz

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

Explanation

String vibrates in five segment so $\frac{5}{2} λ = l$ ⇒ $λ = \frac{2 l}{5}$

Hence $n = \frac{v}{λ} = 5 \times \frac{v}{2 l} = 5 \times \frac{20}{2 \times 10} = 5 H z$

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.