Blog 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

Language English हिंदी

● ● ● ● ●

A wave of the frequency of 500 Hz has a velocity of 360 m/sec. The distance between two nearest points 60° out of phase, is :

0.6 cm

12 cm

60 cm

120 cm

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

Explanation

The distance between two points i.e. path difference between them $Δ = \frac{λ}{2 π} \times ϕ = \frac{λ}{2 π} \times \frac{π}{3} = \frac{λ}{6} = \frac{v}{6 n}$ $(∵ v = n λ)$

⇒ $Δ = \frac{360}{6 \times 500} = 0 .12 m = 12 c m$

● ● ● ● ●

Sound waves of wavelength greater than that of audible sound are called :

Seismic waves

Sonic waves

Ultrasonic waves

Infrasonic waves

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

Explanation

Infrasonic waves have frequency less than (20 Hz) audible sound and wavelength more than audible sound.

● ● ● ● ●

The ratio of masses of nitrogen and oxygen is 14:16. The temperature at which the speed of sound in nitrogen will be same at that in oxygen at 55°C is -

35°C

48°C

65°C

14°C

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

Explanation

$v = \sqrt{\frac{γ R T}{M}}$ ⇒ $\frac{T_{N}}{T_{0}} = \frac{M_{N}}{M_{0}}$ ⇒ $\frac{T_{N}}{273 + 55} = \frac{14}{16} = \frac{7}{8}$

⇒ $T_{N} = 287 K = 14 ° C$

● ● ● ● ●

A wavelength 0.60 cm is produced in air and it travels at a speed of 300 ms^–1. It will be an

Audible wave

Infrasonic wave

Ultrasonic wave

None of the above

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

Explanation

$n = \frac{v}{λ} = \frac{300}{0 .6 \times 10^{- 2}} H z$ $= \frac{3}{6} \times 10^{5} H z = 50, 000 H z$

⇒ Wave is ultrasonic.

● ● ● ● ●

An observer standing near the seashore observes 54 waves per minute. If the wavelength of the water wave is 10m then the velocity of a water wave is :

540 ms^-1

5.4 ms^-1

0.184 ms^-1

9 ms^-1

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

Explanation

Number of waves per minute = 54

∴ Number of waves per second = 54/60

Now v = nλ ⇒ $n = \frac{54}{60} \times 10 = 9 m / s .$

● ● ● ● ●

The equation of a wave is $y = 2 \sin π (0 .5 x - 200 t)$ , where x and y are expressed in cm and t in sec. The wave velocity is :

100 cm/sec

200 cm/sec

300 cm/sec

400 cm/sec

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

Explanation

Comparing given equation with standard equation of progressive wave. The velocity of wave

$v = \frac{ω (Co-efficient of t)}{k (Co-efficient of x)} = \frac{200 π}{0 .5 π} = 400 c m / s$

● ● ● ● ●

Equation of a progressive wave is given by

$y = 0 .2 \cos π (0 .04 t + . 02 x - \frac{π}{6})$

The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of π/2 :

4 cm

8 cm

25 cm

12.5 cm

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

Explanation

Comparing with $y = a \cos (ω t + k x - ϕ)$ ,

We get $k = \frac{2 π}{λ} = 0 .02 π \Rightarrow λ = 100 c m$

Also, it is given that phase difference between particles $Δ ϕ = \frac{π}{2} .$

Hence path difference between them $Δ = \frac{λ}{2 π} \times Δ ϕ = \frac{λ}{2 π} \times \frac{π}{2} = \frac{λ}{4} = \frac{100}{4} = 25 c m$

● ● ● ● ●

Two waves are given by $y_{1} = a \sin (ω t - k x)$ and $y_{2} = a \cos (ω t - k x)$ . The phase difference between the two waves is :

$\frac{π}{4}$

$\frac{π}{8}$

$\frac{π}{2}$

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

Explanation

$y_{1} = a \sin (ω t - k x)$

and $y_{2} = a \cos (ω t - k x) = a \sin (ω t - k x + \frac{π}{2})$

Hence phase difference between these two is $\frac{π}{2} .$

● ● ● ● ●

If the amplitude of waves at distance r from a point source is A, the amplitude at a distance 2r will be :

A/2

A/4

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

Explanation

$I \propto A^{2} \propto \frac{1}{d^{2}}$ ⇒ $a \propto \frac{1}{d}$

● ● ● ● ●

A wave is reflected from a rigid support. The change in phase on reflection will be :

π/4

π/2

2π

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

Explanation

After reflection from rigid support, a wave suffers a phase change of π.

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.