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

● ● ● ● ●

The rate of change of the voltage of a parallel plate capacitor if the instantaneous displacement current of 1 A is established between the two plates of a 1 $μ$ F parallel plate capacitor:

$10^{6} V / s$

10 V/s

$10^{8} V / s$

$10^{- 6} V / s$

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

Explanation

$I_{D} = C \frac{d V}{d t} \frac{d V}{d t} = \frac{1}{1 \times 10^{- 6}} = 10^{6} V / s$

● ● ● ● ●

The relation between electric field E and magnetic field induction B in an electromagnetic waves

$E = \sqrt{\frac{μ_{0}}{ε_{0}}} B$

E = cB

E = $\frac{B}{c}$

$E = \frac{B}{c^{2}}$

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

Explanation

$E = cB$ is the correct relation.

● ● ● ● ●

An electromagnetic wave is propagating along Y-axis. Then:

The oscillating electric field is along X-axis and the oscillating magnetic field is along Y-axis.

The oscillating electric field is along Z-axis and the oscillating magnetic field is along X-axis.

Both oscillating electric and magnetic fields are along Y-axis, but the phase difference between them is $90^{°}$

Both oscillating electric and magnetic fields are mutually perpendicular in arbitrary directions.

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

Explanation

The oscillating electric field is along Z-axis and the oscillating magnetic field is along X-axis.

● ● ● ● ●

In electromagnetic wave the phase difference between electric and magnetic field vectors $\vec{E} a n d \vec{B}$ is :

$π / 2$

$π$

$π / 4$

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

Explanation

The electric and magnetic field components of a linearly polarized electromagnetic wave oscillate in such a way that they peak at the same time and they become zero at the same time but they point to different directions in space, separated by an angle of $90^{0} .$

Since there is no time difference between the peaks of the electric and magnetic oscillations the phase difference between the electric and magnetic field vectors of a linearly polarized electromagnetic wave is zero.

● ● ● ● ●

An electromagnetic wave going through the vacuum is described by $E = E_{0} \sin (k x - ω t)$

Which is the following is/are independent of the wavelength?

$k / ω$

$k ω$

$ω$

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

Explanation

$k = \frac{2 π}{λ} a n d ω = \frac{2 π c}{λ} H e n c e, \frac{k}{2 π} = \frac{ω}{2 π c} \Rightarrow \frac{k}{ω} i n d e p e n d e n t o f w a v e l e n g t h .$

● ● ● ● ●

In a plane E.M. wave, the electric field oscillates sinusoidally at a frequency of 2.5 $\times 10^{10}$ Hz and amplitude 480 V/m. The amplitude of the oscillating magnetic field will be:

$1.52 \times 10^{- 8} W b / m^{2}$

$1.52 \times 10^{- 7} W b / m^{2}$

$1.6 \times 10^{- 6} W b / m^{2}$

$1.6 \times 10^{- 7} W b / m^{2}$

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

Explanation

$(3) A m p l i t u d e o f M a g n e t i c f i e l d = \frac{E_{0}}{c} = \frac{480}{3 \times 10^{8}} = 1.6 \times 10^{- 6} W b / m^{2}$

● ● ● ● ●

The energy density of the electromagnetic wave in vacuum is given by the relation

$\frac{1}{2} . \frac{E^{2}}{ε_{0}} + \frac{B^{2}}{2 μ_{0}}$

$\frac{1}{2} ε_{0} E^{2} + \frac{1}{2} μ_{0} B^{2}$

$\frac{E^{2} + B^{2}}{C}$

$\frac{1}{2} ε_{0} E^{2} + \frac{B^{2}}{2 μ_{0}}$

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

Explanation

$E l e c t r i c e n e r g y d e n s i t y = \frac{1}{2} ε_{0} E^{2} M a g n e t i c e n e r g y d e n s i t y = \frac{B^{2}}{2 μ_{0}} S o, t o t a l e n e r g y = \frac{1}{2} ε_{0} E^{2} + \frac{B^{2}}{2 μ_{0}}$

● ● ● ● ●

A lamp radiates power $P_{0}$ uniformly in all directions, the amplitude of electric field strength $E_{0}$ at a distance r from it is:

$E_{0} = \frac{P_{0}}{2 {πε}_{0} {cr}^{2}}$

$E_{0} = \sqrt{\{\frac{P_{0}}{2 {πε}_{0} c r^{2}}\}}$

$E_{0} = \sqrt{\{\frac{P_{0}}{4 {πε}_{0} {cr}^{2}}\}}$

$E_{0} = \sqrt{\{\frac{P_{0}}{8 {πε}_{0} {cr}^{}}\}}$

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

Explanation

$S_{a v} = \frac{P}{4 {πr}^{2}} = \frac{1}{2} ε_{0} {E_{0}}^{2} c E_{0} = \sqrt{\frac{P}{2 π ε_{0} c r^{2}}}$

● ● ● ● ●

The intensity of visible radiation at a distance of 1 m from a bulb of 100 W which converts only 5% its power into light is :

0.4 $W / m^{2}$

0.5 $W / m^{2}$

0.1 $W / m^{2}$

0.01 $W / m^{2}$

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

Explanation

$I n e n s i t y = \frac{P o w e r}{A r e a} = \frac{0.05 \times 100}{4 {πr}^{2}} = \frac{5}{4 \times 3.14} = 0.4 W$ $/ m^{2}$

● ● ● ● ●

On an EM wave, the amplitude of electric and magnetic fields are 100 v/m and 0.265 A/m. The maximum energy flow is

26.5 $W / m^{2}$

46.7 $W / m^{2}$

66.5 $W / m^{2}$

86.5 $W / m^{2}$

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

Explanation

$M a x i m u m E n e r g y F l o w, S = E_{0} B_{0} = 100 \times 0.265 = 26.5 W / m^{2}$

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.