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Physics MCQs for NEET — Practice Questions with Answers

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A parallel plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. The displacement current through this area is

i/2

i/4

i/8

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Explanation

At any instant, let charge on capacitor be Q

Electric field between plates, $\vec{E} = \frac{Q}{A ε_{0}}$

Flux through area $\frac{A}{2}, ϕ_{E} = \frac{E A}{2}$

Displacement current ,

$i_{d} = ε_{0} \frac{d ϕ_{E}}{d t} = \frac{1}{2} \frac{d Q}{d t} = \frac{i}{2}$

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The sun delivers $10^{4} W / m^{2}$ of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions (10 $\times$ 10) $m^{2}$ will be

$10^{4}$ W

$10^{5} W$

$10^{6} W$

$10^{7} W$

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Explanation

$P = I n t e n s i t y \times A r e a = 10^{4} \times 10 \times 10 = 10^{6} W$

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The sun delivers $10^{3} W / m^{2}$ of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions 8 m $\times$ 20 m will be

$2.56 \times 10^{4} W$

$6.4 \times 10^{5} W$

$4.0 \times 10^{5} W$

1.6 $\times 10^{5} W$

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Explanation

$P = I n t e n s i t y \times A r e a = 10^{3} \times 8 \times 20 = 1.6 \times 10^{5} W$

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The wave of wavelength 5900 $\overset{0}{A}$ emitted by any atom or molecule must have some finite total length which is known as coherence length. For sodium light, this length is 2.4 cm. The number of oscillations in this length will be

$4.068 \times 10^{8}$

$4.068 \times 10^{4}$

$4.068 \times 10^{6}$

$4.068 \times 10^{5}$

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Explanation

No. of oscillation = Wavelengths in coherence length

$n = \frac{2.4 c m}{5900 \overset{0}{A}} = 4.068 \times 10^{4}$

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A radio wave has a maximum electric field intensity of $10^{- 4} V m^{- 1}$ on arrival at a receiving antenna. The maximum magnetic flux density of such a wave is

zero

$3 \times 10^{4} T$

$5.8 \times 10^{- 9} T$

$3.3 \times 10^{- 13} T$

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Explanation

Magnetic flux density, $B = \frac{ϕ_{B}}{A}$

$B_{0} = \frac{E_{0}}{C} = \frac{10^{- 4}}{3 \times 10^{8}} = 3.3 \times 10^{- 13} T$

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The transmitting antenna of a radio station is mounted vertically. At a point 10 km due north of the transmitter , the peak electric field is $10^{- 3}$ V/m. The amplitude of the radiated magnetic field is

$3.33 \times 10^{- 10} T$

$3.33 \times 10^{- 12} T$

$10^{- 3} T$

$3 \times 10^{5} T$

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Explanation

$E_{0} = 10^{- 3} V / m B_{0} = \frac{E_{0}}{C} = \frac{10^{- 3}}{3 \times 10^{8}} = 3.33 \times 10^{- 12} T$

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A plane EM-wave of wave intensity of 10 $W / m^{2}$ strikes a small mirror of area 20 $c m^{2}$ , held perpendicular to the approaching wave. The radiation force on the mirror will be

$6.6 \times 10^{- 11} N$

$1.33 \times 10^{- 11} N$

$1.33 \times 10^{- 10} N$

$6.6 \times 10^{- 10} N$

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Explanation

For perfectly reflecting surface,

$F = \frac{2 U_{a v}}{C} [U_{a v} = A v g . e n e r g y d e n s i t y] = \frac{2 [I A]}{C} = \frac{2 \times 10 \times 20 \times 10^{- 4}}{3 \times 10^{8}} = 1.33 \times 10^{- 10} N$

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In a region of free space, the electric field at some instant of time is $\vec{E} = (80 \hat{i} + 32 \hat{j} - 64 \hat{k})$ V/m and the magnetic field is $\vec{B} = (0.2 \vec{i} + 0.08 \hat{j} + 0.29 \hat{k}) μ T$ . The Poynting vector for these fields is

$- 11.52 \hat{i} + 28.8 \hat{j}$

$- 28.8 \hat{i} + 11.52 \hat{i}$

$28.8 \hat{i} - 11.52 \hat{j}$

$11.52 \hat{i} - 28.8 \hat{j}$

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Explanation

$Pointing vector = (\vec{E} \times \vec{B}) \frac{1}{μ_{0}} = \frac{10^{- 6}}{μ_{0}} |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 80 & 32 & - 64 \\ 0.2 & 0.08 & 0.29 \end{matrix}| = 11.52 \hat{i} - 28.8 \hat{j}$

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A plane electromagnetic wave propagating in the x-direction has wavelength of 60 mm. The electric field is in the y-direction and its maximum magnitude is 33 V/ $m^{- 1}$ . The equation for the electric field as function of x and t is

11 sin $π$ (t - x/c)

33 sin $π \times 10^{10}$ (t - x/c)

33 sin $π$ (t - x/c)

11 sin $π \times 10^{11}$ (t - x/c)

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Explanation

$λ = 60 m m ω = 2 π f = 2 π (\frac{c}{λ}) = 2 π \times \frac{3 \times 10^{8}}{60 \times 10^{- 3}} = π \times 10^{10} rad / s E = E_{0} \sin [ω (t - x / c)] = 33 \sin [π \times 10^{10} (t - x / c)]$

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In an electromagnetic wave, the electric field oscillated sinusoidally with amplitude 48 V $m^{- 1}$ , the RMS value of the oscillating magnetic field will be :

$1.6 \times 10^{- 8} T$

$1.6 \times 10^{- 9} T$

$144 \times 10^{- 8} T$

$11.3 \times 10^{- 8} T$

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Explanation

$B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{- 7} T B_{rms} = \frac{B_{0}}{\sqrt{2}} = \frac{1.6 \times 10^{- 7}}{\sqrt{2}} = 11.3 \times 10^{- 8} T$

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