Physics 4 Marks Questions

1. (c) $\frac{\text{I}}{\text{c}}$

Explanation:

Pressure exerted by an electromagnetic radiation, $\text{P}=\frac{\text{I}}{\text{c}}.$

2. (d) $6\times10^{-4}\frac{\text{N}}{\text{m}^2}$

Explanation:

$\text{p}_\text{rad}=\frac{\text{Energy flux}}{\text{Speed of light}}=\frac{18\frac{\text{W}}{\text{cm}^2}}{3\times8^8\frac{\text{m}}{\text{s}}}$

$=\frac{18\times10^4\frac{\text{W}}{\text{cm}^2}}{3\times8^8\frac{\text{m}}{\text{s}}}=6\times10^{-4}\frac{\text{N}}{\text{m}^2}$

3. (a) 0.166 × 10^-8Nm^-2

Explanation:

$\text{p}=\frac{\text{I}}{\text{c}}=\frac{0.5}{3\times10^8}=0.166\times10^{-8}\text{Nm}^{-2}$

4. (b) $100$

Explanation:

Intensity of EM wave is given by $\text{I}=\frac{\text{P}}{4\pi\text{R}^2}$

$\text{V}_\text{av}=\frac{1}{2}\in_0\text{E}_0^2\times\text{c}$

$\Rightarrow\text{E}_0=\sqrt{\frac{\text{P}}{2\pi\text{R}^2\in_0\text{c}}}=\sqrt{\frac{1500}{2\times3.14(3)^2\times8.85\times10^{-12}\times3\times18^8}}$

$=\sqrt{10,000}=100\text{Vm}^{-1}$

5. (c) $10^{-6}\frac{\text{N}}{\text{m}^2}$

Explanation:

The radiation pressure of visible light $=7\times10^{-6}\frac{\text{N}}{\text{m}^2}$

1. (d) v_g = v_x = v_m

Explanation:

All electromagnetic waves travel in vacuum with the same speed.

2. (c) Cathode rays.

Explanation:

Cathode rays (beamofelectrons) get deflected in an electric field.

3. (c) Ionization chamber.

Explanation:

$\gamma-\text{rays}$ are detected by ionization chamber.

4. (b) 10¹⁴Hz

Explanation:

Size of particle $=\lambda=\frac{\text{c}}{\upsilon}$

$\upsilon=\frac{\text{c}}{\lambda}=\frac{3\times10^10\text{cm}\ \text{s}^{-1}}{3\times10^{-4}\text{cm}}=3\times10^{14}\text{Hz}$

5. (a) The radiation emitted is in the infrared region.

Explanation:

Every body at a temperature T > 0 K emits radiation in the infrared region.

1. (b) $\vec{\text{E}}\times\vec{\text{B}}$

Explanation:

Electromagnetic waves propagate in the direction of $\vec{\text{E}}\times\vec{\text{B}}.$

2. (a) Photon

Explanation:

Photon is the fundamental particle in an electromagnetic wave.

3. (a) Polarisation

Explanation:

Polarisation establishes the wave nature of electromagnetic waves.

4. (c) Frequency

Explanation:

Frequency u remains unchanged when a wave propagates from one medium to another. Both wavelength and velocity get changed.

5. (c) In phase and perpendicular to each other.

Explanation:

The electric and magnetic fields of an electromagnetic wave are in phase and perpendicular to each other.

1. (c) ML^-1T^-2

Explanation:

$\frac{1}{2}\in_0\text{E}^2=\text{energy density}=\frac{\text{Energy}}{\text{Volume}}$

$\therefore\big[\frac{1}{2}\in_0\text{E}^2\big]=\frac{\text{ML}^2\text{T}^{-2}}{\text{L}^3}=[\text{ML}^{-1}\text{T}^{-2}]$

2. (b) $[\in_0]=\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2$

Explanation:

As $\in_0=\frac{\text{q}_1\text{q}_2}{4\pi\text{FR}^2}$ (from Coulomb's law)

$\in_0=\frac{\text{C}_2}{\text{MLT}^{-2}\text{L}^{2}}=\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2$

3. (d) Wavelength is halved and the frequency remains unchanged.

Explanation:

The frequency of the electromagnetic wave remains same when it passes from one medium to another.

Refractive index of the medium, $\text{n}=\sqrt{\frac{\in}{\in_0}}=\sqrt{\frac{4}{1}}=2$

Wavelength of the electromagnetic wave in the medium,

$\lambda_\text{med}=\frac{\lambda}{\text{n}}=\frac{\lambda}{2}$

4. (b) $\gamma-\text{rays}$

Explanation:

$\beta-\text{rays}$ consists of electrons which are not electromagnetic in nature.

5. (b) The speed oflight c = 3 × 10⁸ms^-1 in free space.

Explanation:

The velocity of electromagnetic waves in free space (vacuum) is equal to velocity of light in vacuum.

(i.e., 3 × 10⁸ms^-1).

1. (b) Nfrared waves.

Explanation:

Infrared rays can be converted into electric energy as in solar cell.

2. (d) Radiowaves.

Explanation:

Radiowaves have longest wavelength.

3. (c) Cathode rays.

Explanation:

Cathode rays are invisible fast moving streams ofelectrons emitted by the cathode of a discharge tube which is maintained at a pressure of about 0.01 mm of mercury.

4. (c) $\gamma-\text{rays}.$

Explanation:

$\gamma-\text{rays}$ have minimum wavelength.

5. (a) Microwave, infrared, ultraviolet, gamma rays.

Explanation:

$\lambda_\text{micro}>\lambda_\text{infra}>\lambda_\text{ultra}>\lambda_\text{gamma}$

1. (d) $\vec{\text{E}}=\text{B}_0\text{c}\sin(\text{Kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$

Explanation:

Given: $\vec{\text{B}}=\text{B}_0\text{c}\sin(\text{Kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$

The relation between electric and magnetic field is,

$\text{c}=\frac{\text{E}}{\text{B}}$ or E = cB

The electric 6 eld component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along z-axis is obtained as $\vec{\text{E}}=\text{cB}_0\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}.$

2. (c) $\frac{2\in_0}{\text{c}}\hat{\text{j}}\sin\text{kz}\sin\omega\text{t}$

Explanation:

$\frac{\text{dE}}{\text{dz}}=-\frac{\text{dB}}{\text{dt}}$

$\frac{\text{dE}}{\text{dz}}=-2\text{E}_0\text{k}\sin\text{kz}\cos\omega\text{t}=-\frac{\text{dB}}{\text{dt}}$

${\text{dB}}=+2\text{E}_0\text{k}\sin\text{kz}\cos\omega\text{t}{\text{dt}}$

${\text{B}}=+2\text{E}_0\text{k}\sin\text{kz}\int\cos\omega\text{t}{\text{dt}}$

$=+2\text{E}_0\frac{\text{k}}{\omega}\sin\text{kz}\sin\omega\text{t}$

$\frac{\text{E}_0}{\text{B}_0}=\frac{\omega}{\text{k}}=\text{c}$

$\text{B}=\frac{2\text{E}_0}{\text{c}}\sin\text{kz}\sin\omega\text{t}$

$\therefore\text{B}=\frac{2\text{E}_0}{\text{c}}\sin\text{kz}\sin\omega\text{t}\hat{\text{j}}$

E is along y-direction and the wave propagates along x-axis.

$\therefore$ B should be in a direction perpendicular to both x and y-axis.

3. (d) $0.021\mu\text{T}$

Explanation:

Here, $\vec{\text{E}}=6.3\hat{\text{j}};\text{c}=3\times10^8\frac{\text{m}}{\text{s}}$

The magnitude of B is:

${\text{Bz}}=\frac{\text{E}}{\text{c}}=\frac{6.3}{3\times10^8}$

$=2.1\times10^8\ \text{T}=0.021\mu\text{T}$

4. (d) $\text{E}_\text{y}=66\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),\text{B}_\text{z}=2.2\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$

Explanation:

Here: $\text{E}_0=66\text{Vm}^{-1},\text{E}_\text{y}=66\cos\omega\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),$

$\lambda=3\text{mm}=3\times10^{-3}\text{m},\text{k}=\frac{2\pi}{\lambda}$

$\frac{\omega}{\text{k}}=\text{c}\Rightarrow\omega=\text{ck}=3\times10^8\times\frac{2\pi}{3\times10^{-3}}$

or $\omega=2\pi\times10^{11}$

$\therefore\text{E}_\text{y}=66\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$

${\text{Bz}}=\frac{\text{E}_\text{y}}{\text{c}}=\Big(\frac{66}{3\times10^8}\Big)\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$

$=2.2\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$

5. (a) 3.1 × 10^-8T

Explanation:

At a particular point, E = 9.3Vm^-1

$\therefore$ Magnetic field at the same point $=\frac{9.3}{3\times10^8}$

= 3.1 × 10^-8T