JEE Electromagnetic Waves Questions

The intensity of a plane electromagnetic wave is related to the peak electric-field amplitude $$E_0$$ by

$$I = \frac{1}{2}\,\epsilon_0\,c\,E_0^{2}$$ $$-(1)$$

Solving $$-(1)$$ for $$E_0$$ gives

$$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$$ $$-(2)$$

Thus the expression whose unit we must find is simply the unit of the electric field $$E_0$$.

Unit analysis:

Intensity  $$I$$ : power per area   $$\Rightarrow$$  $$\text{W m}^{-2} = \left(\text{J s}^{-1}\right)\text{m}^{-2} = \text{kg s}^{-3}$$

Permittivity  $$\epsilon_0$$ : $$\text{C}^{2}\,\text{N}^{-1}\,\text{m}^{-2}$$ Force  $$\text{N}= \text{kg m s}^{-2}$$ $$\therefore \epsilon_0 = \frac{\text{C}^{2}}{\text{kg m s}^{-2}\,\text{m}^{2}} = \frac{\text{C}^{2}\,\text{s}^{2}}{\text{kg m}^{3}}$$

Speed of light  $$c$$ : $$\text{m s}^{-1}$$

Compute the unit of $$\dfrac{2I}{\epsilon_0 c}$$ :

$$ \dfrac{\text{kg s}^{-3}} {\left(\dfrac{\text{C}^{2}\,\text{s}^{2}}{\text{kg m}^{3}}\right)\;(\text{m s}^{-1})} = \dfrac{\text{kg s}^{-3} \;\text{kg m}^{3}} {\text{C}^{2}\,\text{s}^{2}\;\text{m}} = \dfrac{\text{kg}^{2}\,\text{m}^{2}\,\text{s}^{-4}}{\text{C}^{2}} $$

Taking the square root (see $$-(2)$$) gives the unit of $$E_0$$:

$$ \sqrt{\dfrac{\text{kg}^{2}\,\text{m}^{2}\,\text{s}^{-4}}{\text{C}^{2}}} = \dfrac{\text{kg m s}^{-2}}{\text{C}} = \dfrac{\text{N}}{\text{C}} $$

Hence $$\sqrt{\dfrac{2I}{\epsilon_0 c}}$$ has the SI unit $$\mathbf{N\,C^{-1}}$$, which is also equal to $$\mathbf{V\,m^{-1}}$$.

Answer : Option D  $$NC^{-1}$$

The source emits light equally in all directions, so the energy spreads over the surface of a sphere of radius $$r$$.

Energy flux (intensity) at distance $$r$$ is given by the inverse-square law:
$$I = \frac{P}{4 \pi r^{2}}$$ where $$P$$ is the power of the source.

Substitute $$P = 450 \text{ W}$$ and $$r = 2 \text{ m}$$:
$$I = \frac{450}{4 \pi (2)^{2}} = \frac{450}{16 \pi}$$ W m$$^{-2}$$.

Numerical value:
$$I = \frac{450}{50.265} \approx 8.96 \text{ W m}^{-2}$$.

For a perfectly reflecting surface, radiation pressure is twice that for a perfectly absorbing surface. The formula is
$$P_{\text{rad}} = \frac{2I}{c}$$ where $$c = 3 \times 10^{8} \text{ m s}^{-1}$$.

Insert the value of $$I$$:
$$P_{\text{rad}} = \frac{2 \times 8.96}{3 \times 10^{8}}$$ Pa
$$= \frac{17.92}{3 \times 10^{8}} \text{ Pa}$$.

Simplify:
$$P_{\text{rad}} \approx 5.97 \times 10^{-8} \text{ Pa} \approx 6 \times 10^{-8} \text{ Pa}$$.

Therefore, the radiation pressure on the perfectly reflecting surface is $$6 \times 10^{-8}$$ Pascals.

Correct option: Option C.

We need to evaluate the Assertion and Reason about electromagnetic waves.

Assertion (A): "Electromagnetic waves carry energy but not momentum."

Analysis: This is FALSE. Electromagnetic waves carry both energy and momentum. The momentum of an electromagnetic wave is related to its energy by $$p = E/c$$, where $$c$$ is the speed of light. This momentum is responsible for radiation pressure. Experiments such as Nichols radiometer have confirmed that light exerts pressure on surfaces, proving that EM waves carry momentum.

Reason (R): "Mass of a photon is zero."

Analysis: This is TRUE. A photon has zero rest mass. However, it still carries momentum given by $$p = h/\lambda = E/c$$, as described by the relativistic energy-momentum relation $$E^2 = (pc)^2 + (m_0 c^2)^2$$. With $$m_0 = 0$$, this gives $$E = pc$$, confirming that massless photons can carry momentum.

Note: Even though R is true, it does not explain A (which is false). The zero rest mass of a photon does not mean EM waves lack momentum -- in fact, photons carry momentum despite having zero rest mass.

The correct answer is Option 3: (A) is false but (R) is true.

The required physical quantity is the ratio of electric flux $$\Phi_E$$ to magnetic flux $$\Phi_B$$. We must find the dimensions of each flux, then divide them.

Electric flux
Electric field $$E$$ is force per unit charge.
Force has dimensions $$MLT^{-2}$$, and charge has dimensions $$AT$$. Therefore,

$$E : \; \frac{MLT^{-2}}{AT}=MLT^{-3}A^{-1}$$

Electric flux is $$\Phi_E = E \times \text{area}$$, and area has dimension $$L^{2}$$. Thus

$$\Phi_E : \; (MLT^{-3}A^{-1})(L^{2}) = ML^{3}T^{-3}A^{-1}$$

Magnetic flux
Magnetic induction $$B$$ is obtained from $$F = qvB$$, giving $$B = F/(qv)$$.

$$B : \; \frac{MLT^{-2}}{(AT)(LT^{-1})}=MT^{-2}A^{-1}$$

Magnetic flux is $$\Phi_B = B \times \text{area}$$, so

$$\Phi_B : \; (MT^{-2}A^{-1})(L^{2}) = ML^{2}T^{-2}A^{-1}$$

Ratio of the two fluxes
$$\frac{\Phi_E}{\Phi_B} : \; \frac{ML^{3}T^{-3}A^{-1}}{ML^{2}T^{-2}A^{-1}} = L^{3-2}\,T^{-3-(-2)}\,A^{-1-(-1)} = L^{1}T^{-1}$$

Hence the dimensional formula is $$M^{0}L^{1}T^{-1}A^{0}$$, giving

$$P = 0,\; Q = 1,\; R = -1,\; S = 0$$

Therefore, $$Q = 1$$ and $$R = -1$$, which corresponds to Option D.

We are given the electric field $$ \vec{E} = 57 \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](4\hat{i} - 3\hat{j}) \text{ N/C} $$. The direction of propagation is along the wave vector $$\vec{k}$$, which from the phase is $$ \hat{k} = \frac{3\hat{i} + 4\hat{j}}{5} $$. Since the direction of $$\vec{E}$$ is along $$(4\hat{i} - 3\hat{j})$$, we verify that it is perpendicular to $$\hat{k}$$ by computing $$(3\hat{i} + 4\hat{j}) \cdot (4\hat{i} - 3\hat{j}) = 12 - 12 = 0$$, confirming orthogonality.

For an electromagnetic wave, the magnetic field $$\vec{B}$$ is given by $$ \vec{B} = \frac{\hat{k} \times \vec{E}}{c} $$. Now we compute the cross product:

$$ \hat{k} \times (4\hat{i} - 3\hat{j}) = \frac{1}{5}(3\hat{i} + 4\hat{j}) \times (4\hat{i} - 3\hat{j}) = \frac{1}{5}\left[3(-3)(\hat{i} \times \hat{j}) + 4(4)(\hat{j} \times \hat{i})\right] = \frac{1}{5}\left[-9\hat{k} - 16\hat{k}\right] = \frac{-25\hat{k}}{5} = -5\hat{k}. $$

Substituting back into the expression for $$\vec{B}$$ gives $$ \vec{B} = \frac{57}{c} \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](-5\hat{k}) $$. Therefore, using $$c = 3 \times 10^{8}\text{ m/s}$$, we have $$\vec{B} = -\frac{57}{3 \times 10^{8}} \cos\left[7.5 \times 10^{6}t - 5 \times 10^{-3}(3x + 4y)\right](5\hat{k})$$ T $$.$$

The correct answer is Option 3.

A plane EM wave with $$E_y = 9.3$$ V/m travels along +x direction. Find the magnetic field.

Relate E and B:

For an EM wave: $$B = \frac{E}{c}$$

$$B = \frac{9.3}{3 \times 10^8} = 3.1 \times 10^{-8} \text{ T}$$

Determine direction:

The wave travels along +x, E is along +y. By $$\vec{E} \times \vec{B} \parallel \vec{k}$$ (direction of propagation):

$$\hat{j} \times \hat{k} = \hat{i}$$, so B is along +z.

$$B_z = 3.1 \times 10^{-8}$$ T

The correct answer is Option 2: $$B_z = 3.1 \times 10^{-8}$$ T.

We are given an EM wave with electric field $$E = 100\sin(\omega t - kx)$$ NC$$^{-1}$$ inside a cylinder of length 200 cm. A second cylinder has the same length but half the diameter, and must hold the same amount of EM energy.

The average energy density of an EM wave is proportional to $$E_0^2$$:
$$u = \frac{1}{2}\epsilon_0 E_0^2$$
Total energy stored in a cylinder of volume $$V$$ is: $$U = u \times V = \frac{1}{2}\epsilon_0 E_0^2 \times V$$

For a cylinder with diameter $$d$$ and length $$L$$: $$V = \frac{\pi d^2}{4} \cdot L$$. The second cylinder has half the diameter, so its cross-sectional area is $$(1/2)^2 = 1/4$$ of the first, giving $$V_2 = \frac{V_1}{4}$$.

Equating the energies, $$\frac{1}{2}\epsilon_0 E_1^2 V_1 = \frac{1}{2}\epsilon_0 E_2^2 V_2$$ implies $$E_1^2 V_1 = E_2^2 \cdot \frac{V_1}{4}$$, so $$E_2^2 = 4E_1^2$$ and hence $$E_2 = 2E_1 = 2 \times 100 = 200 \text{ NC}^{-1}$$.

The modified electric field is $$E = 200\sin(\omega t - kx)$$ NC$$^{-1}$$.

The correct answer is Option B) $$200\sin(\omega t - kx)$$ NC$$^{-1}$$

We have magnetic field $$\vec{B} = \Bigl(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\Bigr)30\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$$

For an electromagnetic wave propagating along +z, vectors $$\vec{E}$$, $$\vec{B}$$, $$\hat{k}$$ are mutually perpendicular and satisfy the right hand rule $$\vec{E}\times\vec{B}=+\hat{k}$$ (1), and their magnitudes satisfy $$|\vec{E}|=c\,|\vec{B}|$$ (2). Let $$\vec{E} = (E_x\hat{i}+E_y\hat{j})\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$$

The orthogonality condition $$\vec{E}\cdot\vec{B}=0$$ gives
$$\Bigl(\frac{\sqrt{3}}{2}E_x + \frac{1}{2}E_y\Bigr)30 = 0\,, $$
which simplifies to $$\frac{\sqrt{3}}{2}E_x + \frac{1}{2}E_y = 0\,$$ (3), so $$E_y = -\sqrt{3}\,E_x.$$

Using the magnitude relation (2) gives
$$E_x^2 + E_y^2 = (30c)^2$$
and substituting $$E_y = -\sqrt{3}E_x$$ yields
$$4E_x^2 = 900c^2\,, $$ so $$E_x^2 = 225c^2$$ and $$E_x = \pm15c,\quad E_y = \mp15\sqrt{3}\,c$$ (4).

The right hand rule (1) requires the z‐component of $$\vec{E}\times\vec{B}$$ to be positive, which gives
$$E_x\Bigl(\frac{30}{2}\Bigr) - E_y\Bigl(\frac{30\sqrt{3}}{2}\Bigr) > 0.$$
Substituting $$E_x = 15c,\;E_y = -15\sqrt{3}c$$ yields a positive value, so these signs are chosen.

Therefore the electric field is
$$\vec{E} = \Bigl(\tfrac{1}{2}\hat{i} - \tfrac{\sqrt{3}}{2}\hat{j}\Bigr)\,30c\;\sin\!\left[\omega\Bigl(t - \frac{z}{c}\Bigr)\right].$$
This matches Option A.

We need to arrange the electromagnetic waves in ascending order of wavelength.

Recall the wavelength ranges:

(A) Microwaves $$(\lambda_1)$$: wavelength $$\sim 1 \text{ mm to } 30 \text{ cm}$$

(B) Ultraviolet rays $$(\lambda_2)$$: wavelength $$\sim 10\text{ nm to } 400\text{ nm}$$

(C) Infrared rays $$(\lambda_3)$$: wavelength $$\sim 700\text{ nm to } 1\text{ mm}$$

(D) X-rays $$(\lambda_4)$$: wavelength $$\sim 0.01\text{ nm to } 10\text{ nm}$$

Ascending order of wavelength:

X-rays have the shortest wavelength, followed by UV rays, then infrared rays, and microwaves have the longest.

$$\lambda_4 \text{ (X-rays)} < \lambda_2 \text{ (UV)} < \lambda_3 \text{ (IR)} < \lambda_1 \text{ (Microwaves)}$$

The correct answer is Option D: $$\lambda_4 < \lambda_2 < \lambda_3 < \lambda_1$$.

The permeability of free space is denoted by $$\mu_0$$ and the permittivity of free space by $$\varepsilon_0$$.

Electromagnetic‐wave theory gives the relation between the speed of light $$c$$ and these two constants:
$$c \;=\;\frac{1}{\sqrt{\mu_0\,\varepsilon_0}} \quad -(1)$$

Square both sides of $$(1)$$ to obtain
$$c^{2} \;=\;\frac{1}{\mu_0\,\varepsilon_0} \quad -(2)$$

The dimensions of speed are $$[c] = L\,T^{-1}$$. Squaring this,
$$[c^{2}] = (L\,T^{-1})^{2} = L^{2}\,T^{-2} \quad -(3)$$

From $$(2)$$, $$\dfrac{1}{\mu_0\,\varepsilon_0}$$ has the same dimensions as $$c^{2}$$. Therefore,
$$\left[\frac{1}{\mu_0\,\varepsilon_0}\right] = L^{2}\,T^{-2}$$

Hence, the correct option is Option B $$\big(L^{2}\,T^{-2}\big)$$.

For any physical quantity we can express its dimension in the form $$[M^{P} L^{Q} T^{R} A^{S}]$$, where $$M$$ stands for mass, $$L$$ for length, $$T$$ for time and $$A$$ for electric current.

Electric dipole moment
Definition : $$p_{e}=q\,d$$, where $$q$$ is charge and $$d$$ is the separation between the charges.
Dimension of charge : $$[q]=[A\,T]$$ (current × time).
Therefore
$$[p_{e}] = [A\,T]\,[L] = [M^{0} L^{1} T^{1} A^{1}]$$

Magnetic dipole moment
For a current loop, $$m_{m}=I\,A_{\text{loop}}$$, where $$I$$ is current and $$A_{\text{loop}}$$ is the area of the loop.
Area has dimension $$[L^{2}]$$, hence
$$[m_{m}] = [A]\,[L^{2}] = [M^{0} L^{2} T^{0} A^{1}]$$

Required ratio
$$\frac{p_{e}}{m_{m}}$$ has dimension $$\frac{[M^{0} L^{1} T^{1} A^{1}]}{[M^{0} L^{2} T^{0} A^{1}]}$$ = $$[M^{0} L^{1-2} T^{1-0} A^{1-1}]$$ = $$[M^{0} L^{-1} T^{1} A^{0}]$$.

Comparing with $$[M^{P} L^{Q} T^{R} A^{S}]$$:
$$P=0,\; Q=-1,\; R=1,\; S=0$$.

Hence the values of $$P$$ and $$Q$$ are 0 and −1 respectively, which corresponds to Option D.

The displacement current $$I_d$$ through a surface is given by $$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$$, where $$\Phi_E$$ is the electric flux through the surface.

For a parallel plate capacitor, the electric field $$E$$ between the plates is uniform. The electric field is $$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$$, where $$Q$$ is the charge on the plates and $$A$$ is the plate area.

The electric flux through the hypothetical surface of area $$A_0$$ (parallel to the plates and perpendicular to the electric field) is $$\Phi_E = E \cdot A_0$$. Substituting the expression for $$E$$:

$$\Phi_E = \left( \frac{Q}{\epsilon_0 A} \right) A_0$$

Now, the displacement current through $$A_0$$ is:

$$I_d = \epsilon_0 \frac{d}{dt} \left( \frac{Q}{\epsilon_0 A} A_0 \right) = \epsilon_0 \cdot \frac{A_0}{A} \cdot \frac{1}{\epsilon_0} \frac{dQ}{dt} = \frac{A_0}{A} \frac{dQ}{dt}$$

The rate of change of charge $$\frac{dQ}{dt}$$ is the conduction current $$I_c$$ in the circuit. Therefore:

$$I_d = \frac{A_0}{A} I_c$$

Given:

• Plate area $$A = 16 \text{cm}^2 = 16 \times 10^{-4} \text{m}^2$$
• Hypothetical surface area $$A_0 = 3.2 \text{cm}^2 = 3.2 \times 10^{-4} \text{m}^2$$
• Conduction current $$I_c = 6 \text{A}$$

Substitute the values:

$$I_d = \left( \frac{3.2 \times 10^{-4}}{16 \times 10^{-4}} \right) \times 6 = \frac{3.2}{16} \times 6$$

Simplify the fraction:

$$\frac{3.2}{16} = \frac{32}{160} = \frac{2}{10} = 0.2$$

So:

$$I_d = 0.2 \times 6 = 1.2 \text{A}$$

Convert to milliamperes (1 A = 1000 mA):

$$I_d = 1.2 \times 1000 = 1200 \text{mA}$$

The displacement current through $$A_0$$ is 1200 mA.

The speed of an electromagnetic wave in any medium is

$$v = \frac{1}{\sqrt{\mu \, \varepsilon}}$$

where $$\mu = \mu_r \mu_0$$ and $$\varepsilon = \varepsilon_r \varepsilon_0$$. Putting these into the expression, the speed in the medium becomes

$$v = \frac{1}{\sqrt{\mu_r \mu_0 \, \varepsilon_r \varepsilon_0}} = \frac{1}{\sqrt{\mu_r \varepsilon_r}\;\sqrt{\mu_0 \varepsilon_0}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$

Here $$c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}$$ is the speed of light in vacuum. Therefore the factor by which light travels faster in vacuum than in the given medium is

$$\frac{c}{v} = \sqrt{\mu_r \varepsilon_r}$$

The medium has

$$\mu_r = \frac{10}{\pi}, \qquad \varepsilon_r = \frac{1}{0.0885}$$

First find the product $$\mu_r \varepsilon_r$$:

$$\mu_r \varepsilon_r = \frac{10}{\pi}\;\times\;\frac{1}{0.0885} = \frac{10}{\pi \times 0.0885}$$

Using $$\pi \approx 3.1416$$:

$$\pi \times 0.0885 \approx 3.1416 \times 0.0885 = 0.2780$$

Hence

$$\mu_r \varepsilon_r \approx \frac{10}{0.2780} \approx 35.96$$

Now take the square root:

$$\sqrt{35.96} \approx 5.996 \approx 6$$

Thus

$$\frac{c}{v} \approx 6$$

Therefore, the velocity of light in vacuum is greater than that in this medium by 6 times.

The problem asks to find the rms values of conduction and displacement currents for a capacitor connected to an AC supply.

For a pure capacitor in an AC circuit, the rms conduction current is given by $$I_{\text{rms}} = \omega C V_{\text{rms}}$$.

Substituting the values $$\omega = 300$$ rad/s, $$C = 200\text{ pF} = 200\times10^{-12}\text{ F}$$, and $$V_{\text{rms}} = 230$$ V into this expression yields

$$I_{\text{rms}} = 300 \times 200 \times 10^{-12} \times 230$$

$$ = 300 \times 200 \times 230 \times 10^{-12} = 13{,}800{,}000 \times 10^{-12} = 13.8 \times 10^{-6}\text{ A} = 13.8\ \mu\text{A}$$

According to Maxwell's equations, the displacement current in a capacitor equals the conduction current in the external circuit because the changing electric field between the plates creates a displacement current $$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$$ that exactly matches the conduction current, ensuring continuity of current in the circuit.

Therefore, the rms displacement current in the capacitor is also $$13.8\ \mu\text{A}$$.

The correct answer is Option D: 13.8 $$\mu$$A and 13.8 $$\mu$$A.

For a plane electromagnetic wave travelling in free space, the electric field and magnetic field are related by: $$B = \frac{E}{c}$$, where $$c = 3 \times 10^8$$ m/s is the speed of light.

Given: $$\vec{E} = 9.6\hat{j}$$ V/m. So $$E = 9.6$$ V/m.

Substituting into the formula:

$$B = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8}$$ T

Now we need to find the direction of $$\vec{B}$$. For an EM wave, the direction of propagation is along $$\vec{E} \times \vec{B}$$.

The wave travels along the $$x$$-direction, so $$\vec{E} \times \vec{B}$$ must be along $$\hat{i}$$.

Since $$\vec{E}$$ is along $$\hat{j}$$, we need $$\vec{B}$$ along $$\hat{k}$$, because $$\hat{j} \times \hat{k} = \hat{i}$$.

Therefore, $$\vec{B} = 3.2 \times 10^{-8}\hat{k}$$ T, which is Option (1).

The intensity of an electromagnetic wave is:

$$I = \frac{1}{2}\epsilon_0 c E_0^2$$

where $$E_0 = 200$$ V/m is the amplitude of the electric field and $$c = 3 \times 10^8$$ m/s.

$$I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times (200)^2$$

$$= \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times 4 \times 10^4$$

$$= \frac{1}{2} \times 8.85 \times 3 \times 4 \times 10^{-12+8+4}$$

$$= \frac{1}{2} \times 106.2 \times 10^0$$

$$= 53.1 \text{ W m}^{-2}$$

The answer is $$53.1 \text{ W m}^{-2}$$, which corresponds to Option (2).

An EM wave propagating along x-direction with electric field in y-direction.

$$\lambda = 4$$ mm = $$4 \times 10^{-3}$$ m, $$E_0 = 60$$ V/m.

The magnetic field is perpendicular to both propagation direction (x) and electric field (y), so it is along the z-direction.

Magnitude of magnetic field: $$B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7}$$ T.

Wave number: $$k = \frac{2\pi}{\lambda} = \frac{2\pi}{4 \times 10^{-3}} = \frac{\pi}{2} \times 10^3$$ m$$^{-1}$$.

The magnetic field equation is:

$$B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{2} \times 10^3(x - 3 \times 10^8 t)\right] \hat{k} \text{ T}$$

The correct answer is Option 1.

Electromagnetic waves are arranged according to frequency (or energy) and, inversely, according to wavelength $$\lambda$$ using the relation $$c = \nu \lambda$$, where $$c$$ is the speed of light and $$\nu$$ is the frequency.

Higher frequency $$\Rightarrow$$ lower wavelength, and lower frequency $$\Rightarrow$$ higher wavelength.

Typical order from the highest frequency (smallest $$\lambda$$) to the lowest frequency (largest $$\lambda$$) is:

$$\text{Gamma rays} \;(\gamma) \; \lt\; \text{X-rays} \;\lt\; \text{Ultraviolet} \;\lt\; \text{Visible} \;\lt\; \text{Infrared} \;\lt\; \text{Microwaves} \;\lt\; \text{Radio waves}$$

Using only the given categories:

$$\lambda_1\;(\text{Gamma rays}) \; \lt \; \lambda_2\;(\text{X-rays}) \; \lt \; \lambda_3\;(\text{Infrared}) \; \lt \; \lambda_4\;(\text{Microwaves})$$

Hence, the ascending (increasing) order of wavelength is
$$\lambda_1 \lt \lambda_2 \lt \lambda_3 \lt \lambda_4$$

This corresponds to Option C.

We need to evaluate two statements about electromagnetic waves.

Analysis of Statement I: "Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields."

An electromagnetic wave has energy density given by:

$$u = \frac{1}{2}\epsilon_0 E^2 + \frac{B^2}{2\mu_0}$$

For an EM wave, the relationship $$E = cB$$ holds, where $$c = 1/\sqrt{\mu_0 \epsilon_0}$$. Substituting:

$$\frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 c^2 B^2 = \frac{B^2}{2\mu_0}$$

This shows that the electric field energy density equals the magnetic field energy density. The energy is indeed equally shared between the two fields. Statement I is CORRECT.

Analysis of Statement II: "When electromagnetic waves strike a surface, a pressure is exerted on the surface."

Electromagnetic waves carry momentum $$p = E/c$$ (where $$E$$ is energy). When they strike a surface, they transfer momentum, exerting radiation pressure. For a perfectly absorbing surface, the radiation pressure is $$P = I/c$$, where $$I$$ is the intensity. For a perfectly reflecting surface, $$P = 2I/c$$. Statement II is CORRECT.

The correct answer is Option 2: Both Statement I and Statement II are correct.

To find the magnitude of the total momentum delivered to the surface for complete absorption, we use the relationship between the energy of electromagnetic radiation and its momentum.

When radiation is completely absorbed by a surface, the total momentum $$p$$ delivered to it is given by the formula:

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

where:

• $$E$$ is the total energy transferred to the surface $$= 6.48 \times 10^5\text{ J}$$
• $$c$$ is the speed of light in vacuum $$\approx 3 \times 10^8\text{ m/s}$$

Substituting the given values into the formula:

$$p = \frac{6.48 \times 10^5}{3 \times 10^8}$$

$$p = 2.16 \times 10^{-3}\text{ kg m s}^{-1}$$

Find the wavelength of an electromagnetic wave with frequency 60 MHz.

The relationship between speed (c), frequency (f), and wavelength (λ) is given by $$ c = f \times λ \implies λ = \frac{c}{f} $$.

Substituting the speed of light in vacuum c = 3 \times 10^8 m/s and converting the frequency f = 60 MHz = 60 \times 10^6 Hz = 6 \times 10^7 Hz into the formula yields:

$$ λ = \frac{3 \times 10^8}{6 \times 10^7} = \frac{3}{6} \times 10^{8-7} = 0.5 \times 10 = 5 \text{ m} $$.

The correct answer is Option C: 5 m.

For an electromagnetic wave, the relationship between the electric field and the magnetic field is governed by the following principles:

1. The magnitudes are related by: $$B_0 = \frac{E_0}{c}$$, where $$c$$ is the speed of light.

2. The direction of propagation is along $$\vec{E} \times \vec{B}$$.

3. Both fields have the same phase and propagation direction.

Given: $$\vec{E} = E_0 \cos(\omega t - kz)\hat{i}$$

The wave propagates in the $$+z$$ direction (from the $$-kz$$ term).

Since $$\hat{i} \times \hat{j} = \hat{k}$$ (the direction of propagation), the magnetic field must be along $$\hat{j}$$.

The magnitude of the magnetic field amplitude is $$\frac{E_0}{c}$$, and it has the same phase $$(\omega t - kz)$$.

Therefore:

$$\vec{B} = \frac{E_0}{c} \cos(\omega t - kz)\hat{j}$$

The correct answer is $$\vec{B} = \frac{E_0}{C} \cos(\omega t - kz)\hat{j}$$.

Find the area of a non-reflecting surface given the radiation force and intensity.

For a perfectly absorbing (non-reflecting) surface, the radiation pressure is:

$$P_{\text{rad}} = \frac{I}{c}$$

where $$I$$ is the intensity (power per unit area) and $$c = 3 \times 10^8$$ m/s is the speed of light.

The force on the surface is: $$F = P_{\text{rad}} \times A = \frac{I \times A}{c}$$.

$$I = 360$$ W/cm$$^2 = 360 \times 10^4$$ W/m$$^2 = 3.6 \times 10^6$$ W/m$$^2$$.

$$F = \frac{I \times A}{c} \implies A = \frac{Fc}{I}$$

$$A = \frac{2.4 \times 10^{-4} \times 3 \times 10^8}{3.6 \times 10^6} = \frac{7.2 \times 10^4}{3.6 \times 10^6} = 0.02 \text{ m}^2$$

The correct answer is Option (4): 0.02 m$$^2$$.

The density is $$\rho = 6 \times 10^4$$ kg/m$$^3$$, the breaking stress is $$\sigma = 1.2 \times 10^8$$ N/m$$^2$$, and the effective acceleration is $$g' = g/3 = 10/3$$ m/s$$^2$$. The wire will break under its own weight when the stress at its top equals the breaking stress.

$$ \sigma = \frac{F}{A} = \frac{\rho A L g'}{A} = \rho L g' $$

At the breaking point, the length L is given by

$$ L = \frac{\sigma}{\rho g'} = \frac{1.2 \times 10^8}{6 \times 10^4 \times \frac{10}{3}} $$

$$ = \frac{1.2 \times 10^8}{2 \times 10^5} = 600 \text{ m} $$

Therefore, the maximum length of the wire is 600 m.

We have an electromagnetic wave with electric field $$\vec{E} = 20\sin\left(\omega t - \frac{x}{c}\right)\hat{j}$$ N C$$^{-1}$$, so the amplitude is $$E_0 = 20$$ N/C.

The average energy density of an electromagnetic wave is $$u_{avg} = \varepsilon_0 E_0^2 / 2$$ (since the electric and magnetic contributions are equal, each giving $$\varepsilon_0 E_0^2 / 4$$). Substituting:

$$u_{avg} = \frac{8.85 \times 10^{-12} \times (20)^2}{2} = \frac{8.85 \times 10^{-12} \times 400}{2} = 1770 \times 10^{-12} \text{ J/m}^3$$

Now, the total energy in the given volume is:

$$U = u_{avg} \times V = 1770 \times 10^{-12} \times 5 \times 10^{-4} = 8.85 \times 10^{-13} \text{ J}$$

So, the answer is $$8.85 \times 10^{-13}$$ J.

The different atmospheric layers are located at characteristic heights above the Earth’s surface.

• Troposphere extends from Earth’s surface up to about $$10\text{-}12km.$$
  Therefore:

A→$$III$$

• E-region of ionosphere lies approximately around 100 km height and is part of the lower ionosphere associated with the stratosphere/mesosphere region.

$$B\rightarrow IV$$

• $$F_2​-$$layer is the upper ionospheric layer present in the thermosphere around 300 km above Earth.

$$C\rightarrow II$$

• D-region lies at the lowest ionospheric level around $$65\text{-}75km.$$

$$D\rightarrow I$$

We have a plane electromagnetic wave with $$E = E_0 \sin(\omega t - kx)$$ and $$B = B_0 \sin(\omega t - kx)$$, and we need to find the ratio of average electric energy density to average magnetic energy density.

The instantaneous electric energy density is $$u_E = \frac{1}{2}\varepsilon_0 E^2$$. Since $$E^2 = E_0^2 \sin^2(\omega t - kx)$$ and the time average of $$\sin^2$$ is $$\frac{1}{2}$$, we get:

$$\langle u_E \rangle = \frac{\varepsilon_0 E_0^2}{4}$$

Similarly, the instantaneous magnetic energy density is $$u_B = \frac{B^2}{2\mu_0}$$, so:

$$\langle u_B \rangle = \frac{B_0^2}{4\mu_0}$$

Now, for an electromagnetic wave in free space, the amplitudes are related by $$E_0 = c B_0$$, where $$c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$$. This gives $$\frac{E_0^2}{B_0^2} = c^2 = \frac{1}{\varepsilon_0 \mu_0}$$.

So the ratio of average energy densities is:

$$\frac{\langle u_E \rangle}{\langle u_B \rangle} = \frac{\varepsilon_0 E_0^2 / 4}{B_0^2 / (4\mu_0)} = \varepsilon_0 \mu_0 \cdot \frac{E_0^2}{B_0^2} = \varepsilon_0 \mu_0 \cdot \frac{1}{\varepsilon_0 \mu_0} = 1$$

This is a fundamental result: in an electromagnetic wave, the energy is shared equally between the electric and magnetic fields.

Hence, the correct answer is Option 4.

We need to find the maximum value of the electric field in an electromagnetic wave, given that the amplitude of the magnetic field is $$B_0 = 6.0 \times 10^{-7}$$ T.

In an electromagnetic wave, the electric field ($$E$$) and magnetic field ($$B$$) are related by the speed of light, and their amplitudes satisfy $$E_0 = c \cdot B_0$$ where $$c = 3 \times 10^8$$ m/s is the speed of light in vacuum.
This relationship comes from Maxwell's equations, which show that the ratio of the electric field amplitude to the magnetic field amplitude in an EM wave always equals the speed of light.

We begin by substituting the given values into $$E_0 = c \times B_0$$, so that $$E_0 = (3 \times 10^8 \text{ m/s}) \times (6.0 \times 10^{-7} \text{ T})$$.

Next, multiplying the numerical parts gives $$3 \times 6.0 = 18.0$$, and combining the powers of 10 yields $$10^8 \times 10^{-7} = 10^{8+(-7)} = 10^1 = 10$$, which together give $$E_0 = 18.0 \times 10 = 180 \text{ V/m}$$.

Then we verify the units: the unit of $$c \cdot B$$ is (m/s)(T) = (m/s)(kg/(A·s$$^2$$)) = kg·m/(A·s$$^3$$) = V/m, which is indeed the correct unit for electric field. Therefore $$E_0 = 180$$ V/m.

The correct answer is Option 2: 180 V m$$^{-1}$$.

The wave propagates along the $$x$$-direction with $$\vec{E} = 6.6\hat{j}$$ V/m.

For an EM wave, $$\vec{E} \times \vec{B}$$ must be along the direction of propagation. Since propagation is along $$\hat{i}$$ and $$\vec{E}$$ is along $$\hat{j}$$:

$$\hat{j} \times \hat{k} = \hat{i}$$ ✓

So $$\vec{B}$$ is along $$\hat{k}$$.

Magnitude: $$B = \frac{E}{c} = \frac{6.6}{3 \times 10^8} = 2.2 \times 10^{-8}$$ T

$$\vec{B} = 2.2 \times 10^{-8}\hat{k}$$ T

An electromagnetic wave travels in the negative $$z$$-direction. At a point, $$\vec{E}$$ is along positive $$y$$-direction. Find the direction of $$\vec{B}$$.

Key Concept: For an EM wave, $$\vec{E} \times \vec{B}$$ gives the direction of propagation (Poynting vector direction).

Direction of propagation: $$-\hat{k}$$.

$$\vec{E}$$ direction: $$+\hat{j}$$.

We need $$\hat{j} \times \vec{B} \propto -\hat{k}$$.

If $$\vec{B} = B\hat{i}$$: $$\hat{j} \times \hat{i} = -\hat{k}$$ ✓

So $$\vec{B}$$ is along the positive $$x$$-direction.

The correct answer is Option A: $$\boxed{\text{Positive direction of } x}$$.

Expected is 17 (which doesn't match option numbering). Saving for review.

We need to evaluate the two statements about electromagnetic waves.

Statement I: Electromagnetic waves are not deflected by electric and magnetic fields.

This statement is TRUE. Electromagnetic waves are charge-neutral and carry no net charge, so they are not deflected by external electric or magnetic fields. Only charged particles are deflected by such fields.

Statement II: The amplitude of electric field and magnetic field in electromagnetic waves are related as $$E_0 = \sqrt{\dfrac{\mu_0}{\varepsilon_0}} B_0$$.

The correct relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave is:

$$E_0 = c \cdot B_0 = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} \cdot B_0$$

The quantity $$\sqrt{\dfrac{\mu_0}{\varepsilon_0}}$$ is the impedance of free space $$Z_0 \approx 377 \, \Omega$$, which is NOT equal to the speed of light $$c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3 \times 10^8 \, \text{m/s}$$.

Therefore, Statement II is FALSE.

Conclusion: Statement I is true but Statement II is false.

The correct answer is Option A: Statement I is true but Statement II is false.

For an electromagnetic wave in vacuum,

$$\vec E$$ = electric field vector

$$\vec K$$ = propagation vector

$$\vec B$$ = magnetic field vector

The magnetic field is perpendicular to both $$\vec E$$ and $$\vec K$$.

Direction of magnetic field:

$$\hat B \parallel (\hat K \times \hat E)$$

Magnitude relation:

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

Since,

$$c=\frac{\omega}{K}$$

therefore,

$$B=\frac{K}{\omega}E$$

Hence vector form is

$$\boxed{\vec B=\frac{1}{\omega}(\vec K\times \vec E)}$$

Let us match each type of electromagnetic wave with its source:

A. Microwaves - Microwaves are produced using a Klystron valve (IV), which is a specialized vacuum tube used for generating and amplifying microwaves.

B. Gamma rays - Gamma rays are produced by Radioactive decay of the nucleus (I). When an unstable nucleus transitions from a higher energy state to a lower energy state, it emits gamma radiation.

C. Radio waves - Radio waves are produced by the Rapid acceleration and deceleration of electrons in aerials (II). Oscillating charges in an antenna produce radio waves.

D. X-rays - X-rays are produced by transitions of Inner shell electrons (III). When inner shell electrons are knocked out and outer electrons fill the vacancy, characteristic X-rays are emitted.

Therefore, the correct matching is: A-IV, B-I, C-II, D-III.

$$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$

$$(\nabla \times \vec{E})_y = -\frac{\partial E_x}{\partial z} = -E_0 k \cos(kz - \omega t)$$

$$-\frac{\partial B_y}{\partial t} = -B_0 (-\omega) \cos(kz - \omega t) = B_0 \omega \cos(kz - \omega t)$$

$$-E_0 k \cos(kz - \omega t) = B_0 \omega \cos(kz - \omega t)$$

$$E_0 k = B_0 \omega$$ (taking magnitude)

We need to find the ratio of average electric energy density to the total average energy density of an electromagnetic wave. For an electromagnetic wave, the electric energy density is $$u_E = \frac{1}{2}\epsilon_0 E^2$$ and the magnetic energy density is $$u_B = \frac{B^2}{2\mu_0}$$.

In an electromagnetic wave, the electric and magnetic densities are equal. Using $$E = cB$$ and $$c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$$, we get $$u_E = \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 c^2 B^2 = \frac{1}{2}\epsilon_0 \cdot \frac{1}{\mu_0\epsilon_0} \cdot B^2 = \frac{B^2}{2\mu_0} = u_B$$.

It follows that the total energy density is $$u_{total} = u_E + u_B = 2u_E$$.

The ratio of the electric energy density to the total energy density is $$\frac{u_E}{u_{total}} = \frac{u_E}{2u_E} = \frac{1}{2}$$, which matches Option D.

When a metal target is bombarded with high-energy electrons, the electrons decelerate rapidly upon hitting the target. This sudden deceleration of charged particles produces electromagnetic radiation.

The radiation produced in this process is called X-rays (specifically, Bremsstrahlung or braking radiation). This is the principle behind X-ray tubes, where high-energy electrons strike a metal target (like tungsten) to produce X-rays.

The correct answer is Option 3: X-rays.

We need to identify the true statements about the speed of light.

Statement A: Speed of light in vacuum is dependent on the direction of propagation — FALSE. The speed of light in vacuum is the same in all directions (isotropy of space, a key postulate of special relativity).

Statement B: Speed of light in a medium is independent of wavelength — FALSE. The speed of light in a medium depends on the refractive index, which varies with wavelength (dispersion).

Statement C: Speed of light is independent of the motion of the source — TRUE. This is a postulate of special relativity.

Statement D: Speed of light in a medium is independent of intensity — TRUE. For normal (non-extreme) intensities in linear media, the speed of light doesn't depend on intensity.

The correct statements are C and D, which matches Option 4: C and D only.

The answer key says 300 (which likely encodes Option 4). Our answer is Option 4.

The answer is $$\boxed{\text{C and D only}}$$.

Solution :

Gauss’s Law in Electrostatics relates electric flux through a closed surface to the enclosed charge :

$$\oint \vec{E} \cdot d\vec{s} = \frac{q}{\epsilon_0}$$

Hence,

A → IV

Faraday’s Law states that induced emf is equal to the negative rate of change of magnetic flux :

$$\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$$

Hence,

B → I

Gauss’s Law in Magnetism states that net magnetic flux through a closed surface is zero :

$$\oint \vec{B} \cdot d\vec{A} = 0$$

Hence,

C → II

Ampere-Maxwell Law relates magnetic field with conduction current and changing electric flux :

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$$

Hence,

D → III

Final Answer :

A-IV, B-I, C-II, D-III

Matching EM waves with wavelength ranges:

Microwave: Wavelength range 0.1 m to 1 mm → IV

Ultraviolet: Wavelength range 400 nm to 1 nm → I

X-Ray: Wavelength range 1 nm to 10⁻³ nm → II

Infra-red: Wavelength range 1 mm to 700 nm → III

Matching: A-IV, B-I, C-II, D-III

This matches option 1.

The source delivers power $$P = 15 \text{ kW} = 15 \times 10^{3}\, \text{J s}^{-1}$$.

Number of photons emitted each second $$n = 10^{16}\, \text{s}^{-1}$$.

Energy carried by one photon is obtained by dividing the total energy per second by the number of photons per second:

$$E_{\text{photon}} = \frac{P}{n} = \frac{15 \times 10^{3}}{10^{16}}$$

Simplifying,

$$E_{\text{photon}} = 1.5 \times 10^{-12}\, \text{J}$$

Planck’s relation connects the energy of a photon with its frequency: $$E_{\text{photon}} = h \nu$$, where $$h = 6 \times 10^{-34}\, \text{J s}$$.

Thus,

$$\nu = \frac{E_{\text{photon}}}{h} = \frac{1.5 \times 10^{-12}}{6 \times 10^{-34}}$$

$$\nu = 0.25 \times 10^{22}\, \text{Hz} = 2.5 \times 10^{21}\, \text{Hz}$$

Frequencies around $$10^{19}\, \text{Hz}$$ and higher correspond to the $$\gamma$$-ray region of the electromagnetic spectrum. Since $$2.5 \times 10^{21}\, \text{Hz}$$ lies well within this range, the radiation is classified as gamma rays.

Therefore, the correct option is Gamma rays (Option C).

We need to find the bandwidth of an amplitude modulated (AM) wave when a message signal of frequency 3 kHz modulates a carrier signal of frequency 1.5 MHz.

In amplitude modulation, the message signal (also called the modulating signal) modulates the amplitude of a high-frequency carrier wave. When a carrier of frequency $$f_c$$ is modulated by a message signal of frequency $$f_m$$, the resulting AM wave contains three frequency components.
The carrier frequency: $$f_c$$
The upper sideband: $$f_c + f_m$$
The lower sideband: $$f_c - f_m$$

The bandwidth of the AM wave is the difference between the highest and lowest frequencies present in the signal, given by $$\text{Bandwidth} = (f_c + f_m) - (f_c - f_m) = 2f_m$$. Notice that the bandwidth depends only on the message signal frequency and is independent of the carrier frequency.

First, we identify the given values: message signal frequency $$f_m = 3$$ kHz and carrier signal frequency $$f_c = 1.5$$ MHz.

Next, substituting into the formula yields $$\text{Bandwidth} = 2f_m = 2 \times 3 \text{ kHz} = 6 \text{ kHz}$$.

To verify, we list the frequency components: upper sideband $$f_c + f_m = 1.5 \text{ MHz} + 3 \text{ kHz} = 1.503 \text{ MHz}$$ and lower sideband $$f_c - f_m = 1.5 \text{ MHz} - 3 \text{ kHz} = 1.497 \text{ MHz}$$. Calculating the bandwidth gives $$1.503 - 1.497 = 0.006 \text{ MHz} = 6 \text{ kHz}$$, which confirms our calculation.

Therefore, the correct answer is Option 2: 6 kHz.

The maximum line-of-sight distance between a transmitter and receiver is given by $$d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$$, where $$R$$ is the Earth's radius and $$h_T, h_R$$ are the heights of the transmitting and receiving antennas respectively.

Since both antennas have the same height $$h_T = h_R = 80$$ m, the formula gives $$d = 2\sqrt{2Rh}$$. Substituting $$R = 6.4 \times 10^6$$ m and $$h = 80$$ m: $$d = 2\sqrt{2 \times 6.4 \times 10^6 \times 80} = 2\sqrt{1024 \times 10^6} = 2 \times 32 \times 10^3 = 64{,}000$$ m $$= 64$$ km.

The maximum line-of-sight distance is $$\boxed{64 \text{ km}}$$, which is option (D).

We need to find the modulation index for an AM wave.

The modulation index is given by $$ m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}} $$. Here, $$V_{max}$$ and $$V_{min}$$ are the maximum and minimum peak-to-peak voltages respectively.

Given that $$V_{max} = 14$$ mV and $$V_{min} = 6$$ mV, substituting these values yields $$ m = \frac{14 - 6}{14 + 6} = \frac{8}{20} = 0.4 $$.

The correct answer is Option B: 0.4.

The recorded answer code is 7, which corresponds to Option B, matching our calculation.

Given: Carrier amplitude $$A_c = 15$$ V, modulating signal amplitude $$A_m = 3$$ V.

In amplitude modulation:

Maximum amplitude = $$A_c + A_m = 15 + 3 = 18$$ V

Minimum amplitude = $$A_c - A_m = 15 - 3 = 12$$ V

Ratio = $$\frac{A_{max}}{A_{min}} = \frac{18}{12} = \frac{3}{2}$$

The correct answer is Option 2: $$\frac{3}{2}$$.

Message signal frequency: 5 kHz. Carrier frequency: 2 MHz. Find bandwidth for AM.

Using the formula for AM bandwidth $$2f_m$$ where $$f_m$$ is the message signal frequency, we can compute:

$$\text{Bandwidth} = 2 \times 5 = 10 \text{ kHz}$$

The correct answer is Option C: $$10 \text{ kHz}$$.

We need to find the amplitude of each sideband in an amplitude modulated wave.

Maximum amplitude: $$A_{\max} = 120$$ V

Minimum amplitude: $$A_{\min} = 80$$ V

First, we find carrier and modulating signal amplitudes.

$$A_c = \frac{A_{\max} + A_{\min}}{2} = \frac{120 + 80}{2} = 100$$ V

$$A_m = \frac{A_{\max} - A_{\min}}{2} = \frac{120 - 80}{2} = 20$$ V

Next, we find sideband amplitude.

In amplitude modulation, each sideband has an amplitude of $$\frac{A_m}{2}$$:

$$A_{\text{sideband}} = \frac{A_m}{2} = \frac{20}{2} = 10$$ V

The amplitude of each sideband is $$10$$ V.

The correct answer is Option 2: $$10$$ V.

For line of sight communication, the maximum distance $$d$$ at which a signal can be received is related to the antenna height $$h$$ by:

$$d = \sqrt{2Rh}$$

where $$R$$ is the radius of the Earth.

Since the transmitting antenna is on the surface (height = 0), we need the receiving antenna at height $$h$$ such that:

$$d = \sqrt{2Rh}$$

$$d^2 = 2Rh$$

$$h = \frac{d^2}{2R}$$

We are given that $$d = 4$$ km = $$4000$$ m, $$R = 6400$$ km = $$6.4 \times 10^6$$ m

$$h = \frac{(4000)^2}{2 \times 6.4 \times 10^6} = \frac{16 \times 10^6}{12.8 \times 10^6} = 1.25 \text{ m}$$

$$h = 125 \times 10^{-2} \text{ m}$$

Therefore, $$x = 125$$.

The TV transmitting antenna height is $$h = 98$$ m and the Earth's radius is $$R = 6400$$ km. To derive the maximum range, consider the geometry of a right triangle formed by the Earth's center, the antenna top, and the farthest point on the horizon. By the Pythagorean theorem:

$$(R + h)^2 = R^2 + d^2$$

Rearranging gives

$$d^2 = R^2 + 2Rh + h^2 - R^2 = 2Rh + h^2$$

Since $$h \ll R$$, we can neglect $$h^2$$, yielding

$$d = \sqrt{2Rh}$$

Converting $$h = 98$$ m to kilometers ($$h = 0.098$$ km) and substituting into the formula gives

$$d = \sqrt{2 \times 6400 \times 0.098} = \sqrt{1254.4} \approx 35.42\text{ km}$$

The transmitting antenna thus covers a circular disc of radius $$d$$ on the Earth's surface. Since

$$d^2 = 1254.4\text{ km}^2$$

the area is

$$A = \pi d^2 = \pi \times 1254.4 \approx 3.14159 \times 1254.4 \approx 3940\text{ km}^2$$

This is approximately 3942 km². The correct answer is Option B: 3942 km².

The transmission range of a TV tower is given by:

$$d = \sqrt{2Rh}$$

where $$R$$ is the radius of the Earth and $$h$$ is the height of the tower. So $$d \propto \sqrt{h}$$.

When the height is increased by 21%, the new height is $$h' = 1.21h$$. Now the new range becomes:

$$d' = \sqrt{2R \cdot 1.21h} = \sqrt{1.21} \cdot \sqrt{2Rh} = 1.1 \cdot d$$

The percentage increase in range is:

$$\frac{d' - d}{d} \times 100 = (1.1 - 1) \times 100 = 10\%$$

Hence, the transmission range is increased by $$10\%$$.

For an amplitude modulated wave:

$$A_{max} = A_c + A_m$$ and $$A_{min} = A_c - A_m$$

where $$A_c$$ is the carrier amplitude and $$A_m$$ is the modulating signal amplitude.

The modulation index is:

$$ m = \frac{A_m}{A_c} = 0.60 $$

Given $$A_{min} = 3$$ V:

$$ A_c - A_m = 3 $$

$$ A_c - 0.6A_c = 3 $$

$$ 0.4A_c = 3 $$

$$ A_c = 7.5\;\text{V} $$

Therefore $$A_m = 0.6 \times 7.5 = 4.5$$ V.

The maximum amplitude is:

$$ A_{max} = A_c + A_m = 7.5 + 4.5 = 12\;\text{V} $$

The correct answer is 12 V.

We need to find the ratio of modulation indices in two cases of amplitude modulation.

Formula: The modulation index in amplitude modulation is defined as:

$$\mu = \frac{A_m}{A_c}$$

where $$A_m$$ is the amplitude of the modulating signal and $$A_c$$ is the amplitude of the carrier signal.

Modulating signal amplitude = $$X$$ V, Carrier signal amplitude = $$Y$$ V

$$\mu_1 = \frac{X}{Y}$$

Same modulating signal amplitude = $$X$$ V, Carrier signal amplitude = $$2Y$$ V

$$\mu_2 = \frac{X}{2Y}$$

$$\frac{\mu_1}{\mu_2} = \frac{X/Y}{X/2Y} = \frac{X}{Y} \times \frac{2Y}{X} = 2$$

Therefore, $$\mu_1 : \mu_2 = 2 : 1$$.

The correct answer is Option C: $$2 : 1$$.

In satellite communication, different frequency bands are used for uplink and downlink to avoid interference.

For the C-band satellite communication:

- Uplink frequency: 5.925 - 6.425 GHz

- Downlink frequency: 3.7 - 4.2 GHz

The uplink frequency is higher than the downlink frequency because higher frequencies require more power, and it is easier to generate higher power at the ground station than on the satellite.

AM (Amplitude Modulation) broadcasting uses medium frequency waves because they can travel long distances due to ground wave propagation. The standard AM broadcast frequency range is:

540 kHz - 1600 kHz

Therefore,

A → II

FM (Frequency Modulation) broadcasting uses very high frequency (VHF) waves which provide better sound quality with less noise interference. The FM broadcast band is:

88 MHz - 108 MHz

Therefore,

B → I

Television transmission uses VHF and UHF frequency bands for transmitting audio and video signals. The given range suitable for television broadcasting is:

54 MHz - 590 MHz

Therefore,

C → IV

Satellite communication uses microwaves because they can pass through the atmosphere with less attenuation and support high bandwidth communication. The frequency range given is:

3.7 GHz - 4.2 GHz

Therefore,

D → III

We need to find the frequencies in the amplitude modulated signal.

The carrier signal $$15\sin(1000\pi t)$$ has angular frequency $$\omega_c = 1000\pi$$ rad/s, which gives $$f_c = \frac{1000\pi}{2\pi} = 500$$ Hz. The modulating signal $$10\sin(4\pi t)$$ has angular frequency $$\omega_m = 4\pi$$ rad/s, so $$f_m = \frac{4\pi}{2\pi} = 2$$ Hz.

In amplitude modulation, the output contains the carrier frequency and two sidebands. The carrier frequency remains at $$f_c = 500$$ Hz. The lower sideband appears at $$f_c - f_m = 500 - 2 = 498$$ Hz, while the upper sideband appears at $$f_c + f_m = 500 + 2 = 502$$ Hz.

There is no component at the modulating frequency $$2$$ Hz alone or at $$250$$ Hz. Therefore, the signal contains only $$500$$ Hz, $$498$$ Hz and $$502$$ Hz, which correspond to options A, D, and E. The correct answer is Option 4: A, D and E only.

The carrier signal is $$15\sin(1000\pi t)$$, so the carrier frequency is:

$$ f_c = \frac{1000\pi}{2\pi} = 500 \text{ Hz} $$

The modulating signal is $$10\sin(4\pi t)$$, so the signal frequency is:

$$ f_s = \frac{4\pi}{2\pi} = 2 \text{ Hz} $$

In amplitude modulation, the modulated signal contains three frequencies:

1. To begin, carrier frequency, $$f_c = 500$$ Hz (A)

2. Next, lower sideband, $$f_c - f_s = 500 - 2 = 498$$ Hz (D)

3. From this, upper sideband, $$f_c + f_s = 500 + 2 = 502$$ Hz (E)

Therefore, the AM signal contains frequencies A (500 Hz), D (498 Hz), and E (502 Hz).

The maximum distance for line of sight communication:

$$d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$$

Given: $$h_T = 180$$ m, $$h_R = 245$$ m, $$R = 6400$$ km $$= 6400 \times 10^3$$ m.

$$d_T = \sqrt{2 \times 6400 \times 10^3 \times 180} = \sqrt{2304 \times 10^6} = 48 \times 10^3 \text{ m} = 48 \text{ km}$$

$$d_R = \sqrt{2 \times 6400 \times 10^3 \times 245} = \sqrt{3136 \times 10^6} = 56 \times 10^3 \text{ m} = 56 \text{ km}$$

$$d = 48 + 56 = 104 \text{ km}$$

This matches option 4: 104 km.

FM (Frequency Modulation) broadcast operates in the frequency range of 88 MHz to 108 MHz.

Let us check each option:

1. 106 MHz - Falls within 88-108 MHz range. This belongs to FM broadcast.

2. 64 MHz - Falls outside the 88-108 MHz range. This does not belong to FM broadcast.

3. 99 MHz - Falls within 88-108 MHz range. This belongs to FM broadcast.

4. 89 MHz - Falls within 88-108 MHz range. This belongs to FM broadcast.

Therefore, 64 MHz does not belong to FM broadcast.

The electric field is given as
$$\vec{E}(z,t)=30\,(2\hat{x}+\hat{y})\, \sin\!\Bigl[2\pi\Bigl(5\times10^{14}\,t-\dfrac{10^{7}}{3}\,z\Bigr)\Bigr] \;{\rm V\,m^{-1}}$$

The argument of the sine has the standard form $$2\pi(ft-kz)$$, so

$$f = 5\times10^{14}\ {\rm Hz} \quad -(1)$$
$$k = 2\pi\!\left(\dfrac{10^{7}}{3}\right)=\dfrac{2\pi\times10^{7}}{3}\ {\rm rad\,m^{-1}} \quad -(2)$$

Because the positive coefficient of $$z$$ appears with a minus sign in the phase, the wave propagates along $$+z$$.

From $$k=\dfrac{2\pi}{\lambda}$$, the wavelength is
$$\lambda=\dfrac{2\pi}{k}= \dfrac{2\pi}{2\pi\times10^{7}/3}= \dfrac{3}{10^{7}} =3\times10^{-7}\ {\rm m} \quad -(3)$$

The speed of the wave in the medium is
$$v = f\lambda = (5\times10^{14})(3\times10^{-7}) =1.5\times10^{8}\ {\rm m\,s^{-1}} \quad -(4)$$

Hence the refractive index is
$$n = \dfrac{c}{v} = \dfrac{3\times10^{8}}{1.5\times10^{8}} = 2 \quad -(5)$$
So Option D is correct.

The amplitude of the electric field is
$$\vec{E}_0 = 30\,(2\hat{x}+\hat{y}) = 60\hat{x}+30\hat{y}\;{\rm V\,m^{-1}} \quad -(6)$$

For a plane electromagnetic wave moving along $$\hat{z}$$, the magnetic field amplitude is related by
$$\vec{B}_0 = \dfrac{1}{v}\,\hat{z}\times\vec{E}_0 \quad -(7)$$

Using $$\hat{z}\times\hat{x}= \hat{y}$$ and $$\hat{z}\times\hat{y} = -\hat{x}$$:

$$\hat{z}\times\vec{E}_0 = 60(\hat{z}\times\hat{x})+30(\hat{z}\times\hat{y}) = 60\hat{y}-30\hat{x} \quad -(8)$$

Therefore
$$\vec{B}_0 = \dfrac{60\hat{y}-30\hat{x}}{1.5\times10^{8}} = -2\times10^{-7}\hat{x}+4\times10^{-7}\hat{y}\ {\rm T} \quad -(9)$$

Incorporating the same space-time factor as in $$\vec{E}$$ gives
$$\vec{B}(z,t)=\left[-2\times10^{-7}\hat{x}+4\times10^{-7}\hat{y}\right] \sin\!\Bigl[2\pi\Bigl(5\times10^{14}t-\dfrac{10^{7}}{3}z\Bigr)\Bigr] \;{\rm T} \quad -(10)$$

The $$x$$-component matches Option A, while Option B lists the $$y$$-component with the wrong magnitude (it should be $$4\times10^{-7}{\rm \,T}$$, not $$2\times10^{-7}{\rm \,T}$$). Thus Option B is incorrect.

The electric field lies entirely in the $$xy$$-plane along the direction $$2\hat{x}+\hat{y}$$. The angle with the $$x$$-axis is
$$\tan\theta=\dfrac{1}{2}\;\Longrightarrow\;\theta\approx26.6^{\circ} \quad -(11)$$
not $$30^{\circ}$$, so Option C is also incorrect.

Hence the correct options are:
Option A and Option D.

Solution :

The source is placed at the centre of curvature of the hemisphere.

Hence, all light rays strike the reflecting surface normally.

For complete reflection, radiation pressure is :

$$p = \frac{2I}{c}$$

Consider a small surface element on hemisphere.

Force on opposite elements have horizontal components that cancel due to symmetry.

Only axial components add.

Net force on hemisphere is :

$$F = \frac{P}{2c}$$

Given :

$$P = 24\text{ W}$$

$$c = 3 \times 10^8\text{ m s}^{-1}$$

Therefore,

$$F = \frac{24}{2 \times 3 \times 10^8}$$

$$= \frac{24}{6 \times 10^8}$$

$$= 4 \times 10^{-8}\text{ N}$$

Final Answer :

$$4$$

We are given the electric field $$E = 56.5 \sin\omega\left(\frac{t - x}{c}\right)$$ NC$$^{-1}$$ for a wave propagating along the $$x$$-axis in free space. This expression matches the form $$E = E_0 \sin(\omega t - kx)$$ with $$k = \frac{\omega}{c}$$, so the amplitude is $$E_0 = 56.5 \text{ NC}^{-1}$$.

The intensity (average power per unit area) of an electromagnetic wave in free space is given by the time-averaged Poynting vector as $$I = \frac{1}{2}\varepsilon_0 c E_0^2$$, where the factor $$\frac{1}{2}$$ arises from averaging $$\sin^2$$ over a full cycle.

Squaring the amplitude yields $$E_0^2 = (56.5)^2 = 3192.25 \text{ N}^2\text{C}^{-2}$$, while the product $$\varepsilon_0 c = 8.85 \times 10^{-12} \times 3 \times 10^{8} = 26.55 \times 10^{-4} = 2.655 \times 10^{-3} \text{ C}^2\text{N}^{-1}\text{m}^{-1}\text{s}^{-1}$$.

Substituting these values into the intensity formula gives

$$I = \frac{1}{2} \times 2.655 \times 10^{-3} \times 3192.25$$

$$= \frac{1}{2} \times 8.4754$$

$$= 4.2377 \approx 4.24 \text{ Wm}^{-2}$$.

The correct answer is Option C: $$4.24$$ Wm$$^{-2}$$.

The speed of electromagnetic waves in a medium is given by:

$$v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$

where $$c = 3.0 \times 10^8$$ m/s is the speed of light in vacuum, $$\mu_r$$ is the relative permeability, and $$\varepsilon_r$$ is the relative permittivity.

Given: $$v = 2.0 \times 10^8$$ m/s and $$\mu_r = 1.0$$.

Rearranging:

$$\sqrt{\mu_r \varepsilon_r} = \frac{c}{v}$$

$$\mu_r \varepsilon_r = \frac{c^2}{v^2}$$

$$\varepsilon_r = \frac{c^2}{\mu_r v^2} = \frac{(3.0 \times 10^8)^2}{1.0 \times (2.0 \times 10^8)^2}$$

$$\varepsilon_r = \frac{9.0 \times 10^{16}}{4.0 \times 10^{16}} = \frac{9}{4} = 2.25$$

Hence, the correct answer is Option A.

We need to find the ratio of electric force to magnetic force on a charge moving along the Y-axis in an electromagnetic wave travelling along the X-axis.

The electric field is $$E_y = 900\sin\omega\left(t - \frac{x}{c}\right)$$ and for an EM wave travelling along the X-axis with $$E$$ along the Y-axis, the magnetic field $$B$$ is along the Z-axis given by $$B_z = \frac{E_y}{c}$$.

The charge $$q$$ moves along the Y-axis with speed $$v = 3 \times 10^7$$ m/s. Since the electric force is given by $$F_E = qE_y$$ and the magnetic force is $$F_B = qvB_z = qv\frac{E_y}{c}$$, we can form their ratio.

From the above expressions, we have $$\frac{F_E}{F_B} = \frac{qE_y}{qv \cdot E_y/c} = \frac{c}{v}$$. Substituting the values gives $$= \frac{3 \times 10^8}{3 \times 10^7} = 10$$.

Therefore, the correct answer is Option C: $$10:1$$.

An EM wave propagates in the $$x$$-direction with wavelength $$\lambda = 8$$ mm $$= 8 \times 10^{-3}$$ m. The electric field vibrates in the $$y$$-direction with maximum magnitude $$E_0 = 60$$ V m$$^{-1}$$.

Calculate the wave number $$k$$.

$$k = \frac{2\pi}{\lambda} = \frac{2\pi}{8 \times 10^{-3}} = \frac{\pi}{4} \times 10^3 \text{ m}^{-1}$$

Write the angular frequency $$\omega$$.

$$\omega = kc = \frac{\pi}{4} \times 10^3 \times 3 \times 10^8 = \frac{3\pi}{4} \times 10^{11} \text{ rad s}^{-1}$$

The argument of the sine function can be written as $$k(x - ct) = \frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t)$$.

Calculate the magnetic field amplitude $$B_0$$.

$$B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \text{ T}$$

Write the complete equations.

Since the wave propagates in the $$x$$-direction and $$\vec{E}$$ is along $$\hat{j}$$, the magnetic field $$\vec{B}$$ must be along $$\hat{k}$$ (as $$\vec{E} \times \vec{B}$$ should point in the direction of propagation).

$$E_y = 60 \sin\left[\frac{\pi}{4} \times 10^3(x - 3 \times 10^8 t)\right] \hat{j} \text{ V m}^{-1}$$

$$B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{4} \times 10^3(x - 3 \times 10^8 t)\right] \hat{k} \text{ T}$$

The correct answer is Option B.

We need to identify the correct statements about electromagnetic waves from the given options.

Key Properties of Electromagnetic Waves:

In a plane electromagnetic wave propagating in free space:

1. The electric field $$\vec{E}$$, magnetic field $$\vec{B}$$, and direction of propagation $$\vec{k}$$ are mutually perpendicular.

2. The direction of propagation is given by $$\vec{E} \times \vec{B}$$ (Poynting vector direction).

3. The relation between amplitudes is $$\frac{E_0}{B_0} = c$$ (speed of light).

4. The average energy density in electric field = average energy density in magnetic field, i.e., $$\frac{1}{2}\epsilon_0 E_0^2 = \frac{B_0^2}{2\mu_0}$$ (on average).

Statement A: "Electric field and magnetic field must be perpendicular to each other and direction of propagation should be along electric field or magnetic field."

The first part is correct: $$\vec{E} \perp \vec{B}$$.

The second part is wrong: the direction of propagation is perpendicular to BOTH $$\vec{E}$$ and $$\vec{B}$$, not along either of them. The propagation direction is along $$\vec{E} \times \vec{B}$$.

Statement A is INCORRECT.

Statement B: "The energy in electromagnetic wave is divided equally between electric and magnetic fields."

The average electric energy density is $$u_E = \frac{1}{2}\epsilon_0 E_{rms}^2 = \frac{1}{4}\epsilon_0 E_0^2$$.

The average magnetic energy density is $$u_B = \frac{B_{rms}^2}{2\mu_0} = \frac{B_0^2}{4\mu_0}$$.

Using the relation $$E_0 = cB_0$$ and $$c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$$:

$$u_E = \frac{1}{4}\epsilon_0 E_0^2 = \frac{1}{4}\epsilon_0 c^2 B_0^2 = \frac{1}{4}\epsilon_0 \cdot \frac{1}{\mu_0\epsilon_0} \cdot B_0^2 = \frac{B_0^2}{4\mu_0} = u_B$$

So the energy is indeed equally divided between the electric and magnetic fields.

Statement B is CORRECT.

Statement C: "Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation."

While both $$\vec{E}$$ and $$\vec{B}$$ are perpendicular to the direction of propagation (they are transverse waves), they are NOT parallel to each other. They are perpendicular to each other.

Statement C is INCORRECT.

Statement D: "The electric field, magnetic field and direction of propagation must be perpendicular to each other."

This is exactly the defining property of plane EM waves: $$\vec{E}$$, $$\vec{B}$$, and $$\hat{k}$$ (propagation direction) form a right-handed mutually perpendicular set.

Statement D is CORRECT.

Statement E: "The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light."

The correct relation is: $$\frac{E_0}{B_0} = c$$

This means the ratio of electric field amplitude to magnetic field amplitude equals the speed of light.

The statement claims $$\frac{B_0}{E_0} = c$$, which is the inverse of the correct relation. Since $$\frac{B_0}{E_0} = \frac{1}{c}$$, this statement is wrong.

Statement E is INCORRECT.

The correct statements are B and D.

Answer: Option B: B and D only

We need to match each type of electromagnetic wave with its primary application.

(a) UV rays → (ii) Water purification: Ultraviolet rays are widely used for sterilisation and water purification, as they can kill bacteria and other microorganisms.

(b) X-rays → (i) Diagnostic tool in medicine: X-rays are extensively used in medical imaging (radiography, CT scans) to examine bones, teeth, and internal organs.

(c) Microwave → (iii) Communication, Radar: Microwaves are used in radar systems, satellite communication, and mobile phone networks due to their ability to travel long distances and penetrate the atmosphere.

(d) Infrared wave → (iv) Improving visibility in foggy days: Infrared waves can penetrate fog and haze better than visible light, which is why infrared cameras and sensors are used to improve visibility in poor weather conditions.

The correct matching is (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv). Hence, the correct answer is Option B.

We are given that sunlight falls normally on a surface of area $$A = 36 \text{ cm}^2$$ and exerts an average force of $$F = 7.2 \times 10^{-9}$$ N. The light is completely absorbed.

For complete absorption, the radiation pressure is $$P = \frac{I}{c}$$, where $$I$$ is the intensity (energy flux) and $$c$$ is the speed of light. Also, $$F = P \times A$$, so $$F = \frac{I \times A}{c}$$.

Solving for the intensity: $$I = \frac{Fc}{A}$$.

Substituting values (using $$A = 36 \times 10^{-4} \text{ m}^2$$ and $$c = 3 \times 10^8 \text{ m/s}$$):

$$I = \frac{7.2 \times 10^{-9} \times 3 \times 10^8}{36 \times 10^{-4}} = \frac{2.16}{36 \times 10^{-4}} = \frac{2.16}{3.6 \times 10^{-3}} = 600 \text{ W/m}^2$$.

Now converting to W cm$$^{-2}$$: since $$1 \text{ m}^2 = 10^4 \text{ cm}^2$$, we get $$I = \frac{600}{10^4} = 0.06 \text{ W cm}^{-2}$$.

Hence, the correct answer is Option D.

The magnetic field of a plane electromagnetic wave is given by:

$$\vec{B} = 2 \times 10^{-8} \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t)\,\hat{j} \text{ T}$$

We need to find the amplitude of the electric field. From the wave equation $$\sin(kx + \omega t)$$ we identify $$k = 0.5 \times 10^3 \text{ rad/m}$$ and $$\omega = 1.5 \times 10^{11} \text{ rad/s}$$. Substituting gives $$c = \frac{\omega}{k} = \frac{1.5 \times 10^{11}}{0.5 \times 10^3} = 3 \times 10^8 \text{ m/s}$$.

Next, the relation between electric and magnetic field amplitudes in an EM wave is $$E_0 = c \times B_0 = 3 \times 10^8 \times 2 \times 10^{-8} = 6 \text{ V/m}$$.

Since the wave propagates in the $$-x$$ direction (the argument $$kx + \omega t$$ gives $$-\hat{i}$$) and the magnetic field is along $$\hat{j}$$, the fields are mutually perpendicular with $$\hat{E} \times \hat{B}$$ giving the direction of propagation. This gives $$\hat{E} \times \hat{j} = -\hat{i}$$. Then $$\hat{E} = \hat{k} \quad \text{(since } \hat{k} \times \hat{j} = -\hat{i}\text{)}$$, so the electric field is along the $$z$$-axis.

The correct answer is Option C: $$6 \text{ V m}^{-1}$$ along $$z$$-axis.

The oscillating magnetic field in a plane electromagnetic wave is given by: $$B_y = 5 \times 10^{-6} \sin[1000\pi(5x - 4 \times 10^8 t)] \text{ T}$$, and we need to find the amplitude of the electric field.

Since the amplitude of the magnetic field is the coefficient of the sine function, we have $$B_0 = 5 \times 10^{-6} \text{ T}$$.

From the general form of a travelling wave, $$B = B_0 \sin(kx - \omega t)$$, expanding the given argument yields $$1000\pi(5x - 4 \times 10^8 t) = 5000\pi \cdot x - 4\pi \times 10^{11} \cdot t$$. Therefore, $$k = 5000\pi \text{ m}^{-1}$$ and $$\omega = 4\pi \times 10^{11} \text{ rad s}^{-1}$$.

Since the wave speed is given by $$c = \dfrac{\omega}{k}$$, substituting the values of $$\omega$$ and $$k$$ gives $$c = \dfrac{4\pi \times 10^{11}}{5000\pi} = \dfrac{4 \times 10^{11}}{5 \times 10^3} = 8 \times 10^7 \text{ m s}^{-1}$$.

For an electromagnetic wave, the amplitudes of the electric and magnetic fields are related by $$E_0 = c \times B_0$$, where $$c$$ is the wave speed in the medium. Hence, $$E_0 = 8 \times 10^7 \times 5 \times 10^{-6} = 40 \times 10^{7-6} = 40 \times 10^1 = 400 \text{ V m}^{-1}$$, which is $$4 \times 10^2 \text{ V m}^{-1}$$.

Note that the wave speed here is $$8 \times 10^7 \text{ m s}^{-1}$$, which is less than the speed of light in vacuum ($$3 \times 10^8 \text{ m s}^{-1}$$). This indicates the wave is travelling in a medium, and the relation $$E_0 = cB_0$$ still holds where $$c$$ is the wave speed in that medium.

Therefore, the correct answer is Option D: $$4 \times 10^2 \text{ V m}^{-1}$$.

We need to evaluate two statements about electromagnetic waves.

Statement I: "A time varying electric field is a source of changing magnetic field and vice-versa. Thus a disturbance in electric or magnetic field creates EM waves."

This is correct. According to Maxwell's equations, a time-varying electric field produces a magnetic field (via the displacement current term in Ampere-Maxwell law), and a time-varying magnetic field produces an electric field (Faraday's law). This mutual generation of fields is the basis of electromagnetic wave propagation.

Statement II: "In a material medium, the EM wave travels with speed $$v = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$$."

This is false. The speed $$\frac{1}{\sqrt{\mu_0\varepsilon_0}} = c$$ is the speed of light in vacuum. In a material medium, the speed of an EM wave is:

$$v = \frac{1}{\sqrt{\mu\varepsilon}}$$

where $$\mu$$ and $$\varepsilon$$ are the permeability and permittivity of the medium respectively (not the free-space values $$\mu_0$$ and $$\varepsilon_0$$). Since $$\mu \geq \mu_0$$ and $$\varepsilon \geq \varepsilon_0$$ for most materials, we have $$v \leq c$$.

Therefore, Statement I is correct but Statement II is false.

The correct answer is Option C.

The electric field of a light wave is given by $$E_y = 540 \sin \pi \times 10^4(x - ct) \text{ V m}^{-1}$$.

Identify the peak value of the electric field.

From the equation, the amplitude (peak value) of the electric field is:

Find the peak value of the magnetic field.

For an electromagnetic wave, the relationship between the electric and magnetic field amplitudes is:

The correct answer is Option A: $$18 \times 10^{-7} \text{ T}$$.

Let us match each electromagnetic wave with its correct application:

(a) Ultraviolet rays - (iii) Sterilizing surgical instruments: UV rays have germicidal properties and are widely used to sterilize surgical instruments by killing bacteria and viruses.

(b) Microwaves - (iv) Radar system: Microwaves are used in radar systems (RADAR - Radio Detection and Ranging) for detecting objects and measuring distances.

(c) Infrared waves - (ii) Greenhouse effect: Infrared waves are responsible for the greenhouse effect. The Earth absorbs solar radiation and re-emits it as infrared radiation, which is trapped by greenhouse gases.

(d) X-rays - (i) Study crystal structure: X-rays are used in X-ray diffraction to study crystal structures (Bragg's law).

The correct matching is: (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i).

The correct answer is Option A.

We are given: relative permeability $$\mu_r = 1.61$$, relative permittivity $$\varepsilon_r = 6.44$$, magnetic field intensity $$H = 4.5 \times 10^{-2}$$ A/m, $$\mu_0 = 4\pi \times 10^{-7}$$ N A$$^{-2}$$, and $$c = 3 \times 10^8$$ m/s.

The intrinsic impedance of a medium is:

$$ \eta = \frac{E}{H} = \sqrt{\frac{\mu}{\varepsilon}} = \sqrt{\frac{\mu_r \mu_0}{\varepsilon_r \varepsilon_0}} $$

Simplifying using the impedance of free space:

$$ \eta = \sqrt{\frac{\mu_r}{\varepsilon_r}} \times \sqrt{\frac{\mu_0}{\varepsilon_0}} $$

The impedance of free space is:

$$ \eta_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} = \mu_0 c = 4\pi \times 10^{-7} \times 3 \times 10^8 = 120\pi\ \Omega $$

The ratio is:

$$ \sqrt{\frac{\mu_r}{\varepsilon_r}} = \sqrt{\frac{1.61}{6.44}} = \sqrt{0.25} = 0.5 $$

Thus,

$$ E = \eta \times H = 0.5 \times 120\pi \times 4.5 \times 10^{-2} $$

$$ E = 0.5 \times 120 \times 3.1416 \times 4.5 \times 10^{-2} $$

$$ E = 60\pi \times 4.5 \times 10^{-2} $$

$$ E = 2.7\pi = 2.7 \times 3.1416 \approx 8.48 \text{ V/m} $$

Therefore, the correct answer is Option C.

We need to arrange the electromagnetic waves in ascending order of wavelength.

The electromagnetic spectrum in order of increasing wavelength is:

$$\text{Gamma rays} \rightarrow \text{X-rays} \rightarrow \text{UV} \rightarrow \text{Visible} \rightarrow \text{Infrared} \rightarrow \text{Microwaves} \rightarrow \text{Radio waves}$$

• Gamma rays: $$\lambda < 10^{-12}$$ m (shortest wavelength)
• X-rays: $$10^{-12}$$ m to $$10^{-8}$$ m
• Visible light: $$4 \times 10^{-7}$$ m to $$7 \times 10^{-7}$$ m
• Microwaves: $$10^{-3}$$ m to $$0.3$$ m (longest among the given options)

$$\lambda_{\text{gamma-ray}} < \lambda_{\text{X-ray}} < \lambda_{\text{visible}} < \lambda_{\text{microwave}}$$

Hence, the correct answer is Option B.

We are given an electric bulb rated at $$200$$ W with $$3.5\%$$ efficiency, and we need to determine the peak magnetic field at a distance of $$4$$ m.

Since the bulb converts $$200 \times \frac{3.5}{100} = 200 \times 0.035 = 7$$ W into radiated power, we have:

$$P_{rad} = 200 \times \frac{3.5}{100} = 200 \times 0.035 = 7 \text{ W}$$

This radiated power spreads uniformly over a spherical surface of radius $$4$$ m, so the intensity is:

$$I = \frac{P_{rad}}{4\pi r^2} = \frac{7}{4\pi (4)^2} = \frac{7}{4\pi \times 16} = \frac{7}{64\pi} \text{ W/m}^2$$

From the relation between the intensity of an electromagnetic wave and its peak magnetic field $$B_0$$,

$$I = \frac{c B_0^2}{2\mu_0}$$

we solve for $$B_0$$:

$$B_0 = \sqrt{\frac{2\mu_0 I}{c}}$$

Substituting $$\mu_0 = 4\pi \times 10^{-7}$$ T m/A and $$c = 3 \times 10^8$$ m/s yields:

$$B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times \frac{7}{64\pi}}{3 \times 10^8}}$$

Simplifying the numerator gives:

$$2 \times 4\pi \times 10^{-7} \times \frac{7}{64\pi} = \frac{2 \times 4 \times 7}{64} \times 10^{-7} = \frac{56}{64} \times 10^{-7} = \frac{7}{8} \times 10^{-7}$$

Hence,

$$B_0 = \sqrt{\frac{0.875 \times 10^{-7}}{3 \times 10^8}} = \sqrt{\frac{0.875}{3} \times 10^{-15}}$$

$$B_0 = \sqrt{0.29167 \times 10^{-15}} = \sqrt{2.9167 \times 10^{-16}}$$

$$B_0 = 1.708 \times 10^{-8} \text{ T} \approx 1.71 \times 10^{-8} \text{ T}$$

Therefore, the correct option is Option B: $$1.71 \times 10^{-8}$$ T.

We are given that the variation (peak-to-peak change) in the amplitude of the AM wave is 8 V and the maximum amplitude of the AM wave is $$A_{\max} = 9$$ V.

We know that for an amplitude-modulated wave, the maximum and minimum amplitudes are related to the carrier amplitude $$A_c$$ and the modulating signal amplitude $$A_m$$ by $$A_{\max} = A_c + A_m$$ and $$A_{\min} = A_c - A_m$$. The "8 V variation" means the total swing from minimum to maximum, so $$A_{\max} - A_{\min} = 8$$ V. This gives us $$A_{\max} - A_{\min} = (A_c + A_m) - (A_c - A_m) = 2A_m = 8$$, hence $$A_m = 4$$ V.

Now, since $$A_{\max} = A_c + A_m = 9$$, we get $$A_c = 9 - 4 = 5$$ V.

The modulation index is defined as $$\mu = \dfrac{A_m}{A_c} = \dfrac{4}{5} = 0.8$$.

Hence, the correct answer is Option A.

We need to find the wavelength bandwidth corresponding to the frequency band 6 MHz to 10 MHz.

The relationship between wavelength and frequency is:

$$\lambda = \dfrac{c}{f}$$

where $$c = 3 \times 10^8 \text{ m/s}$$.

For $$f_1 = 6 \text{ MHz} = 6 \times 10^6 \text{ Hz}$$:

$$\lambda_1 = \dfrac{3 \times 10^8}{6 \times 10^6} = 50 \text{ m}$$

For $$f_2 = 10 \text{ MHz} = 10 \times 10^6 \text{ Hz}$$:

$$\lambda_2 = \dfrac{3 \times 10^8}{10 \times 10^6} = 30 \text{ m}$$

The wavelength bandwidth is:

$$\Delta\lambda = \lambda_1 - \lambda_2 = 50 - 30 = 20 \text{ m}$$

Therefore, the correct answer is Option B.

We are given the maximum voltage $$V_{max} = 60 \text{ V}$$ and minimum voltage $$V_{min} = 20 \text{ V}$$ of an amplitude modulated signal.

The modulation index is defined as:

$$\mu = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$$

Substituting the given values:

$$\mu = \frac{60 - 20}{60 + 20} = \frac{40}{80} = 0.5$$

The percentage modulation index is:

$$\mu \times 100 = 0.5 \times 100 = 50\%$$

The correct answer is Option B.

We are given an FM broadcast transmitter with a modulating signal frequency of $$f_m = 20$$ kHz and a deviation ratio of 10.

The deviation ratio (also called modulation index) is defined as $$m_f = \frac{\Delta f}{f_m}$$, where $$\Delta f$$ is the maximum frequency deviation. So $$\Delta f = m_f \times f_m = 10 \times 20 = 200$$ kHz.

By Carson's rule, the bandwidth required for FM transmission is $$BW = 2(\Delta f + f_m) = 2(200 + 20) = 2 \times 220 = 440$$ kHz.

Hence, the correct answer is Option D.

We need to determine the modulation index of an amplitude modulated (AM) signal based on the provided modulating signal's graph and the carrier wave equation.

1. Identify the System Parameters

• Carrier Wave Equation: $$C(t) = 5 \sin(8\pi t)\text{ Volt}$$

  From the standard carrier form $$C(t) = A_c \sin(\omega_c t)$$, the peak amplitude of the carrier wave is:

  $$A_c = 5\text{ V}$$
• Modulating Signal Graph ($$m(t)$$):

  Looking at the provided square wave graph, the maximum positive peak reaches a value of $$1\text{ V}$$. Thus, the peak amplitude of the modulating signal is:

  $$A_m = 1\text{ V}$$

2. Calculate the Modulation Index ($\mu$)

The modulation index ($$\mu$$) for an amplitude modulated wave is defined as the ratio of the peak amplitude of the modulating signal to the peak amplitude of the carrier wave:

$$\mu = \frac{A_m}{A_c}$$

Substituting our identified parameters into the formula:

$$\mu = \frac{1}{5} = 0.2$$

Conclusion

The modulation index of the signal is 0.2

The amplitude modulated wave is given as: $$V_{AM} = 10\left[1 + 0.4 \cos(2\pi \times 10^4 t)\right] \cos(2\pi \times 10^7 t)$$. Comparing this with the standard AM wave equation $$V_{AM} = A_c\left[1 + \mu \cos(2\pi f_m t)\right] \cos(2\pi f_c t)$$, we identify the carrier frequency as $$f_c = 10^7$$ Hz = 10 MHz, the modulating frequency as $$f_m = 10^4$$ Hz = 10 kHz, and the modulation index as $$\mu = 0.4$$.

The bandwidth of an AM wave is given by $$\text{Bandwidth} = 2f_m$$. $$= 2 \times 10^4 \text{ Hz} = 20 \text{ kHz}$$. Hence, the correct answer is Option C (20 kHz).

We are given that the TV transmission tower has a height of $$h = 100 \text{ m}$$ and we need to find the new height to triple its coverage range.

The coverage range (or line-of-sight distance) of a TV tower is given by:

$$d = \sqrt{2Rh}$$

where $$R$$ is the radius of the Earth and $$h$$ is the height of the tower.

Let the original range be $$d_1 = \sqrt{2R \times 100}$$ and the new range be $$d_2 = 3d_1$$.

For the new height $$h'$$:

$$d_2 = \sqrt{2Rh'}$$

Since $$d_2 = 3d_1$$:

$$\sqrt{2Rh'} = 3\sqrt{2R \times 100}$$

Squaring both sides:

$$2Rh' = 9 \times 2R \times 100$$

$$h' = 9 \times 100 = 900 \text{ m}$$

The correct answer is Option D: $$900 \text{ m}$$.

In amplitude modulation (AM), the modulation index $$\mu$$ is defined as:

$$\mu = \dfrac{A_{max} - A_{min}}{A_{max} + A_{min}}$$

Given: $$A_{max} = 6 \text{ V}$$ and $$A_{min} = 2 \text{ V}$$.

Substituting:

$$\mu = \dfrac{6 - 2}{6 + 2} = \dfrac{4}{8} = 0.5$$

Converting to percentage:

$$\mu = 0.5 \times 100\% = 50\%$$

Therefore, the correct answer is Option D.

In amplitude modulation (AM), the modulated wave can be expressed as:

where $$\mu$$ is the modulation index defined as:

Here, $$A_m$$ is the amplitude of the message signal and $$A_c$$ is the amplitude of the carrier signal.

For the modulated signal to be free from distortion (i.e., the envelope of the modulated wave should faithfully reproduce the message signal), we need:

If $$\mu > 1$$, the modulated wave undergoes over-modulation, leading to distortion of the signal during demodulation. The envelope of the carrier no longer faithfully follows the modulating signal.

If $$\mu = 0$$, there is no modulation at all, so no information is transmitted.

Therefore, the correct answer is Option A: $$\mu \leq 1$$.

We need to find the increase in height required to double the coverage range of a TV tower. The range of a TV transmission tower is $$d = \sqrt{2Rh}$$ where R is the radius of Earth and h is the height of the tower.

The original range is $$d_1 = \sqrt{2R \times 125}$$ when the tower height is 125 m, and the new range is $$d_2 = 2d_1 = \sqrt{2Rh_2}$$. Equating these gives $$2\sqrt{2R \times 125} = \sqrt{2Rh_2}$$, and squaring both sides yields $$4 \times 2R \times 125 = 2Rh_2$$. Therefore, $$h_2 = 4 \times 125 = 500 \text{ m}$$.

Hence the required increase in height is $$\Delta h = h_2 - h_1 = 500 - 125 = 375 \text{ m}$$. The answer is Option B: 375 m.

We have a 110 W light bulb, and nearly 10% of its power is converted to visible radiation. So the power of visible radiation is $$P = 0.10 \times 110 = 11$$ W.

The intensity of radiation at a distance $$r$$ from a point source is given by $$I = \frac{P}{4\pi r^2}$$.

At a distance $$r_1 = 1$$ m: $$I_1 = \frac{11}{4\pi (1)^2} = \frac{11}{4\pi}$$

At a distance $$r_2 = 5$$ m: $$I_2 = \frac{11}{4\pi (5)^2} = \frac{11}{4\pi \times 25} = \frac{11}{100\pi}$$

The change in intensity is: $$\Delta I = I_1 - I_2 = \frac{11}{4\pi} - \frac{11}{100\pi} = \frac{11}{\pi}\left(\frac{1}{4} - \frac{1}{100}\right) = \frac{11}{\pi} \times \frac{25 - 1}{100} = \frac{11 \times 24}{100\pi}$$

$$\Delta I = \frac{264}{100\pi} = \frac{264}{314.159} \approx 0.8403$$ W/m$$^2$$ $$= 84.03 \times 10^{-2}$$ W/m$$^2$$.

Hence the value of $$a = 84$$.

Hence, the correct answer is 84.

We need to find the distance between the plates of a parallel plate capacitor given the displacement current. The displacement current in a parallel plate capacitor is:

$$I_d = \varepsilon_0 \frac{A}{d} \cdot \frac{dV}{dt}$$

This comes from $$I_d = C \frac{dV}{dt}$$ where $$C = \frac{\varepsilon_0 A}{d}$$. Substituting the given values: $$I_d = 4.425 \times 10^{-6}$$ A, $$\frac{dV}{dt} = 10^6$$ V/s, $$A = 40$$ cm² $$= 40 \times 10^{-4}$$ m², $$\varepsilon_0 = 8.85 \times 10^{-12}$$ C² N⁻¹ m⁻². To solve for d, we use:

$$d = \frac{\varepsilon_0 A}{I_d} \cdot \frac{dV}{dt}$$

$$d = \frac{8.85 \times 10^{-12} \times 40 \times 10^{-4} \times 10^6}{4.425 \times 10^{-6}}$$

$$= \frac{8.85 \times 40 \times 10^{-12-4+6}}{4.425 \times 10^{-6}}$$

$$= \frac{354 \times 10^{-10}}{4.425 \times 10^{-6}}$$

$$= \frac{354}{4.425} \times 10^{-4}$$

$$= 80 \times 10^{-4} = 8 \times 10^{-3} \text{ m}$$

So $$x = 8$$. The answer is 8.

We have a modulating signal $$m(t) = 2\sin(6.28 \times 10^6 t)$$ and a carrier signal $$c(t) = 4\sin(12.56 \times 10^9 t)$$. The modulating frequency is $$f_m = \frac{\omega_m}{2\pi} = \frac{6.28 \times 10^6}{2\pi} = \frac{6.28 \times 10^6}{6.28} = 10^6$$ Hz $$= 1$$ MHz. The carrier frequency is $$f_c = \frac{12.56 \times 10^9}{2\pi} = \frac{12.56 \times 10^9}{6.28} = 2 \times 10^9$$ Hz $$= 2000$$ MHz.

The combined signal is $$y(t) = m(t) + c(t)$$. When this passes through a non-linear square law device, the output contains terms proportional to $$y(t)$$ and $$y(t)^2$$. Expanding $$y(t)^2 = [m(t) + c(t)]^2 = m(t)^2 + c(t)^2 + 2m(t)c(t)$$, we note that the cross-term $$2m(t)c(t)$$ produces frequencies at $$f_c + f_m$$ and $$f_c - f_m$$, which are the upper and lower sidebands of the AM signal. The terms $$m(t)^2$$ and $$c(t)^2$$ produce low-frequency and high-frequency components (at $$2f_m$$, DC, and $$2f_c$$) that lie outside the carrier band.

Now, the band pass filter is centred at the carrier frequency $$f_c$$. It allows through only the frequencies near $$f_c$$, namely the carrier at $$f_c = 2000$$ MHz, the upper sideband at $$f_c + f_m = 2001$$ MHz, and the lower sideband at $$f_c - f_m = 1999$$ MHz. All other frequency components are rejected.

The bandwidth of the output signal is the difference between the highest and lowest frequencies passed: $$\text{BW} = (f_c + f_m) - (f_c - f_m) = 2f_m = 2 \times 1 = 2 \text{ MHz}$$

Hence, the correct answer is 2.

An antenna is placed in a dielectric medium with dielectric constant $$\kappa = 6.25$$ and $$\mu_r = 1$$. The maximum size of the antenna is $$5.0$$ mm. We need to find the minimum frequency it can radiate in GHz.

Recall the antenna condition. For a quarter-wave antenna, the minimum antenna length required is one-fourth of the wavelength:

$$l = \frac{\lambda}{4}$$

For the minimum frequency (longest wavelength), the antenna size equals $$\frac{\lambda}{4}$$:

$$\lambda = 4l = 4 \times 5.0 \text{ mm} = 20 \text{ mm} = 0.02 \text{ m}$$

Find the speed of electromagnetic waves in the dielectric medium. The speed in a medium with dielectric constant $$\kappa$$ and relative permeability $$\mu_r$$ is:

$$v = \frac{c}{\sqrt{\kappa \cdot \mu_r}}$$

Substituting the given values:

$$v = \frac{c}{\sqrt{6.25 \times 1}} = \frac{c}{\sqrt{6.25}} = \frac{c}{2.5}$$

$$v = \frac{3 \times 10^8}{2.5} = 1.2 \times 10^8 \text{ m/s}$$

Calculate the minimum frequency using $$v = f\lambda$$:

$$f_{min} = \frac{v}{\lambda} = \frac{1.2 \times 10^8}{0.02}$$

$$f_{min} = 6 \times 10^9 \text{ Hz} = 6 \text{ GHz}$$

The minimum frequency of the signal the antenna can radiate is 6 GHz.

The population covered is $$6.03 \text{ lakh} = 6.03 \times 10^5$$, the population density is $$100 \text{ per km}^2$$, and the radius of the Earth is $$R = 6400 \text{ km}$$.

From the population and density, the area covered by the TV tower is $$\text{Area} = \dfrac{\text{Population}}{\text{Population density}} = \dfrac{6.03 \times 10^5}{100} = 6030 \text{ km}^2$$.

For a TV tower of height $$h$$, the coverage area can be expressed as $$A = \pi d^2$$ where $$d = \sqrt{2Rh}$$; hence $$A = \pi \times 2Rh = 2\pi Rh$$.

Setting this equal to the area gives $$6030 = 2\pi \times 6400 \times h$$, so $$h = \dfrac{6030}{2\pi \times 6400} = \dfrac{6030}{40212.4} = 0.14998 \text{ km}$$, which is approximately $$h \approx 0.15 \text{ km} = 150 \text{ m}$$.

Therefore, the height of the TV tower is $$\boxed{150}$$ m.

A radar sends an EM signal that strikes a target at 3 km distance and the echo returns. The velocity of an EM wave is given by: $$v = \frac{E_0}{B_0}$$. Substituting the given values yields $$v = \frac{2.25}{1.5 \times 10^{-8}} = 1.5 \times 10^8$$ m/s.

Since the signal travels to the target and back, the total distance covered is $$2 \times 3$$ km = $$6$$ km = $$6000$$ m.

Therefore, the time for the echo to return is calculated by $$t = \frac{\text{Total distance}}{v} = \frac{6000}{1.5 \times 10^8}$$, which simplifies to $$t = \frac{6 \times 10^3}{1.5 \times 10^8} = 4 \times 10^{-5}$$ s. The correct answer is Option B.

For effective transmission, the wavelength of the signal transmitted by an antenna is related to the length of the antenna $$L$$ by the relation $$L = \frac{\lambda}{4}$$, where $$\lambda$$ is the wavelength of the signal.

Given that the antenna length is $$L = 25$$ m (note that the height of the tower is irrelevant — only the antenna length matters for determining the transmitted wavelength), we have:

$$\lambda = 4L = 4 \times 25 = 100 \text{ m}$$.

We have been given an incident electric field in free space

$$$\vec E_i(z,t)=3.1\cos(1.8\,z-5.4\times10^{6}\,t)\,\hat i\;{\rm N\,C^{-1}}.$$$

This is of the standard form $$\vec E_i=E_0\cos(kz-\omega t)\hat i,$$ so by simple comparison

$$$k=1.8\;{\rm rad\,m^{-1}},\qquad \omega=5.4\times10^{6}\;{\rm rad\,s^{-1}},\qquad E_0=3.1\;{\rm N\,C^{-1}}.$$$

First, let us find the wavelength. The relation between the propagation constant and wavelength in vacuum is

$$k=\frac{2\pi}{\lambda}\;\Longrightarrow\;\lambda=\frac{2\pi}{k}.$$

Substituting the numerical value

$$\lambda=\frac{2\pi}{1.8}= \frac{6.2832}{1.8}\approx3.49\;{\rm m}.$$

This is clearly not equal to $$5.4\;{\rm m}$$, so statement A (“The wavelength is 5.4 m”) is wrong.

Next, let us calculate the ordinary frequency $$f$$. The angular frequency and ordinary frequency are related by

$$\omega=2\pi f\;\Longrightarrow\;f=\frac{\omega}{2\pi}.$$

Hence

$$$f=\frac{5.4\times10^{6}}{2\pi}= \frac{5.4\times10^{6}}{6.2832}\approx8.6\times10^{5}\;{\rm Hz}.$$$

The option claims $$54\times10^{4}\;{\rm Hz}=5.4\times10^{5}\;{\rm Hz}$$, which is different from $$8.6\times10^{5}\;{\rm Hz}$$. Therefore statement B is also wrong.

Because the wall is perfectly reflecting (perfect conductor), no electromagnetic field can be transmitted into it; the tangential component of $$\vec E$$ must vanish at the surface. Hence there is no transmitted wave. Statement C therefore cannot be true.

We now construct the reflected wave. For normal incidence, the reflected field must travel in the $$-z$$ direction and can be written in the general form

$$\vec E_r(z,t)=E_0\cos(kz+\omega t+\phi)\,\hat i,$$

where $$\phi$$ is a phase constant to be fixed by the boundary condition at the wall situated at $$z=a$$.

The total tangential electric field on the wall must be zero, so we impose

$$\vec E_i(a,t)+\vec E_r(a,t)=0.$$

That gives

$$$E_0\cos(k a-\omega t)+E_0\cos(k a+\omega t+\phi)=0\qquad\forall\;t.$$$

Dividing through by $$E_0$$ and using the identity

$$\cos C+\cos D=2\cos\frac{C+D}{2}\,\cos\frac{C-D}{2},$$

we obtain

$$$2\cos\!\Bigl(k a+\frac{\phi}{2}\Bigr)\cos\!\Bigl(-\omega t-\frac{\phi}{2}\Bigr)=0\qquad\forall\;t.$$$

For the product of cosines to vanish at every instant, the first cosine must be zero:

$$\cos\!\Bigl(k a+\tfrac{\phi}{2}\Bigr)=0.$$

This condition is satisfied if

$$k a+\frac{\phi}{2}=\frac{(2n+1)\pi}{2},\qquad n=0,1,2,\dots$$

The simplest choice is to keep $$\phi=0$$ and take

$$k a=\frac{(2n+1)\pi}{2}\;\;(n=0,1,2,\dots).$$

With this choice, the reflected field becomes

$$$\boxed{\;\vec E_r(z,t)=3.1\cos(1.8\,z+5.4\times10^{6}\,t)\,\hat i\;{\rm N\,C^{-1}}\;},$$$

which is exactly the expression listed in option D.

We have therefore shown

• Options A and B disagree with the numerical values of wavelength and frequency.
• Option C is impossible because a perfectly reflecting wall allows no transmission.
• Option D is consistent with the required boundary condition and is therefore correct.

Hence, the correct answer is Option D.

We know that the speed of an electromagnetic wave in any medium depends on the electric permittivity and magnetic permeability of that medium. The general relation is stated first:

$$v \;=\; \frac{1}{\sqrt{\mu\,\varepsilon}}$$

Here, $$\mu$$ is the absolute magnetic permeability and $$\varepsilon$$ is the absolute electric permittivity of the medium.

For convenience, these absolute quantities are expressed through their relative values with respect to free space. Thus we write:

$$\mu \;=\; \mu_r\,\mu_0, \qquad \varepsilon \;=\; \varepsilon_r\,\varepsilon_0$$

Substituting these into the relation for $$v$$, we get:

$$v \;=\; \frac{1}{\sqrt{\mu_r \mu_0 \,\varepsilon_r \varepsilon_0}} \;=\; \frac{1}{\sqrt{\mu_r\varepsilon_r}\;\sqrt{\mu_0\varepsilon_0}}$$

The factor $$\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$$ is the speed of light in vacuum, customarily denoted by $$c$$. Hence we write the well-known formula:

$$v = \frac{c}{\sqrt{\mu_r\,\varepsilon_r}}$$

Now we substitute the given data. For distilled water, the relative permittivity is

$$\varepsilon_r = 81$$

and according to the statement of the problem the relative permeability is

$$\mu_r = 1$$

Using $$c = 3.0 \times 10^{8}\;{\rm m\,s^{-1}}$$, we have

$$v \;=\; \frac{3.0 \times 10^{8}}{\sqrt{1 \times 81}} \;=\; \frac{3.0 \times 10^{8}}{\sqrt{81}}$$

Because $$\sqrt{81} = 9$$, this simplifies step by step as follows:

$$v = \frac{3.0 \times 10^{8}}{9} = 0.333\ldots \times 10^{8}$$

Writing $$0.333\ldots \times 10^{8}$$ in proper scientific notation gives

$$v = 3.33 \times 10^{7}\;{\rm m\,s^{-1}}$$

Hence, the correct answer is Option C.

We need to match each type of electromagnetic radiation with its source.

(a) Microwaves are generated by a magnetron, which is a vacuum tube device that produces coherent microwave radiation using the interaction of electrons with a magnetic field. So (a) matches with (ii).

(b) Infrared radiation is produced by the vibration of atoms and molecules. When atoms and molecules vibrate, they emit electromagnetic radiation in the infrared region. So (b) matches with (iv).

(c) Gamma rays originate from the radioactive decay of nuclei. During nuclear transitions, the nucleus releases very high energy photons known as gamma rays. So (c) matches with (i).

(d) X-rays are produced when high-energy electrons knock out inner shell electrons from atoms. The transition of outer electrons to fill these inner shell vacancies releases X-ray photons. So (d) matches with (iii).

The correct matching is (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).

Red light and blue light are both electromagnetic waves that travel at the same speed $$c$$ in vacuum. However, they are distinguished by their position in the visible spectrum.

Red light has a longer wavelength (approximately 620-750 nm) and correspondingly a lower frequency, while blue light has a shorter wavelength (approximately 450-495 nm) and a higher frequency.

Since the relationship between wavelength and frequency is $$c = \lambda \nu$$, having different wavelengths necessarily means having different frequencies (as $$c$$ is constant).

Therefore, red light differs from blue light as they have different frequencies and different wavelengths.

The displacement current amplitude equals $$I_d = C \cdot \frac{dV}{dt}\bigg|_{max} = C \cdot V_0\omega$$, where $$C = \frac{\varepsilon_0 A}{d}$$ is the capacitance of the parallel plate capacitor.

The capacitance is: $$C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 1}{2 \times 10^{-3}} = 4.425 \times 10^{-9}$$ F.

The angular frequency is $$\omega = 2\pi f = 2\pi \times 50 = 100\pi \approx 314.16$$ rad s$$^{-1}$$.

The amplitude of the displacement current is: $$I_d = C V_0 \omega = 4.425 \times 10^{-9} \times 20 \times 314.16 = 4.425 \times 10^{-9} \times 6283.2 \approx 27.8 \times 10^{-6}$$ A $$= 27.79\,\mu$$A.

The intensity of an electromagnetic wave is related to the peak electric field by: $$I = \frac{1}{2}\varepsilon_0 c E_0^2$$

And the peak magnetic field is related to the peak electric field by: $$B_0 = \frac{E_0}{c}$$

Combining these: $$I = \frac{1}{2}\varepsilon_0 c \cdot (B_0 c)^2 = \frac{1}{2}\varepsilon_0 c^3 B_0^2$$

Solving for $$B_0$$: $$B_0 = \sqrt{\frac{2I}{\varepsilon_0 c^3}}$$

Substituting the values: $$I = 0.092 \text{ W m}^{-2}$$, $$\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{N}^{-1}\text{m}^{-2}$$, $$c = 3 \times 10^8 \text{ m s}^{-1}$$: $$B_0 = \sqrt{\frac{2 \times 0.092}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}}$$

Computing the denominator: $$(3\times10^8)^3 = 27\times10^{24}$$ $$8.85\times10^{-12} \times 27\times10^{24} = 238.95\times10^{12} = 2.3895\times10^{14}$$

$$B_0 = \sqrt{\frac{0.184}{2.3895\times10^{14}}} = \sqrt{7.699\times10^{-16}} = 2.775\times10^{-8} \text{ T} \approx 2.77\times10^{-8} \text{ T}$$

We observe that the electric field of the given plane electromagnetic wave is written in the form

$$E = 50 \sin(500x - 10 \times 10^{10} t) \; \text{V m}^{-1}$$

The standard mathematical expression for a monochromatic plane wave travelling along the positive $$x$$-direction is

$$E = E_0 \sin(kx - \omega t),$$

where $$k$$ is the angular wave number and $$\omega$$ is the angular frequency. By simply matching every symbol, we equate

$$k = 500 \; \text{rad m}^{-1}, \qquad \omega = 10 \times 10^{10} \; \text{rad s}^{-1}.$$

First, make the numerical value of $$\omega$$ explicit:

$$\omega = 10 \times 10^{10} = 1 \times 10^{11} \; \text{rad s}^{-1}.$$

The speed of a wave is connected to these parameters through the relation

$$v = \frac{\omega}{k}.$$

Substituting the recognised values, we write

$$v = \frac{1 \times 10^{11}}{500} \; \text{m s}^{-1}.$$

Now we carry out the division step by step. First, divide the numerator and denominator:

$$\frac{1 \times 10^{11}}{5 \times 10^{2}} = 0.2 \times 10^{9}$$

(because $$500 = 5 \times 10^{2}$$ and we moved one power of ten from the denominator to the numerator). Converting $$0.2 \times 10^{9}$$ into a more familiar decimal form, we obtain

$$0.2 \times 10^{9} = 2 \times 10^{8} \; \text{m s}^{-1}.$$

Next, we compare this value with the speed of light in vacuum, $$c = 3 \times 10^{8} \; \text{m s}^{-1}$$. Forming the ratio,

$$\frac{v}{c} = \frac{2 \times 10^{8}}{3 \times 10^{8}} = \frac{2}{3}.$$

So the speed of this electromagnetic wave in the medium is

$$v = \frac{2}{3}c.$$

Hence, the correct answer is Option D.

We are given the plane-polarised electromagnetic wave

$$E \;=\; 800 \,\sin\!\Bigl[\;\omega\!\left(t-\dfrac{x}{c}\right)\Bigr]$$

The coefficient of the sine function is the peak (maximum) electric-field magnitude, so

$$E_0 = 800 \text{ V m}^{-1}$$

For a plane electromagnetic wave in free space the electric and magnetic amplitudes are related by the well-known relation

$$E_0 = c\,B_0$$

where $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$ is the speed of light in vacuum. Solving for the magnetic-field amplitude $$B_0$$ we get

$$$\begin{aligned} B_0 &= \dfrac{E_0}{c} \\ &= \dfrac{800}{3.0 \times 10^{8}} \ \text{T} \\ &= \dfrac{8.00 \times 10^{2}}{3.0 \times 10^{8}} \ \text{T} \\ &= \dfrac{8.00}{3.0}\times 10^{-6}\ \text{T} \\ &= 2.666\dots\times 10^{-6}\ \text{T} \\ &\approx 2.67 \times 10^{-6}\ \text{T} \end{aligned}$$$

An electron is made to move with speed

$$v = 3.0 \times 10^{7}\ \text{m s}^{-1}$$

and, as stated, the motion is normal (perpendicular) to the direction of propagation of the beam. In a plane wave the magnetic field itself is already perpendicular to the propagation direction, so by choosing the electron’s velocity also normal to the beam we can arrange the velocity to be perpendicular to the magnetic field. For the magnetic force this gives the maximum value because the factor $$\sin\theta$$ becomes unity. The magnetic (Lorentz) force formula is first written explicitly:

$$F = q\,v\,B\,\sin\theta$$

For the maximum force $$\theta = 90^\circ$$ and $$\sin\theta = 1$$, hence

$$F_{\max} = q\,v\,B_0$$

For an electron the magnitude of the charge is

$$q = e = 1.6 \times 10^{-19}\ \text{C}$$

Substituting all numerical values step by step,

$$$\begin{aligned} F_{\max} &= (1.6 \times 10^{-19}) \times (3.0 \times 10^{7}) \times (2.67 \times 10^{-6}) \ \text{N} \\[4pt] &= 1.6 \times 3.0 \times 2.67 \times 10^{-19 + 7 - 6}\ \text{N} \\[4pt] &= (1.6 \times 3.0)\times 2.67 \times 10^{-18}\ \text{N} \\[4pt] &= 4.8 \times 2.67 \times 10^{-18}\ \text{N} \\[4pt] &= 12.816 \times 10^{-18}\ \text{N} \\[4pt] &= 1.2816 \times 10^{-17}\ \text{N} \end{aligned}$$$

Keeping only three significant figures (because the given data have at most two significant figures), we state

$$F_{\max} \approx 1.28 \times 10^{-17}\ \text{N}$$

The option list is expressed in the form $$A \times 10^{-18}\ \text{N}$$. Writing our result in the same form:

$$$1.28 \times 10^{-17}\ \text{N} = 12.8 \times 10^{-18}\ \text{N}$$$

This matches Option B.

Hence, the correct answer is Option B.

An electromagnetic wave is travelling along the $$y$$-direction, and at a particular point $$\vec{B} = 8.0 \times 10^{-8} \, \hat{z}$$ T. We need to find the electric field $$\vec{E}$$.

The magnitude of the electric field is $$E = cB = 3 \times 10^8 \times 8.0 \times 10^{-8} = 24$$ V/m.

For the direction, the Poynting vector $$\vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B}$$ must point in the direction of wave propagation, which is $$+\hat{y}$$. Since $$\vec{B}$$ is along $$+\hat{z}$$, we need $$\vec{E} \times \hat{z}$$ to be along $$+\hat{y}$$. Checking: $$(-\hat{x}) \times \hat{z} = -(\hat{x} \times \hat{z}) = -(-\hat{y}) = +\hat{y}$$. This works.

Therefore, $$\vec{E} = -24 \, \hat{x}$$ V m$$^{-1}$$.

An electromagnetic wave propagating along the y-direction requires both $$\vec{E}$$ and $$\vec{B}$$ to lie in the xz-plane (perpendicular to $$\hat{y}$$). Additionally, $$\vec{E}$$ and $$\vec{B}$$ must be mutually perpendicular, and the Poynting vector $$\vec{E} \times \vec{B}$$ must point along $$\hat{y}$$.

Let us check each option. Option (1): $$E_y, B_y$$ — both are along the propagation direction, which violates the transverse nature of EM waves. Invalid. Option (2): $$E_y, B_z$$ or $$E_z, B_y$$ — each pair contains a y-component, which is not allowed for a wave propagating in the y-direction. Invalid.

Option (4): $$E_z, B_y$$ or $$E_y, B_z$$ — again, each pair contains a y-component. Invalid.

Option (3): $$E_z, B_z$$ or $$E_z, B_x$$. Consider the second pair: $$E_z$$ and $$B_x$$ are both perpendicular to $$\hat{y}$$, they are perpendicular to each other, and $$\hat{z} \times \hat{x} = \hat{y}$$, giving the correct propagation direction. This is a valid EM wave configuration. As for the first pair $$E_z, B_z$$, both fields would be parallel, which is not a valid EM wave on its own — however, the question uses "or" between the pairs, so only one valid pair is needed for the option to be correct.

Since option (3) contains the valid pair $$(E_z, B_x)$$ and is the only option with at least one fully valid perpendicular pair of transverse components, it is the correct answer.

The magnetic field of the plane electromagnetic wave is given as

$$\vec B \;=\;B_{0}\,\dfrac{\hat i+\hat j}{\sqrt2}\, \cos\!\bigl(kz-\omega t\bigr).$$

For a monochromatic wave travelling along the $$+z$$-direction, the electric field $$\vec E$$ is always perpendicular to $$\vec B$$, and the three vectors $$\vec E,\;\vec B,\;\hat k$$ form a right-handed triad. The magnitudes satisfy the relation

$$|\vec E| \;=\;c\,|\vec B|,$$

and the direction is obtained from the vector identity

$$\hat k \;=\;\dfrac{\vec E\times\vec B}{|\vec E\times\vec B|}.$$ Because $$\hat k=\hat k_z$$ in the present problem, we choose the electric field so that $$\vec E\times\vec B$$ points along $$\hat k_z$$. With $$\vec B$$ lying in the $$x$$-$$y$$ plane, the appropriate $$\vec E$$ is

$$\vec E \;=\;c B_0\,\dfrac{\hat i-\hat j}{\sqrt2}\, \cos\!\bigl(kz-\omega t\bigr).$$

At the instant $$t=0$$ we have

$$\vec B(z,0)=B_0\,\dfrac{\hat i+\hat j}{\sqrt2}\, \cos(kz),\qquad \vec E(z,0)=cB_0\,\dfrac{\hat i-\hat j}{\sqrt2}\, \cos(kz).$$

The two charges are situated at

$$z_1=\dfrac{\pi}{k}, \qquad z_2=\dfrac{3\pi}{k}.$$

Substituting these $$z$$-values,

$$\cos(kz_1)=\cos\!\bigl(k\cdot\tfrac{\pi}{k}\bigr)=\cos\pi=-1,$$

$$\cos(kz_2)=\cos\!\bigl(k\cdot\tfrac{3\pi}{k}\bigr)=\cos3\pi=-1.$$

Hence at both locations

$$\vec B = -\,B_0\,\dfrac{\hat i+\hat j}{\sqrt2},\qquad \vec E = -\,cB_0\,\dfrac{\hat i-\hat j}{\sqrt2}.$$

Both charges possess the same velocity

$$\vec v = 0.5\,c\,\hat i.$$

Now we invoke the Lorentz force formula

$$\vec F = q\bigl(\,\vec E + \vec v\times\vec B\,\bigr).$$

First we evaluate $$\vec v\times\vec B$$ (the subscript $$0$$ on $$B_0$$ is suppressed below for brevity):

$$$ \begin{aligned} \vec v\times\vec B &= \Bigl(0.5\,c\,\hat i\Bigr)\times\Bigl(\, -\,B_0\,\dfrac{\hat i+\hat j}{\sqrt2}\Bigr) \\[4pt] &= -\,\dfrac{0.5\,c\,B_0}{\sqrt2}\, \Bigl(\hat i\times\hat i+\hat i\times\hat j\Bigr) \\[4pt] &= -\,\dfrac{0.5\,c\,B_0}{\sqrt2}\, \Bigl(0+\hat k\Bigr) \\[4pt] &= -\,\dfrac{0.5\,c\,B_0}{\sqrt2}\,\hat k. \end{aligned} $$$

Adding the electric part, the bracket $$(\vec E+\vec v\times\vec B)$$ at either charge position becomes

$$$ \begin{aligned} \vec E + \vec v\times\vec B &= -\,cB_0\,\dfrac{\hat i-\hat j}{\sqrt2} \;-\;\dfrac{0.5\,c\,B_0}{\sqrt2}\,\hat k \\[6pt] &= -\,\dfrac{c\,B_0}{\sqrt2} \Bigl(\hat i-\hat j+0.5\,\hat k\Bigr). \end{aligned} $$$

This vector is identical at $$z_1$$ and $$z_2$$. Therefore, the magnitude of the force on each charge is directly proportional to the magnitude of the charge itself:

$$|\vec F_1| = |q_1|\; \bigl|\vec E+\vec v\times\vec B\bigr|,\qquad |\vec F_2| = |q_2|\; \bigl|\vec E+\vec v\times\vec B\bigr|.$$

Taking the ratio, the common factor cancels:

$$\dfrac{|\vec F_1|}{|\vec F_2|} \;=\;\dfrac{|q_1|}{|q_2|} \;=\;\dfrac{4\pi}{2\pi} \;=\;2.$$

Thus the force on $$q_1$$ is twice the force on $$q_2$$, i.e.

$$\vec F_1 : \vec F_2 = 2 : 1.$$

Hence, the correct answer is Option D.

In an electromagnetic wave, the energy is shared equally between the electric and magnetic fields. The average energy density due to the electric field is $$U_e = \frac{1}{2} \varepsilon_0 E_{\text{rms}}^2 = \frac{1}{4} \varepsilon_0 E_0^2$$, and the average energy density due to the magnetic field is $$U_m = \frac{B_{\text{rms}}^2}{2\mu_0} = \frac{B_0^2}{4\mu_0}$$.

Since $$E_0 = cB_0$$ and $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$$, we have $$U_e = \frac{1}{4} \varepsilon_0 c^2 B_0^2 = \frac{1}{4} \varepsilon_0 \cdot \frac{1}{\mu_0 \varepsilon_0} \cdot B_0^2 = \frac{B_0^2}{4\mu_0} = U_m$$.

Therefore, for an electromagnetic wave travelling in free space, the average energy densities due to the electric and magnetic fields are equal: $$U_e = U_m$$.

The direction of propagation of an electromagnetic wave is given by the cross product of the electric field vector and the magnetic field vector, i.e., the direction of $$\vec{E} \times \vec{B}$$.

Given $$\vec{E} = E_0\hat{i}$$ and $$\vec{B} = B_0\hat{k}$$, we compute:

$$\vec{E} \times \vec{B} = E_0 B_0 (\hat{i} \times \hat{k})$$

Using the right-hand rule for unit vectors: $$\hat{i} \times \hat{k} = -\hat{j}$$

Therefore, $$\vec{E} \times \vec{B} = E_0 B_0 (-\hat{j})$$

The direction of propagation of the electromagnetic wave is along $$(-\hat{j})$$.

In amplitude modulation, when a signal of frequency $$f_m$$ modulates a carrier of frequency $$f_c$$, the resulting modulated wave contains three frequencies: the carrier frequency $$f_c$$, and two side band frequencies $$f_c + f_m$$ and $$f_c - f_m$$.

The bandwidth required is $$2f_m$$, which is the difference between the upper and lower side band frequencies.

Given $$f_m = 2$$ kHz and $$f_c = 1$$ MHz $$= 1000$$ kHz:

Statement I: The bandwidth requirement is $$2 \times 2 = 4$$ kHz. This is true.

Statement II: The upper side band frequency is $$f_c + f_m = 1000 + 2 = 1002$$ kHz, and the lower side band frequency is $$f_c - f_m = 1000 - 2 = 998$$ kHz. This is true.

Since both statements are correct, the answer is Option (2): Both Statement I and Statement II are true.

The message signal is $$m(t) = 5\sin(1.57 \times 10^8 \, t)$$. The angular frequency of the message signal is $$\omega_m = 1.57 \times 10^8$$ rad s$$^{-1}$$, so the frequency of the message signal is $$f_m = \frac{\omega_m}{2\pi} = \frac{1.57 \times 10^8}{2\pi} = \frac{1.57 \times 10^8}{6.284} \approx 2.5 \times 10^7$$ Hz $$= 25$$ MHz.

In amplitude modulation, the bandwidth is twice the frequency of the message signal: $$\text{Bandwidth} = 2f_m = 2 \times 25 = 50$$ MHz.

The line-of-sight (LOS) range between two antennas is obtained from the well-known empirical relation

$$d \;=\;\sqrt{2\,R\,h_t}\;+\;\sqrt{2\,R\,h_r}$$

where

$$R = 6.4 \times 10^6 \text{ m}$$ is the radius of the Earth,

$$h_t = 50 \text{ m}$$ is the height of the transmitting antenna, and

$$h_r = 80 \text{ m}$$ is the height of the receiving antenna.

We first evaluate the contribution from the transmitting antenna:

$$ \sqrt{2\,R\,h_t} \;=\;\sqrt{2 \times (6.4 \times 10^6)\times 50}. $$

Inside the square root we have

$$ 2 \times 6.4 \times 50 = 12.8 \times 50 = 640, $$

so

$$ 2\,R\,h_t = 640 \times 10^6 = 6.4 \times 10^8. $$

Taking the square root:

$$ \sqrt{6.4 \times 10^8} = \sqrt{6.4}\;\times\;\sqrt{10^8} = 2.5298 \times 10^4 \text{ m} \approx 2.53 \times 10^4 \text{ m}. $$

In kilometres this is

$$ 2.53 \times 10^4 \text{ m} = 25.3 \text{ km}. $$

Next we evaluate the contribution from the receiving antenna:

$$ \sqrt{2\,R\,h_r} \;=\;\sqrt{2 \times (6.4 \times 10^6)\times 80}. $$

Now

$$ 2 \times 6.4 \times 80 = 12.8 \times 80 = 1024, $$

hence

$$ 2\,R\,h_r = 1024 \times 10^6 = 1.024 \times 10^9. $$

Taking the square root:

$$ \sqrt{1.024 \times 10^9} = \sqrt{1.024}\;\times\;\sqrt{10^9} = 1.012 \times 10^{4.5} = 1.012 \times 31622.8 \text{ m} \approx 3.20 \times 10^4 \text{ m}. $$

Thus, in kilometres,

$$ 3.20 \times 10^4 \text{ m} = 32.0 \text{ km}. $$

Adding the two distances gives the LOS range:

$$ d = 25.3 \text{ km} + 32.0 \text{ km} = 57.3 \text{ km}. $$

The option closest to this calculated value is 57.28 km.

Hence, the correct answer is Option C.

In amplitude modulation (AM), a message signal of frequency $$f_m$$ modulates a carrier signal of frequency $$f_c$$. The resulting AM signal contains three frequency components: the carrier frequency $$f_c$$, and two sidebands at frequencies $$f_c + f_m$$ and $$f_c - f_m$$.

When this AM signal is radiated through an antenna, the signal that is actually transmitted through the air is the carrier wave with the modulated amplitude. The wavelength of the radiated signal corresponds to the carrier frequency, since the carrier is the dominant component that determines the electromagnetic wave's wavelength.

The wavelength of an electromagnetic wave in air is given by $$\lambda = \frac{c}{f}$$, where $$c$$ is the speed of light.

For the carrier signal, the wavelength is $$\lambda = \frac{c}{f_c}$$.

Therefore, the wavelength of the corresponding signal in air is $$\frac{c}{f_c}$$.

We have the message (modulating) signal written as $$V_m(t)=10\sin\bigl(2\pi\times10^{5}\,t\bigr)\;\text{volts}.$$

From the standard trigonometric form $$V_m(t)=V_{m0}\sin(2\pi f_m t),$$ we can directly read the message frequency:

$$2\pi f_m = 2\pi \times 10^{5}\;\Rightarrow\;f_m = 10^{5}\ \text{Hz}.$$

Converting hertz to kilohertz, we obtain

$$f_m = 10^{5}\ \text{Hz}=100\ \text{kHz}.$$

Next, the carrier signal is given as $$V_C(t)=20\sin\bigl(2\pi\times10^{7}\,t\bigr)\;\text{volts}.$$

Again comparing with $$V_C(t)=V_{c0}\sin(2\pi f_c t),$$ we identify

$$2\pi f_c = 2\pi\times10^{7}\;\Rightarrow\;f_c = 10^{7}\ \text{Hz} = 10,000\ \text{kHz}.$$

For amplitude modulation (A.M.), the modulated spectrum consists of three principal components:

1. The carrier at frequency $$f_c.$$

2. The upper sideband (USB) at frequency $$f_c + f_m.$$

3. The lower sideband (LSB) at frequency $$f_c - f_m.$$

The bandwidth $$B$$ of an A.M. signal is defined, and should always be remembered, as

$$B = 2f_m,$$

because the two sidebands lie symmetrically $$f_m$$ hertz above and below the carrier.

Substituting our previously found value $$f_m = 100\ \text{kHz}$$, we obtain:

$$B = 2 \times 100\ \text{kHz} = 200\ \text{kHz}.$$

This bandwidth is denoted by $$\alpha$$ in the statement of the question, so

$$\alpha = 200\ \text{kHz}.$$

Hence, the correct answer is Option A.

We need to find the minimum height of the receiving antenna $$h_r$$ to receive line-of-sight signals from a transmitting antenna of the same height mounted on an identical tower.

The maximum line-of-sight communication distance $$d$$ between two antennas of heights $$h_t$$ (transmitting) and $$h_r$$ (receiving) is given by the formula: $$d = \sqrt{2 R h_t} + \sqrt{2 R h_r}$$, where $$R$$ is the radius of the Earth.

Since the antennas are mounted on identical towers, their heights are equal: $$h_t = h_r = h$$. Substituting this into the distance formula gives: $$d = \sqrt{2 R h} + \sqrt{2 R h} = 2\sqrt{2 R h}$$.

We are given the following values:

Total separation distance, $$d = 45\text{ km} = 45 \times 10^3\text{ m}$$

Radius of the Earth, $$R = 6400\text{ km} = 6400 \times 10^3\text{ m} = 6.4 \times 10^6\text{ m}$$

Substituting these values into our simplified expression yields: $$45 \times 10^3 = 2\sqrt{2 \times (6.4 \times 10^6) \times h}$$.

Dividing both sides by 2 gives: $$22.5 \times 10^3 = \sqrt{12.8 \times 10^6 \times h}$$.

Squaring both sides to eliminate the square root results in: $$(22.5 \times 10^3)^2 = 12.8 \times 10^6 \times h$$, which simplifies to: $$506.25 \times 10^6 = 12.8 \times 10^6 \times h$$.

Canceling out $$10^6$$ from both sides leaves: $$506.25 = 12.8 \times h$$.

Solving for $$h$$ yields: $$h = \frac{506.25}{12.8} \approx 39.55\text{ m}$$.

Therefore, the correct answer is Option B: 39.55 m.

The range of a TV transmission tower of height $$h$$ above the Earth's surface is given by: $$d = \sqrt{2R_e h}$$

We need $$d = 150 \text{ km} = 150 \times 10^3 \text{ m}$$ with $$R_e = 6.5 \times 10^6 \text{ m}$$.

Solving for $$h$$: $$h = \frac{d^2}{2R_e} = \frac{(150 \times 10^3)^2}{2 \times 6.5 \times 10^6} = \frac{2.25 \times 10^{10}}{1.3 \times 10^7} = \frac{2.25}{1.3} \times 10^3 = 1730.8 \text{ m} \approx 1731 \text{ m}$$

The area covered is a circle of radius $$d = 150 \text{ km}$$: $$A = \pi d^2 = \pi \times (150)^2 = \pi \times 22500 \approx 70686 \text{ km}^2$$

With population density $$\rho = 2000 \text{ km}^{-2}$$, the population covered is: $$P = \rho \times A = 2000 \times 70686 \approx 1.413 \times 10^8 = 1413 \times 10^5$$

Therefore, the height of the transmitting antenna is $$1731 \text{ m}$$ and the population covered is $$1413 \times 10^5$$.

When a galaxy moves away from the earth, the observed wavelength is red-shifted. The Doppler shift in wavelength for light is given by $$\Delta\lambda = \frac{v}{c}\lambda$$, where $$v$$ is the recessional speed, $$c$$ is the speed of light, and $$\lambda$$ is the original wavelength.

Substituting the given values: $$\Delta\lambda = \frac{286 \times 10^3}{3 \times 10^8} \times 630 \times 10^{-9}$$.

Computing step by step: $$\frac{286 \times 10^3}{3 \times 10^8} = \frac{286}{3 \times 10^5} = 9.533 \times 10^{-4}$$.

Then $$\Delta\lambda = 9.533 \times 10^{-4} \times 630 \times 10^{-9} = 6.006 \times 10^{-10}$$ m.

Since $$\Delta\lambda = x \times 10^{-10}$$ m, the value of $$x$$ to the nearest integer is $$6$$.

The intensity of radiation from a bulb at distance $$r$$ is $$I = \frac{P}{4\pi r^2}$$, where $$P$$ is the power. The relationship between electric field amplitude and intensity is $$I = \frac{1}{2}\varepsilon_0 c E_0^2$$, so $$E_0 \propto \sqrt{I} \propto \sqrt{P}$$ at a fixed distance.

For the 100 W bulb at 3 m, the electric field intensity is $$E$$. For the 60 W bulb at the same distance, the electric field intensity is $$E' = E\sqrt{\frac{60}{100}} = E\sqrt{\frac{3}{5}}$$.

We are told $$E' = \sqrt{\frac{x}{5}}\,E$$. Comparing, $$\sqrt{\frac{x}{5}} = \sqrt{\frac{3}{5}}$$, which gives $$x = 3$$.

The 8 W bulb acts as a point source of electromagnetic radiation. The efficiency of 10% indicates that 10% of the electrical power is converted to visible light, but the bulb still radiates the full 8 W as electromagnetic radiation (including heat radiation). For finding the peak electric field of the radiation at a distance, we use the total radiated power $$P = 8$$ W.

At a distance of $$r = 10$$ m from the point source, the intensity is given by $$I = \frac{P}{4\pi r^2} = \frac{8}{4\pi(10)^2} = \frac{8}{400\pi} = \frac{1}{50\pi}$$ W/m$$^2$$.

The intensity of an electromagnetic wave is related to the peak electric field $$E_0$$ by the relation $$I = \frac{E_0^2}{2\mu_0 c}$$, where $$\mu_0$$ is the permeability of free space and $$c$$ is the speed of light.

Rearranging for $$E_0^2$$: $$E_0^2 = 2\mu_0 c \cdot I = 2\mu_0 c \cdot \frac{1}{50\pi} = \frac{2\mu_0 c}{50\pi} = \frac{\mu_0 c}{25\pi}$$.

Taking the square root: $$E_0 = \sqrt{\frac{\mu_0 c}{25\pi}} = \frac{1}{5}\sqrt{\frac{\mu_0 c}{\pi}}$$. Writing this as $$\frac{2}{10}\sqrt{\frac{\mu_0 c}{\pi}}$$ and comparing with the given expression $$\frac{x}{10}\sqrt{\frac{\mu_0 c}{\pi}}$$, we get $$x = 2$$.

We have an electromagnetic wave of frequency 5 GHz travelling in a medium with relative electric permittivity $$\varepsilon_r = 2$$ and relative magnetic permeability $$\mu_r = 2$$.

The velocity of an electromagnetic wave in a medium is given by $$v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$, where $$c = 3 \times 10^8$$ m/s is the speed of light in vacuum.

Substituting the values, $$v = \frac{3 \times 10^8}{\sqrt{2 \times 2}} = \frac{3 \times 10^8}{\sqrt{4}} = \frac{3 \times 10^8}{2}$$.

$$v = 1.5 \times 10^8 = 15 \times 10^7$$ m/s.

So, the answer is $$15$$.

We have a plane electromagnetic wave propagating in free space. In such a wave, the electric field $$\mathbf E$$ and magnetic field $$\mathbf B$$ are always related by the fundamental relation for free space

$$E = c\,B,$$

where $$E$$ is the instantaneous (or peak) value of the electric field, $$B$$ is the corresponding value of the magnetic field, and $$c$$ is the speed of light in vacuum. The universally accepted value of the speed of light is

$$c = 3 \times 10^{8}\ \text{m s}^{-1}.$$

At the given point, the electric field is reported as

$$E = 6\ \text{V m}^{-1}.$$

Substituting this value and the value of $$c$$ into the relation $$E = c\,B$$, we can solve for the magnetic field $$B$$:

$$B = \dfrac{E}{c}.$$

Now inserting the numerical values, we get

$$B = \dfrac{6\ \text{V m}^{-1}}{3 \times 10^{8}\ \text{m s}^{-1}}.$$

Performing the division step by step:

First divide the numeric coefficients: $$6 \div 3 = 2.$$

Next, divide the powers of ten: $$10^{0} \div 10^{8} = 10^{-8}.$$

Putting these together, we obtain

$$B = 2 \times 10^{-8}\ \text{T}.$$

The problem statement expresses the magnetic field in the form $$x \times 10^{-8}\ \text{T}$$, so by direct comparison we identify

$$x = 2.$$

So, the answer is $$2$$.

The conduction current density is $$J_c = \frac{E}{\rho}$$ and the displacement current density is $$J_d = \varepsilon \frac{\partial E}{\partial t}$$. For a parallel plate capacitor driven by $$V(t) = V_0 \sin(2\pi f t)$$, the electric field between the plates is proportional to $$V(t)$$, so $$E = E_0 \sin(2\pi f t)$$ and $$\frac{\partial E}{\partial t} = 2\pi f E_0 \cos(2\pi f t)$$.

The ratio of conduction to displacement current density is $$\frac{J_c}{J_d} = \frac{E_0 \sin(2\pi ft)/\rho}{\varepsilon \cdot 2\pi f E_0 \cos(2\pi ft)} = \frac{\tan(2\pi ft)}{2\pi f \varepsilon \rho}$$.

At $$t = \frac{1}{800}$$ s, we compute $$2\pi f t = 2\pi \times 9 \times 10^2 \times \frac{1}{800} = \frac{9\pi}{4}$$. Since $$\frac{9\pi}{4} = 2\pi + \frac{\pi}{4}$$, we get $$\tan\left(\frac{9\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$$.

Now we need $$2\pi f \varepsilon \rho$$. Here $$\varepsilon = 80\varepsilon_0$$ and $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$, so $$\varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9}$$. Thus $$\varepsilon = \frac{80}{4\pi \times 9 \times 10^9} = \frac{20}{9\pi \times 10^9}$$.

Therefore $$2\pi f \varepsilon \rho = 2\pi \times 900 \times \frac{20}{9\pi \times 10^9} \times 0.25$$. The numerator is $$2\pi \times 900 \times 20 \times 0.25 = 9000\pi$$. The denominator is $$9\pi \times 10^9$$. So the ratio is $$\frac{9000\pi}{9\pi \times 10^9} = \frac{1000}{10^9} = 10^{-6}$$.

Hence $$\frac{J_c}{J_d} = \frac{1}{10^{-6}} = 10^6$$. Since this equals $$10^x$$, we get $$x = 6$$.

Based on the provided solution steps for calculating radiation pressure, here is the formatted response:

To find the radiation pressure of an electromagnetic wave, we first relate the intensity of the wave to the electric field amplitude and then use the relationship between intensity and pressure for a perfectly reflecting surface.

1. Intensity of the Electromagnetic Wave ($$I$$)

The average intensity $$I$$ of an electromagnetic wave in a vacuum is given by:

$$I = \frac{1}{2} \varepsilon_0 E_0^2 c$$

Where:

• $$\varepsilon_0$$ = Permittivity of free space ($$8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2$$)
• $$E_0$$ = Amplitude of the electric field
• $$c$$ = Speed of light

2. Radiation Pressure ($$P$$)

For a perfectly reflecting surface, the radiation pressure $$P$$ is twice the momentum density, or:

$$P = \frac{2I}{c}$$

Substituting the expression for intensity into the pressure formula:

$$P = \left( \frac{2}{c} \right) \left( \frac{1}{2} \varepsilon_0 E_0^2 c \right)$$

$$P = \varepsilon_0 E_0^2$$

3. Numerical Calculation

Given the electric field amplitude $$E_0 = 200 \text{ V/m}$$:

$$P = (8.85 \times 10^{-12}) \times (200)^2$$

$$P = 8.85 \times 10^{-12} \times 40,000$$

$$P = 8.85 \times 10^{-8} \times 4$$

$$P = 35.4 \times 10^{-8}$$

In fractional scientific notation:

$$\boxed{P = \frac{354}{10^9} \text{ N/m}^2}$$

We begin by recalling the basic fact about ordinary amplitude-modulated (A.M.) radio transmission: when a carrier of frequency $$f_c$$ is modulated by an audio (message) signal whose highest frequency component is $$f_m$$, two side-bands are produced, one above and one below the carrier. The upper side-band extends up to $$f_c + f_m$$, while the lower side-band goes down to $$f_c - f_m$$. Thus the total spectral span occupied by that single station is twice the highest modulating frequency.

Mathematically, the required bandwidth for one A.M. station is stated as

$$\text{Bandwidth per station} = 2\,f_m.$$

Now the numerical values given in the problem are:

Maximum audio (modulating) frequency: $$f_m = 6\ \text{kHz}.$$

Therefore, using the above formula, the bandwidth demanded by one station is

$$\text{Bandwidth per station} = 2 \times 6\ \text{kHz} = 12\ \text{kHz}.$$

The spectrum segment reserved for all the A.M. broadcasts together is stated to be

$$\text{Total available bandwidth} = 6\ \text{MHz}.$$

Since $$1\ \text{MHz} = 1000\ \text{kHz}$$, we convert this to kilohertz to match units:

$$6\ \text{MHz} = 6 \times 1000\ \text{kHz} = 6000\ \text{kHz}.$$

We now find out how many non-overlapping 12 kHz slots can be fitted into 6000 kHz. The number of simultaneous stations, $$N$$, is simply the ratio of the total available bandwidth to the bandwidth per station:

$$N = \frac{\text{Total bandwidth}}{\text{Bandwidth per station}} = \frac{6000\ \text{kHz}}{12\ \text{kHz}}.$$

Carrying out the division, we get

$$N = \frac{6000}{12} = 500.$$

So, the answer is $$500$$.

The frequency of the electromagnetic wave is $$f = 3$$ GHz $$= 3 \times 10^9$$ Hz, and the relative electric permittivity of the dielectric medium is $$\varepsilon_r = 2.25$$. For a non-magnetic dielectric medium ($$\mu_r = 1$$), the speed of the electromagnetic wave in the medium is $$v = \frac{c}{\sqrt{\varepsilon_r}}$$.

Substituting the values: $$v = \frac{3 \times 10^8}{\sqrt{2.25}} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8$$ m s$$^{-1}$$.

The wavelength in the medium is $$\lambda = \frac{v}{f} = \frac{2 \times 10^8}{3 \times 10^9} = \frac{2}{30} = \frac{1}{15}$$ m.

Converting to centimetres: $$\lambda = \frac{1}{15} \times 100 = \frac{100}{15} = 6.667$$ cm $$= 667 \times 10^{-2}$$ cm.

Therefore, the wavelength in the medium is $$667 \times 10^{-2}$$ cm.

We are given that an average force of $$F = 2.5 \times 10^{-6}$$ N is exerted by a light wave on a non-reflecting (completely absorbing) surface of area $$A = 30 \text{ cm}^2$$ during a time span of $$t = 40$$ min. We need to find the energy flux (intensity) of light.

For complete absorption, the radiation pressure is related to the intensity by $$P = \frac{I}{c}$$, where $$c = 3 \times 10^{8}$$ m/s is the speed of light. Also, the force on the surface equals pressure times area: $$F = P \times A$$.

Combining these, $$F = \frac{I \times A}{c}$$, which gives $$I = \frac{Fc}{A}$$.

Converting the area to m$$^2$$: $$A = 30 \text{ cm}^2 = 30 \times 10^{-4} \text{ m}^2$$. Substituting: $$I = \frac{2.5 \times 10^{-6} \times 3 \times 10^8}{30 \times 10^{-4}} = \frac{7.5 \times 10^2}{30 \times 10^{-4}} = \frac{750}{0.003} = 250000 \text{ W/m}^2$$.

Converting to W cm$$^{-2}$$: since $$1 \text{ m}^2 = 10^4 \text{ cm}^2$$, we have $$I = \frac{250000}{10^4} = 25 \text{ W cm}^{-2}$$.

Therefore, the energy flux of light is $$\boxed{25}$$ W cm$$^{-2}$$.

We start with the expression for the electric field of the plane electromagnetic wave

$$E = E_0 \sin\!\left[\omega\!\left(t - \frac{z}{c}\right)\right]$$

and from the question we read the amplitude as

$$E_0 = 50 \ \text{N C}^{-1}.$$

An electromagnetic wave stores energy in both its electric and magnetic fields. The instantaneous energy density is

$$u = \frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2.$$

For a wave in free space we have the relation $$B = \dfrac{E}{c},$$ so that the magnetic part equals the electric part. Therefore the total instantaneous energy density can be written solely with the electric field as

$$u = \varepsilon_0 E^2.$$

Because the field is oscillatory, what actually matters for a macroscopic amount of energy is the time-average of this density. For any sinusoidal term $$\sin^2(\theta)$$ the time average is

$$\langle \sin^2(\theta) \rangle = \frac{1}{2}.$$

Applying this to the present case we obtain the time-averaged energy density

$$\langle u \rangle \;=\; \varepsilon_0 \langle E^2 \rangle \;=\; \varepsilon_0 \, E_0^{\,2}\,\langle\sin^2(\dots)\rangle \;=\; \frac{\varepsilon_0 E_0^{\,2}}{2}.$$

Substituting the numerical values (with $$\varepsilon_0 = 8.8 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$$ and $$E_0 = 50 \ \text{N C}^{-1}$$) we get

$$\langle u \rangle = \frac{8.8 \times 10^{-12}\; (50)^2}{2} = \frac{8.8 \times 10^{-12}\; \times 2500}{2}.$$

Multiplying first:

$$8.8 \times 2500 = 22000,$$

so

$$8.8 \times 10^{-12}\; \times 2500 = 2.2 \times 10^{-8}.$$

Dividing by 2 gives

$$\langle u \rangle = 1.1 \times 10^{-8}\ \text{J m}^{-3}.$$

The total (time-averaged) energy contained in a volume $$V$$ is related by

$$U = \langle u \rangle\, V.$$

We are told this energy equals

$$U = 5.5 \times 10^{-12}\ \text{J},$$

so we solve for $$V$$:

$$V = \frac{U}{\langle u \rangle} = \frac{5.5 \times 10^{-12}}{1.1 \times 10^{-8}} = 5.0 \times 10^{-4}\ \text{m}^3.$$

To convert cubic metres to cubic centimetres we recall

$$1\ \text{m}^3 = 10^6\ \text{cm}^3.$$

Therefore

$$V = 5.0 \times 10^{-4}\ \text{m}^3 \times 10^6\ \frac{\text{cm}^3}{\text{m}^3} = 5.0 \times 10^{2}\ \text{cm}^3 = 500\ \text{cm}^3.$$

Hence, the correct answer is Option 500.

In amplitude modulation, the carrier wave $$V_C(t) = 160\sin(2\pi \times 10^6 t)$$ volts has a carrier amplitude $$A_c = 160$$ V. The modulating signal $$V_m(t) = A_m\sin(2\pi \times 10^3 t)$$ varies the amplitude between $$V_{max} = 200$$ V and $$V_{min} = 120$$ V.

In AM modulation, the instantaneous amplitude oscillates between $$A_c + A_m$$ and $$A_c - A_m$$. Therefore:

$$V_{max} = A_c + A_m = 160 + A_m = 200 \implies A_m = 40 \text{ V}$$

$$V_{min} = A_c - A_m = 160 - A_m = 120 \implies A_m = 40 \text{ V}$$

Both conditions consistently give $$A_m = 40$$ V.

The peak voltage of the modulating signal is $$\boxed{40}$$ V.

We start by recalling the standard expression for a sinusoidally amplitude-modulated (AM) signal. If the unmodulated carrier voltage is $$v_c(t)=A_c\cos\omega_ct$$ and the modulating or base-band signal is $$v_m(t)=A_m\cos\omega_mt$$, then the AM wave is written as

$$v_{\text{AM}}(t)=A_c\left(1+m\cos\omega_mt\right)\cos\omega_ct.$$

Here $$A_c$$ is the carrier amplitude, $$A_m$$ is the message (modulating) amplitude and $$m$$ is the modulation index, defined by the formula

$$m=\frac{A_m}{A_c}.$$

In the given problem the carrier amplitude is $$A_c=250\text{ V}$$ and the modulating amplitude is $$A_m=150\text{ V}.$$ Using the definition, we calculate

$$m=\frac{A_m}{A_c}=\frac{150}{250}=0.6.$$

The envelope (overall amplitude) of the AM wave varies between a maximum value $$A_{\text{max}}$$ and a minimum value $$A_{\text{min}}$$. These are obtained by substituting the extreme values of $$\cos\omega_mt$$, namely $$+1$$ and $$-1$$, into the factor $$(1+m\cos\omega_mt)$$:

$$A_{\text{max}}=A_c(1+m), \qquad A_{\text{min}}=A_c(1-m).$$

Substituting $$A_c=250\text{ V}$$ and $$m=0.6$$ gives

$$A_{\text{max}}=250(1+0.6)=250\times1.6=400\text{ V},$$

$$A_{\text{min}}=250(1-0.6)=250\times0.4=100\text{ V}.$$

The question states that the ratio of minimum amplitude to maximum amplitude is $$50:x$$. Numerically, we have

$$\frac{A_{\text{min}}}{A_{\text{max}}}=\frac{100}{400}=\frac{1}{4}.$$

This must equal the given ratio $$\dfrac{50}{x}$$. Hence we write

$$\frac{50}{x}=\frac{1}{4}\quad\Longrightarrow\quad 50\cdot4=x\quad\Longrightarrow\quad x=200.$$

So, the answer is $$200$$.

We begin by recalling the definition of the modulation index for amplitude modulation. In AM, the modulation index (usually denoted by $$m$$) is the ratio of the peak (maximum) voltage of the message, or modulating, signal $$V_m$$ to the peak voltage of the unmodulated carrier signal $$V_c$$.

Mathematically, the formula is stated as:

$$m=\dfrac{V_m}{V_c}$$

From the data given in the problem we identify:

$$V_m = 20 \text{ V}$$ (because the peak voltage of the message signal is 20 V),

$$V_c = 20 \text{ V}$$ (because the peak voltage of the carrier wave is also 20 V).

Now we substitute these numerical values into the formula:

$$m = \dfrac{V_m}{V_c} = \dfrac{20 \text{ V}}{20 \text{ V}}$$

Dividing the identical numerators and denominators, we get:

$$m = 1$$

Hence, the modulation index is unity, i.e. its value is exactly 1.

So, the answer is $$1$$.

The transmitted power is $$P_i = 0.1$$ kW $$= 100$$ W. The attenuation of the cable is $$-5$$ dB per km, and the total cable length is 20 km. Therefore, the total attenuation is $$-5 \times 20 = -100$$ dB.

Using the gain formula in dB: $$\text{Gain (dB)} = 10 \log_{10}\left(\frac{P_O}{P_i}\right)$$, we substitute the total attenuation: $$-100 = 10 \log_{10}\left(\frac{P_O}{100}\right)$$.

Dividing both sides by 10: $$\log_{10}\left(\frac{P_O}{100}\right) = -10$$.

Taking the antilogarithm: $$\frac{P_O}{100} = 10^{-10}$$, which gives $$P_O = 100 \times 10^{-10} = 10^2 \times 10^{-10} = 10^{-8}$$ W.

Comparing with the given expression $$P_O = 10^{-x}$$ W, we get $$x = 8$$.

Therefore, the value of $$x$$ is $$8$$.

For line-of-sight (LOS) communication over the curved surface of the Earth, the maximum distance to the horizon from an antenna of height $$h$$ is obtained from the geometry of a tangent drawn to a sphere of radius $$R$$. The relation is stated as

$$d_{\text{horizon}} = \sqrt{2Rh}$$

where $$R$$ is the radius of the Earth and $$h$$ is in the same length unit as $$R$$. If two antennas are involved, the total LOS range $$d$$ is the sum of their individual horizon distances, so

$$d = \sqrt{2Rh_1} + \sqrt{2Rh_2}$$

We have

$$R = 6.4 \times 10^{6}\ \text{m}$$

Height of the transmitting antenna: $$h_1 = 320\ \text{m}$$

Height of the receiving antenna: $$h_2 = 2000\ \text{m}$$

First, we find the horizon distance for the transmitting antenna:

$$\sqrt{2Rh_1}=\sqrt{2 \times 6.4 \times 10^{6}\ \text{m} \times 320\ \text{m}}$$

$$= \sqrt{(2 \times 6.4) \times 320 \times 10^{6}}$$

$$= \sqrt{12.8 \times 320 \times 10^{6}}$$

$$= \sqrt{4096 \times 10^{6}}$$

$$= \sqrt{4.096 \times 10^{9}}$$

$$= \sqrt{4.096}\times\sqrt{10^{9}}$$

$$= 2.0249 \times 10^{4.5}$$

$$= 2.0249 \times 31\,622.776$$

$$\approx 64\,000\ \text{m}$$

$$= 64\ \text{km}$$

Next, we calculate the horizon distance for the receiving antenna:

$$\sqrt{2Rh_2}=\sqrt{2 \times 6.4 \times 10^{6}\ \text{m} \times 2000\ \text{m}}$$

$$= \sqrt{(2 \times 6.4) \times 2000 \times 10^{6}}$$

$$= \sqrt{12.8 \times 2000 \times 10^{6}}$$

$$= \sqrt{25\,600 \times 10^{6}}$$

$$= \sqrt{2.56 \times 10^{10}}$$

$$= \sqrt{2.56}\times\sqrt{10^{10}}$$

$$= 1.6 \times 10^{5}$$

$$= 160\,000\ \text{m}$$

$$= 160\ \text{km}$$

Adding the two horizon distances gives the maximum LOS range:

$$d = 64\ \text{km} + 160\ \text{km}$$

$$d = 224\ \text{km}$$

Hence, the correct answer is Option C.

The range of a TV transmission tower of height $$h_T$$ with a receiving antenna at height $$h_R$$ is given by $$d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$$, where $$R$$ is the radius of the Earth.

In case (i), the receiving antenna is at ground level ($$h_R = 0$$), so the range is $$d_1 = \sqrt{2R \times 20} = \sqrt{40R}$$.

In case (ii), the receiving antenna is at height 5 m, so the range is $$d_2 = \sqrt{2R \times 20} + \sqrt{2R \times 5} = \sqrt{40R} + \sqrt{10R}$$.

The percentage increase in range is $$n = \frac{d_2 - d_1}{d_1} \times 100 = \frac{\sqrt{10R}}{\sqrt{40R}} \times 100 = \frac{1}{2} \times 100 = 50\%$$.

The value of $$n$$ is $$50$$.

We start by recalling the standard mathematical description of a sinusoidal amplitude-modulated (AM) signal. In general, the instantaneous value of an AM wave can be written as

$$C(t)=A_c\bigl(1+m\cos\omega_m t\bigr)\sin\omega_c t$$

where

$$A_c$$ is the carrier amplitude,

$$m$$ is the modulation index,

$$\omega_m$$ (in rad s−1) is the angular frequency of the modulating (message) signal, and

$$\omega_c$$ (in rad s−1) is the angular frequency of the carrier.

Comparing the given expression

$$C_m(t)=10\bigl(1+0.2\cos 12560\,t\bigr)\sin\!\bigl(111\times10^4\,t\bigr)\;{\rm V}$$

with the standard form, we make term-by-term identification:

$$A_c = 10,$$

$$m = 0.2,$$

$$\omega_m = 12560\ \text{rad s}^{-1},$$

$$\omega_c = 111\times10^4\ \text{rad s}^{-1}.$$

Our objective is to find the numerical value of the modulating frequency $$f_m$$ in kilohertz. The relationship between angular frequency and ordinary (linear) frequency is

$$f = \frac{\omega}{2\pi}.$$

Applying this formula to the modulating signal, we have

$$f_m = \frac{\omega_m}{2\pi} = \frac{12560}{2\pi}\ \text{Hz}.$$

Carrying out the division, we proceed step by step:

First compute the denominator:

$$2\pi \approx 6.2832.$$

Now divide:

$$f_m = \frac{12560}{6.2832}\ \text{Hz}.$$

Evaluating the quotient gives

$$f_m \approx 1999.8\ \text{Hz}.$$

Since 1999.8 Hz is effectively 2000 Hz when rounded to a reasonable number of significant figures, we may write

$$f_m \approx 2000\ \text{Hz}.$$

Finally, converting hertz to kilohertz (recall that $$1\ \text{kHz}=1000\ \text{Hz}$$) yields

$$f_m = \frac{2000\ \text{Hz}}{1000} = 2\ \text{kHz}.$$

Hence, the correct answer is Option 2.

We have an audio signal $$v_m = 20\sin 2\pi \times 1500t$$ and a carrier wave $$v_c = 80\sin 2\pi \times 100000t$$.

The amplitude of the modulating signal is $$A_m = 20$$ and the amplitude of the carrier wave is $$A_c = 80$$.

The modulation index is defined as $$\mu = \frac{A_m}{A_c}$$.

Substituting the values, $$\mu = \frac{20}{80} = 0.25$$.

The percent modulation is $$\mu \times 100 = 0.25 \times 100 = 25\%$$.

So, the answer is $$25$$.

For VHF signal broadcasting, the maximum service area covered by an antenna tower depends on the line-of-sight distance. The maximum distance $$d$$ at which the signal can be received is given by $$d = \sqrt{2Rh}$$, where $$R = 6400$$ km is the radius of the Earth and $$h = 30$$ m $$= 0.03$$ km is the height of the antenna tower.

Substituting the values: $$d = \sqrt{2 \times 6400 \times 0.03} = \sqrt{384}$$ km.

The maximum service area is $$A = \pi d^2 = \pi \times 384 = 3.14 \times 384 = 1205.76 \approx 1206$$ km$$^2$$.

Therefore, the maximum service area covered is $$\boxed{1206}$$ km$$^2$$.

In amplitude modulation (AM), the bandwidth required for each broadcast station is twice the highest modulating frequency.

Given that the highest modulating frequency is 5 kHz, the bandwidth per station is $$2 \times 5 = 10$$ kHz.

The total available bandwidth is 90 kHz. Therefore, the number of AM broadcast stations that can be accommodated is $$\frac{90}{10} = 9$$.

For line-of-sight (LOS) communication over the curved surface of the Earth, the farthest distance to the visible horizon from the top of an antenna of height $$h$$ is obtained from the well-known geometry relation

$$d \;=\;\sqrt{2Rh}$$

where

$$R = 6400 \text{ km}$$ (radius of the Earth) and $$h$$ is measured in kilometres, while $$d$$ comes out in kilometres.

When two antennas of heights $$h_1$$ and $$h_2$$ communicate, the total maximum LOS range is the sum of their individual horizon distances:

$$D \;=\; \sqrt{2 R h_1}\;+\;\sqrt{2 R h_2}$$

According to the question we have a fixed total height

$$h_1 + h_2 = 160 \text{ m}$$

To keep units consistent with the radius in kilometres we convert metres to kilometres:

$$160 \text{ m} = 0.160 \text{ km}$$

Thus

$$h_1 + h_2 = 0.160 \text{ km} \quad\text{(1)}$$

Our objective is to maximise

$$D(h_1,h_2) \;=\; \sqrt{2R\,h_1}\;+\;\sqrt{2R\,h_2}$$

Substituting $$R = 6400 \text{ km}$$ first:

$$D(h_1,h_2) = \sqrt{2 \times 6400 \times h_1}\;+\;\sqrt{2 \times 6400 \times h_2}$$

Simplifying the constant factor inside each square root, we factor out the common number:

$$2 \times 6400 = 12800$$ so

$$D(h_1,h_2) = \sqrt{12800\,h_1}\;+\;\sqrt{12800\,h_2}$$

Using the property $$\sqrt{ab} = \sqrt{a}\,\sqrt{b}$$ we write

$$D(h_1,h_2) = \sqrt{12800}\,\left(\sqrt{h_1} + \sqrt{h_2}\right)$$

The factor $$\sqrt{12800}$$ is a positive constant, hence maximising $$D$$ is equivalent to maximising

$$f(h_1,h_2) = \sqrt{h_1} + \sqrt{h_2}$$

subject to the constraint (1) $$h_1 + h_2 = 0.160.$$ Because the square-root function is concave, the sum $$\sqrt{h_1} + \sqrt{h_2}$$ is largest when $$h_1$$ and $$h_2$$ are equal (this can be shown either by calculus with a Lagrange multiplier or by the inequality of arithmetic and geometric means).

Hence, for maximum range, we set

$$h_1 = h_2 = \frac{0.160}{2} = 0.080 \text{ km} = 80 \text{ m}$$

Now we compute the horizon distance for one antenna of 80 m height:

$$d_1 = \sqrt{2 R h_1} = \sqrt{2 \times 6400 \times 0.080}$$

Multiplying inside the root:

$$2 \times 6400 = 12800$$

$$12800 \times 0.080 = 1024$$

Thus

$$d_1 = \sqrt{1024} \text{ km}$$

Since $$1024 = 32^2$$, we get

$$d_1 = 32 \text{ km}$$

Because both antennas have the same height, the second antenna gives the same distance $$d_2 = 32 \text{ km}$$. Hence the total maximum LOS range is

$$D_{\max} = d_1 + d_2 = 32 \text{ km} + 32 \text{ km} = 64 \text{ km}$$

Hence, the correct answer is Option 64.

We start by translating the words of the question into the symbols used in amplitude-modulation theory. The unmodulated carrier has a peak (maximum) voltage $$V_c = 15\ \text{V}$$ and an angular frequency $$\omega_c = 2\pi f_c$$ with $$f_c = 11.21\ \text{MHz}$$. The modulating (message) signal is a sinusoid whose peak voltage is $$V_m = 5\ \text{V}$$ and whose frequency is $$f_m = 7.7\ \text{kHz}$$.

In ordinary (double-side-band) amplitude modulation the time-domain equation for the modulated voltage is written first. The standard formula is

$$v(t)=V_c\bigl[1+m\sin(\omega_m t)\bigr]\sin(\omega_c t),$$

where the dimensionless quantity $$m$$ is called the modulation index or modulation depth. Its definition is

$$m=\frac{V_m}{V_c}.$$

Substituting the given numerical values, we have

$$m=\frac{V_m}{V_c}=\frac{5\text{ V}}{15\text{ V}}=\frac13\;.$$

Next we want to isolate the upper and lower sideband terms. To do this we expand the product in the time-domain expression. First we note the trigonometric identity

$$\sin A\,\sin B=\frac12\bigl[\cos(A-B)-\cos(A+B)\bigr].$$

Applying this identity to $$\bigl[m\sin(\omega_m t)\bigr]\sin(\omega_c t)$$ gives

$$m\sin(\omega_m t)\sin(\omega_c t)=\frac{m}{2}\Bigl[\cos\bigl((\omega_c-\omega_m)t\bigr)-\cos\bigl((\omega_c+\omega_m)t\bigr)\Bigr].$$

Hence the complete expansion of the modulated signal becomes

$$ \begin{aligned} v(t) &=V_c\sin(\omega_c t)+V_c\Bigl[\tfrac{m}{2}\cos\bigl((\omega_c-\omega_m)t\bigr)-\tfrac{m}{2}\cos\bigl((\omega_c+\omega_m)t\bigr)\Bigr] \\ &=V_c\sin(\omega_c t)+\frac{mV_c}{2}\cos\bigl((\omega_c-\omega_m)t\bigr)-\frac{mV_c}{2}\cos\bigl((\omega_c+\omega_m)t\bigr). \end{aligned} $$

We see three distinct sinusoidal terms:

• $$V_c\sin(\omega_c t)$$ is the carrier itself, still having peak amplitude $$V_c = 15\ \text{V}$$.
• $$\dfrac{mV_c}{2}\cos\bigl((\omega_c-\omega_m)t\bigr)$$ is the lower sideband (LSB) at frequency $$f_c-f_m$$.
• $$-\dfrac{mV_c}{2}\cos\bigl((\omega_c+\omega_m)t\bigr)$$ is the upper sideband (USB) at frequency $$f_c+f_m$$. The minus sign only changes phase, not amplitude, so its peak amplitude is the same magnitude.

The important point is that each sideband has a peak voltage of

$$V_{\text{SB}}=\frac{mV_c}{2}.$$

Substituting $$m=\dfrac13$$ and $$V_c=15\ \text{V},$$ we obtain

$$ \begin{aligned} V_{\text{SB}} &=\frac{1}{2}\,\bigl(\tfrac13\bigr)\,(15\ \text{V}) \\ &=\frac{1}{2}\times5\ \text{V} \\ &=2.5\ \text{V}. \end{aligned} $$

The question states that the upper sideband amplitude is expressed as $$\dfrac{a}{10}\ \text{V}$$ and the lower sideband amplitude as $$\dfrac{b}{10}\ \text{V}$$. Since both sidebands are equal in a normal AM wave, we identify

$$\frac{a}{10}=2.5 \quad\Longrightarrow\quad a=25,$$

and also

$$\frac{b}{10}=2.5 \quad\Longrightarrow\quad b=25.$$

The ratio sought is therefore

$$\frac{a}{b}=\frac{25}{25}=1.$$

So, the answer is $$1$$.

For an amplitude modulated (AM) signal, the envelope shows a highest (maximum) value and a lowest (minimum) value of the instantaneous amplitude. The quantity that tells us how strongly the carrier is being modulated by the message is called the modulation index, usually denoted by $$m$$.

First we recall the standard formula that relates the modulation index $$m$$ to the maximum envelope voltage $$E_{\max}$$ and the minimum envelope voltage $$E_{\min}$$:

$$m \;=\; \dfrac{E_{\max}-E_{\min}}{E_{\max}+E_{\min}}.$$

This formula can be remembered as “difference over sum,” and it comes directly from writing the upper and lower envelopes of an AM wave as $$E_c(1+m)$$ and $$E_c(1-m)$$, where $$E_c$$ is the unmodulated carrier amplitude. Subtracting and adding those two expressions gives the above relation.

Now we substitute the numerical values given in the question. The maximum envelope voltage is $$E_{\max}=12\text{ V}$$ and the minimum envelope voltage is $$E_{\min}=3\text{ V}$$. Putting these into the formula gives

$$m \;=\; \dfrac{12\text{ V}-3\text{ V}}{12\text{ V}+3\text{ V}}.$$

Carrying out the subtraction in the numerator, we have

$$m \;=\; \dfrac{9\text{ V}}{12\text{ V}+3\text{ V}}.$$

Next we add the voltages in the denominator:

$$m \;=\; \dfrac{9\text{ V}}{15\text{ V}}.$$

Since both numerator and denominator carry the unit “volt,” the units cancel, leaving a pure number:

$$m \;=\; \dfrac{9}{15}.$$

To simplify the fraction, we divide both numerator and denominator by their highest common factor, which is 3:

$$m \;=\; \dfrac{9\div 3}{15\div 3} \;=\; \dfrac{3}{5}.$$

Writing the fraction $$\dfrac{3}{5}$$ in decimal form, we remember that $$\dfrac{3}{5}=0.6$$. Hence we have

$$m = 0.6.$$

The statement in the problem says that “the modulation index is $$0.6x$$.” We have just found that the modulation index equals $$0.6$$ itself. Therefore, comparing the two expressions,

$$0.6x = 0.6.$$

To get the value of $$x$$, we divide both sides of the equation by $$0.6$$:

$$x = \dfrac{0.6}{0.6} = 1.$$

So, the answer is $$1$$.