MHT CET 2024 Physics Question Paper with Answer and Solution

(A) Let $n$ be the frequency of tuning fork $C$.
Given that the frequency of $A$ is $1.4 \%$ more than $C$, so $n_A = n + 0.014n = 1.014n$.
Given that the frequency of $B$ is $2.6 \%$ less than $C$, so $n_B = n - 0.026n = 0.974n$.
When $A$ and $B$ are sounded together, the number of beats produced per second is the difference in their frequencies: $|n_A - n_B| = 10$.
Substituting the values: $1.014n - 0.974n = 10$.
$0.04n = 10$.
$n = \frac{10}{0.04} = \frac{1000}{4} = 250 \text{ Hz}$.

(B) The beat frequency is defined as the difference between the frequencies of the two waves: $f_{beat} = |f_2 - f_1|$.
Given $f_1 = 340 \ Hz$ and $f_2 = 335 \ Hz$.
$f_{beat} = 340 \ Hz - 335 \ Hz = 5 \ Hz$.
The time interval between two successive maxima is the reciprocal of the beat frequency:
$T = \frac{1}{f_{beat}} = \frac{1}{5} \ s = 0.2 \ s$.

(B) For an open organ pipe of length $L$,the fundamental frequency is given by $f = \frac{V}{2L}$.
For the first pipe of length $l$,the fundamental frequency is $f_1 = \frac{V}{2l}$.
For the second pipe of length $(l+l_1)$,the fundamental frequency is $f_2 = \frac{V}{2(l+l_1)}$.
The beat frequency $f_b$ is the difference between the two frequencies: $f_b = f_1 - f_2 = \frac{V}{2l} - \frac{V}{2(l+l_1)}$.
Simplifying the expression: $f_b = \frac{V}{2} \left[ \frac{(l+l_1) - l}{l(l+l_1)} \right] = \frac{V l_1}{2l(l+l_1)}$.
Given the condition $l_1 \ll l$,we can approximate $(l+l_1) \approx l$ in the denominator.
Therefore,$f_b \approx \frac{V l_1}{2l^2}$.

(D) The beat frequency is given by $f_b = |f_2 - f_1| = |256 \,Hz - 250 \,Hz| = 6 \,Hz$.
The time period of the beat wave is $T_b = \frac{1}{f_b} = \frac{1}{6} \,s$.
The intensity is maximum at $t=0$. The intensity becomes minimum at half the time period of the beat, which corresponds to the interval between a maximum and the subsequent minimum.
Therefore, the time required is $t = \frac{T_b}{2} = \frac{1}{6 \times 2} = \frac{1}{12} \,s$.

(C) Let the velocity of sound be $v$ and the frequency of the source be $f$. The observer moves towards the stationary source with a velocity $v_o = v/5$.
According to the Doppler effect,the apparent frequency $f'$ heard by the observer is given by the formula: $f' = f \left( \frac{v + v_o}{v} \right)$.
Substituting the given values: $f' = f \left( \frac{v + v/5}{v} \right) = f \left( \frac{6v/5}{v} \right) = 1.2f$.
The increase in frequency is $\Delta f = f' - f = 1.2f - f = 0.2f$.
The percentage increase is given by $\left( \frac{\Delta f}{f} \right) \times 100 = \left( \frac{0.2f}{f} \right) \times 100 = 20 \%$.
Therefore,the correct option is $C$.

(C) Step $1$: The sound from the siren travels towards the wall. The wall acts as a stationary observer. The frequency $n_1$ received by the wall is given by the Doppler effect formula for a moving source and stationary observer: $n_1 = n \left( \frac{V}{V - V_1} \right)$.
Step $2$: The wall reflects this sound. Now,the wall acts as a stationary source of frequency $n_1$,and the driver acts as an observer moving towards the wall with speed $V_1$. The frequency $n_2$ heard by the driver is given by: $n_2 = n_1 \left( \frac{V + V_1}{V} \right)$.
Step $3$: Substituting the value of $n_1$ into the equation for $n_2$: $n_2 = \left( n \frac{V}{V - V_1} \right) \left( \frac{V + V_1}{V} \right) = n \left( \frac{V + V_1}{V - V_1} \right)$.

(D) Let the original frequency of the whistle be $n$ and the observed frequency be $n^{\prime}$.
Given that the frequency drops by $30 \%$,the observed frequency is $n^{\prime} = n - 0.30n = 0.7n$.
According to the Doppler effect,when a source moves away from a stationary observer,the observed frequency is given by $n^{\prime} = n \left( \frac{v}{v + v_s} \right)$,where $v = 350 \ m/s$ is the speed of sound and $v_s$ is the speed of the engine.
Substituting the values: $0.7n = n \left( \frac{350}{350 + v_s} \right)$.
Dividing both sides by $n$: $0.7 = \frac{350}{350 + v_s}$.
$0.7(350 + v_s) = 350$.
$245 + 0.7v_s = 350$.
$0.7v_s = 350 - 245 = 105$.
$v_s = \frac{105}{0.7} = 150 \ m/s$.

(A) Given:
Frequency of source $(n_0) = 510 \,Hz$.
Velocity of source $(v_s) = 72 \,km/hr = 72 \times \frac{5}{18} = 20 \,m/s$.
Velocity of sound in air $(v) = 320 \,m/s$.
$(1)$ When the source is approaching a stationary listener,the apparent frequency $(n_1)$ is given by:
$n_1 = n_0 \left( \frac{v}{v - v_s} \right)$
$n_1 = 510 \times \left( \frac{320}{320 - 20} \right) = 510 \times \left( \frac{320}{300} \right) = 544 \,Hz$.
$(2)$ When the source is moving away from a stationary listener,the apparent frequency $(n_2)$ is given by:
$n_2 = n_0 \left( \frac{v}{v + v_s} \right)$
$n_2 = 510 \times \left( \frac{320}{320 + 20} \right) = 510 \times \left( \frac{320}{340} \right) = 480 \,Hz$.
Thus,the frequencies are $544 \,Hz$ and $480 \,Hz$.

(B) According to the Doppler effect,when the listener moves towards a stationary source with velocity $V_1$,the apparent frequency $F_1$ is given by:
$F_1 = F \left( \frac{V + V_1}{V} \right)$
When the listener moves away from the stationary source with velocity $V_1$,the apparent frequency $F_2$ is given by:
$F_2 = F \left( \frac{V - V_1}{V} \right)$
Given the ratio $\frac{F_1}{F_2} = 2$,we substitute the expressions:
$\frac{F_1}{F_2} = \frac{V + V_1}{V - V_1} = 2$
$V + V_1 = 2(V - V_1)$
$V + V_1 = 2V - 2V_1$
$3V_1 = V$
Therefore,$\frac{V}{V_1} = 3$.

(D) Applying Doppler's effect to sound waves,the observed frequency $n^{\prime}$ is given by $n^{\prime} = n \left( \frac{v + v_0}{v} \right) = n \left( 1 + \frac{v_0}{v} \right)$,where $v$ is the velocity of sound and $v_0$ is the velocity of the observer moving towards the source.
Given that the observed frequency is triple the source frequency,$n^{\prime} = 3n$.
Substituting this into the equation: $3n = n \left( 1 + \frac{v_0}{v} \right)$.
Dividing both sides by $n$: $3 = 1 + \frac{v_0}{v}$.
Solving for $v_0$: $\frac{v_0}{v} = 2$,which implies $v_0 = 2v$.
Therefore,the observer must move with twice the velocity of sound towards the source.

(A) The wall acts as a stationary source of reflected sound. The frequency of sound received by the wall is $n_w = n \left(\frac{V}{V-V_1}\right)$.
This reflected sound acts as a source for the driver. Since the driver is moving towards the wall (source),the frequency heard by the driver is $n' = n_w \left(\frac{V+V_1}{V}\right)$.
Substituting $n_w$,we get $n' = n \left(\frac{V}{V-V_1}\right) \left(\frac{V+V_1}{V}\right) = n \left(\frac{V+V_1}{V-V_1}\right)$.

(B) According to the Doppler effect,the observed frequency $n^{\prime}$ is given by the formula:
$n^{\prime} = \left( \frac{V + V_{L}}{V - V_{S}} \right) n$
Here,the speed of the listener $V_{L} = \frac{V}{10}$ and the speed of the source $V_{S} = \frac{V}{10}$.
Substituting these values into the formula:
$n^{\prime} = \left( \frac{V + \frac{V}{10}}{V - \frac{V}{10}} \right) n$
$n^{\prime} = \left( \frac{\frac{11V}{10}}{\frac{9V}{10}} \right) n$
$n^{\prime} = \frac{11}{9} n$
$n^{\prime} \approx 1.22 n$

(C) The resultant amplitude '$R$' of two waves with amplitudes '$A_1$' and '$A_2$' and a phase difference '$\phi$' is given by the formula: $R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\phi)}$.
Given: $A_1 = A$,$A_2 = A$,and $\phi = \frac{\pi}{2}$.
Substituting these values into the formula:
$R = \sqrt{A^2 + A^2 + 2(A)(A) \cos(\frac{\pi}{2})}$
Since $\cos(\frac{\pi}{2}) = 0$,the expression simplifies to:
$R = \sqrt{A^2 + A^2 + 0} = \sqrt{2A^2} = \sqrt{2} A$.
Therefore,the maximum amplitude of the resultant wave is $\sqrt{2} A$.

(A) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi = \phi_1 - \phi_2$ is given by the formula:
$A^2 = a_1^2 + a_2^2 + 2a_1 a_2 \cos \phi$
Given that $a_1 = a_2 = a$ and the resultant amplitude $A = a$,we substitute these into the equation:
$a^2 = a^2 + a^2 + 2(a)(a) \cos \phi$
$a^2 = 2a^2 + 2a^2 \cos \phi$
Subtracting $2a^2$ from both sides:
$-a^2 = 2a^2 \cos \phi$
Dividing by $2a^2$:
$\cos \phi = -\frac{1}{2}$
Therefore,the phase difference is:
$\phi = \phi_1 - \phi_2 = \cos ^{-1}\left(-\frac{1}{2}\right)$

(A) The path difference is given as $\Delta x = \lambda/3$.
The phase difference $\phi$ is related to path difference by the formula $\phi = \frac{2\pi}{\lambda} \times \Delta x$.
Substituting the value,$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} = 120^{\circ}$.
The resultant amplitude $R$ of two waves with equal amplitude $A$ is given by $R = \sqrt{A^2 + A^2 + 2A^2 \cos \phi}$.
Substituting $\phi = 120^{\circ}$ and $\cos(120^{\circ}) = -0.5$:
$R = \sqrt{A^2 + A^2 + 2A^2(-0.5)}$
$R = \sqrt{2A^2 - A^2}$
$R = \sqrt{A^2} = A$.

(B) In a closed organ pipe,the fundamental frequency is $f_1 = \frac{v}{4L}$.
The frequencies of the overtones are given by $f_n = (2n-1)f_1$,where $n$ is the harmonic number.
The first overtone is the $3^{\text{rd}}$ harmonic $(n=2)$,and the second overtone is the $5^{\text{th}}$ harmonic $(n=3)$.
For the $n^{\text{th}}$ harmonic in a closed pipe,the number of nodes is $n$ and the number of antinodes is $n$.
Since the second overtone corresponds to the $5^{\text{th}}$ harmonic ($n=3$ in the overtone sequence,but it is the $3^{\text{rd}}$ mode of vibration),we look at the mode $n=3$.
In the $3^{\text{rd}}$ mode of vibration for a closed pipe,there are $3$ nodes and $3$ antinodes.

(D) The fundamental frequency of an open pipe of length $L$ is given by $f_1 = \frac{V}{2L}$.
When $80\%$ of the pipe is immersed in water,the length of the air column remaining above the water level is $l = L - 0.8L = 0.2L = \frac{L}{5}$.
Since the pipe is now closed at one end (by the water surface),it behaves as a closed organ pipe of length $l$.
The fundamental frequency of a closed pipe is $f_2 = \frac{V}{4l}$.
Substituting $l = \frac{L}{5}$,we get $f_2 = \frac{V}{4(L/5)} = \frac{5V}{4L}$.
Now,the ratio $f_1:f_2$ is $\frac{f_1}{f_2} = \frac{V/2L}{5V/4L} = \frac{V}{2L} \times \frac{4L}{5V} = \frac{4}{10} = \frac{2}{5}$.

(A) For an open organ pipe,the effective length $(L)$ considering the end correction $(e)$ at both ends is given by:
$L = l + 2e$
Since the end correction for an open pipe is $e = 0.6r$,we have:
$L = l + 2(0.6r) = l + 1.2r$
The fundamental frequency $(f)$ of an open pipe is given by the formula:
$f = \frac{v}{2L}$
Substituting the value of $L$:
$f = \frac{v}{2(l + 1.2r)}$

(C) Let the shortest resonating length be $l_1 = 15 \ cm$ and the end correction be $e = 1 \ cm$.
For the first resonance (fundamental mode) in a tube closed at one end:
$l_1 + e = \frac{\lambda}{4}$
Substituting the values:
$15 + 1 = \frac{\lambda}{4} \implies 16 = \frac{\lambda}{4} \implies \lambda = 64 \ cm$
For the next resonating length $l_2$ (first overtone):
$l_2 + e = \frac{3\lambda}{4}$
Substituting the values:
$l_2 + 1 = \frac{3 \times 64}{4}$
$l_2 + 1 = 3 \times 16$
$l_2 + 1 = 48$
$l_2 = 47 \ cm$

(A) The tuning fork is in resonance with an air column in a pipe closed at one end. The resonant frequency is given by $n = \frac{(2N-1)v}{4l}$,where $N = 1, 2, 3, \dots$ corresponds to different modes of vibration.
Substituting $n = 340 \ Hz$ and $v = 340 \ m/s$ (speed of sound in air),the length of the air column $l$ is:
$l = \frac{(2N-1)v}{4n} = \frac{(2N-1) \times 340}{4 \times 340} = \frac{2N-1}{4} \ m = (2N-1) \times 25 \ cm$.
For $N = 1, 2, 3, \dots$,the possible lengths of the air column are $l = 25 \ cm, 75 \ cm, 125 \ cm, \dots$.
Since the tube is only $120 \ cm$ long,the possible lengths of the air column are $25 \ cm$ and $75 \ cm$.
The corresponding height of the water column is $h = \text{Total height} - l$.
For $l = 25 \ cm$,$h = 120 - 25 = 95 \ cm$.
For $l = 75 \ cm$,$h = 120 - 75 = 45 \ cm$.
The minimum height of water required for resonance is $45 \ cm$.

(B) When a string is fixed at both ends (rigid supports),the displacement at these points must be zero because they cannot move.
Points with zero displacement are called nodes.
Therefore,nodes are formed at both ends.
For the string to vibrate,there must be at least one point of maximum displacement between the two fixed ends,which is called an antinode.
Thus,the simplest mode of vibration (the fundamental mode) consists of nodes at both ends and at least one antinode in between.

(A) The frequency of vibration of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Here,$T$ is the tension in the wire,which is provided by the weight of a mass $M$ suspended from it,so $T = Mg$.
Thus,$n = \frac{1}{2l} \sqrt{\frac{Mg}{m}}$.
For the frequency $n$ to remain constant with the same tuning fork,we have $\frac{1}{l} \sqrt{g} = \text{constant}$.
This implies $l \propto \sqrt{g}$.
Since the acceleration due to gravity $g$ is greater at the poles $(g_p)$ than at the equator $(g_e)$,i.e.,$g_p > g_e$,it follows that the length $l$ must be adjusted.
To maintain the same frequency at the equator where $g$ is smaller,the length $l$ must be decreased to compensate for the decrease in $g$ such that the ratio $\frac{\sqrt{g}}{l}$ remains constant.
Therefore,the vibrating length of the wire should be decreased.

(D) The wavelength $\lambda$ of the sound wave is given by $\lambda = \frac{v}{f} = \frac{340}{340} = 1 \ m = 100 \ cm$.
For a tube closed at one end,resonance occurs when the length of the air column $L$ is an odd multiple of $\frac{\lambda}{4}$,i.e.,$L = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \dots$
The possible lengths of the air column are $25 \ cm, 75 \ cm, 125 \ cm, \dots$
The total height of the tube is $150 \ cm$.
To obtain resonance,the air column length must be one of the values above. To have the minimum height of the water column,we must choose the maximum possible air column length that fits in the tube,which is $125 \ cm$.
Minimum height of water column $= \text{Total height} - \text{Maximum air column length} = 150 \ cm - 125 \ cm = 25 \ cm$.

(B) For a pipe closed at one end,the frequency of the $n^{th}$ overtone is given by $f = \frac{(2n+1)V}{4L_c}$.
For the third overtone,$n=3$,so $f = \frac{(2 \times 3 + 1)V}{4L_c} = \frac{7V}{4L_c}$.
For a pipe open at both ends,the frequency of the $n^{th}$ overtone is given by $f = \frac{(n+1)V}{2L_o}$.
For the sixth overtone,$n=6$,so $f = \frac{(6+1)V}{2L_o} = \frac{7V}{2L_o}$.
Equating the two frequencies: $\frac{7V}{4L_c} = \frac{7V}{2L_o}$.
Simplifying,we get $\frac{1}{4L_c} = \frac{1}{2L_o}$,which implies $\frac{L_c}{L_o} = \frac{2}{4} = \frac{1}{2}$.

(B) Given: Length of the pipe $l = 60 \ cm = 0.6 \ m$,Frequency of source $f = 2.2 \ kHz = 2200 \ Hz$,Speed of sound $v = 330 \ m/s$.
For an open pipe,the frequency of the $n^{\text{th}}$ harmonic is given by $f_n = n \frac{v}{2l}$.
Substituting the given values into the formula:
$2200 = n \left[ \frac{330}{2 \times 0.6} \right]$
$2200 = n \left[ \frac{330}{1.2} \right]$
$2200 = n \times 275$
$n = \frac{2200}{275} = 8$.
Therefore,the $8^{\text{th}}$ harmonic mode of the pipe resonates with the source.

(B) The energy $E$ of a sound wave is proportional to the square of its amplitude $A$ and the square of its frequency $n$. Mathematically,$E \propto A^2 n^2$.
Since the energies of the waves produced by $P$ and $Q$ are equal,we have $E_P = E_Q$.
This implies $A_P^2 n_P^2 = A_Q^2 n_Q^2$.
Given $n_P = n$ and $n_Q = \frac{n}{4}$,we substitute these values into the equation:
$A_P^2 n^2 = A_Q^2 (\frac{n}{4})^2$.
$A_P^2 n^2 = A_Q^2 (\frac{n^2}{16})$.
Dividing both sides by $n^2$,we get $A_P^2 = \frac{A_Q^2}{16}$.
$A_Q^2 = 16 A_P^2$.
Taking the square root of both sides,we get $A_Q = 4 A_P$.

(C) The loudness of two sounds is given as $L_2 = 60 \ dB$ and $L_1 = 30 \ dB$.
The formula for the loudness of sound is $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.
The difference in loudness is given by $L_2 - L_1 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
Substituting the values: $60 - 30 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
$30 = 10 \log_{10} \left( \frac{I_2}{I_1} \right)$.
$\log_{10} \left( \frac{I_2}{I_1} \right) = 3$.
Therefore,$\frac{I_2}{I_1} = 10^3 = 1000$.
Thus,the $60 \ dB$ sound is $1000$ times more intense than the $30 \ dB$ sound.

(D) The velocity of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$.
For the same temperature $T$,the ratio of the velocity of sound in hydrogen $(v_H)$ to helium $(v_{He})$ is:
$\frac{v_H}{v_{He}} = \sqrt{\frac{\gamma_H R T / M_H}{\gamma_{He} R T / M_{He}}} = \sqrt{\frac{\gamma_H M_{He}}{\gamma_{He} M_H}}$.
Given:
$\gamma_H = 7/5$,$\gamma_{He} = 5/3$,
$M_H = 2 \text{ g/mol}$,$M_{He} = 4 \text{ g/mol}$.
Substituting these values:
$\frac{v_H}{v_{He}} = \sqrt{\frac{(7/5) \times 4}{(5/3) \times 2}} = \sqrt{\frac{28/5}{10/3}} = \sqrt{\frac{28}{5} \times \frac{3}{10}} = \sqrt{\frac{84}{50}} = \sqrt{\frac{42}{25}} = \frac{\sqrt{42}}{5}$.
Thus,the ratio is $\sqrt{42}: 5$.

(C) The given equation for the stationary wave is $y = 12 \cos \left(\frac{\pi}{6} x\right) \sin (8 \pi t)$.
Comparing this with the standard form $y = A_0 \cos(kx) \sin(\omega t)$,we identify the wave number $k = \frac{2\pi}{\lambda}$.
From the equation,$k = \frac{\pi}{6}$.
Equating the two,we get $\frac{2\pi}{\lambda} = \frac{\pi}{6}$.
Solving for $\lambda$,we find $\lambda = 12 \ cm$.
The distance between two successive antinodes in a stationary wave is given by $\frac{\lambda}{2}$.
Therefore,the distance $= \frac{12 \ cm}{2} = 6 \ cm$.

(C) The length of the string is $L = 90 \ cm$.
There are $3$ nodes $(N)$ in the stationary wave.
The distance between two consecutive nodes is $\frac{\lambda}{2}$.
Since there are $3$ nodes,there are $2$ such segments (loops) between them.
Therefore,the total length $L = 2 \times \frac{\lambda}{2} = \lambda$.
Given $L = 90 \ cm$,we have $\lambda = 90 \ cm$.

(C) The resonant frequencies are given as $f_1 = 90 \,Hz$, $f_2 = 150 \,Hz$, and $f_3 = 210 \,Hz$.
The difference between consecutive resonant frequencies is $\Delta f = 150 - 90 = 60 \,Hz$ and $210 - 150 = 60 \,Hz$.
For a string fixed at both ends, the resonant frequencies are integer multiples of the fundamental frequency $f_0$. Thus, $\Delta f = f_0 = 60 \,Hz$.
The given frequencies are $f_1 = 1.5 f_0$ (not possible for both ends fixed) or they are harmonics of a string fixed at one end. However, if we assume the fundamental frequency $f_0 = 30 \,Hz$ (since $90 = 3 \times 30$, $150 = 5 \times 30$, $210 = 7 \times 30$), these are odd harmonics of a string fixed at both ends where the fundamental is $30 \,Hz$.
Using $f_n = \frac{n v}{2L}$, for $n=3$, $f_3 = \frac{3v}{2L} = 90 \,Hz$.
Given $L = 0.8 \,m$, we have $90 = \frac{3v}{2 \times 0.8}$.
$90 = \frac{3v}{1.6} \implies 3v = 144 \implies v = 48 \,m/s$.

(C) The resultant displacement $Y$ is given by the superposition principle: $Y = Y_1 + Y_2$.
Using the trigonometric identity $\sin(A - B) + \sin(A + B) = 2 \sin A \cos B$,where $A = \frac{2 \pi t}{0.4}$ and $B = \frac{2 \pi x}{4}$,we get:
$Y = 2 \sin \left( \frac{2 \pi t}{0.4} \right) \cos \left( \frac{2 \pi x}{4} \right)$.
The amplitude $R$ of the standing wave at any position $x$ is given by $R = |2 \cos \left( \frac{2 \pi x}{4} \right)|$.
Substituting $x = 0.5 \ m$:
$R = 2 \cos \left( \frac{2 \pi \times 0.5}{4} \right) = 2 \cos \left( \frac{\pi}{4} \right)$.
Since $\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$,we have $R = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \ m$.

(B) The standard equation for a progressive wave is $y = a \sin(\omega t + kx)$.
Comparing the given equation $y = 0.002 \sin(100t + x)$ with the standard form,we get:
Angular frequency $\omega = 100 \ rad/s$
Wave number $k = 1 \ rad/m$
Amplitude $a = 0.002 \ m$
Frequency $f = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} \ Hz$.
Wave velocity $v = \frac{\omega}{k} = \frac{100}{1} = 100 \ m/s$.
Since the sign between $\omega t$ and $kx$ is positive,the wave is travelling in the negative $x$-direction.
Therefore,the wave is travelling with a velocity of $100 \ m/s$ in the negative $x$-direction.

(A) Given: Frequency $(n) = 400 \ Hz$,Velocity $(v) = 336 \ m/s$,Phase difference $(\phi) = 60^{\circ} = \frac{\pi}{3} \ rad$.
First,calculate the wavelength $(\lambda)$ using the formula $v = n \lambda$:
$\lambda = \frac{v}{n} = \frac{336}{400} = 0.84 \ m$.
The relationship between path difference $(\Delta x)$ and phase difference $(\phi)$ is given by $\Delta x = \frac{\lambda}{2\pi} \times \phi$.
Substituting the values:
$\Delta x = \frac{0.84}{2\pi} \times \frac{\pi}{3} = \frac{0.84}{6} = 0.14 \ m$.
Therefore,the two points are $0.14 \ m$ apart.

(C) The wavelength $\lambda$ of the wave is given by the formula $\lambda = \frac{v}{f}$,where $v$ is the speed and $f$ is the frequency.
Substituting the given values,$\lambda = \frac{30}{250} = 0.12 \ m$.
The phase difference $\Delta\phi$ between two points separated by a path difference $\Delta x$ is given by $\Delta\phi = \frac{2\pi}{\lambda} \times \Delta x$.
Given the path difference $\Delta x = 10 \ cm = 0.1 \ m$.
Substituting the values,$\Delta\phi = \frac{2\pi}{0.12} \times 0.1$.
$\Delta\phi = \frac{0.2\pi}{0.12} = \frac{20\pi}{12} = \frac{5\pi}{3} \ radian$.

(B) Mass per unit length $(m)$:
$m = \frac{M}{l} = \frac{0.2 \ kg}{2 \ m} = 0.1 \ kg/m$
Velocity of the transverse wave $(v)$:
$v = \sqrt{\frac{T}{m}} = \sqrt{\frac{2.5 \ N}{0.1 \ kg/m}} = \sqrt{25} = 5 \ m/s$
Time taken $(t)$ by the wave to reach the other end:
$t = \frac{l}{v} = \frac{2 \ m}{5 \ m/s} = 0.4 \ s$

(B) The distance between $N$ consecutive crests is given by $(N-1) \lambda$,where $\lambda$ is the wavelength.
Given that the distance between $5$ consecutive crests is $0.8 \ m$,we have:
$4 \lambda = 0.8 \ m$
$\lambda = \frac{0.8}{4} = 0.2 \ m$
Using the wave equation $v = n \lambda$,where $v$ is the velocity and $n$ is the frequency:
$n = \frac{v}{\lambda}$
Substituting the given values $v = 56 \ m/s$ and $\lambda = 0.2 \ m$:
$n = \frac{56}{0.2} = 280 \ Hz$
Therefore,the frequency of the tuning fork is $280 \ Hz$.

(C) For a string fixed at both ends,the wave forms a stationary wave pattern.
Let $L$ be the length of the string and $\lambda$ be the wavelength.
The number of loops formed in the string is $n$,where $n = x - 1$ (since there are $x$ nodes including the two fixed ends).
The length of the string is given by $L = n \frac{\lambda}{2}$.
Substituting $n = x - 1$,we get $L = (x - 1) \frac{\lambda}{2}$.

(B) When a string is fixed at both ends,waves travel along the string and reflect back from the fixed boundaries. The superposition of the incident and reflected waves results in the formation of stationary (standing) waves. Since the displacement of the string particles is perpendicular to the direction of wave propagation,these waves are transverse in nature. Therefore,the waves are stationary transverse waves.

(C) The fundamental frequency of a string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \rho \pi r^2$.
For string $B$: $n_B = \frac{1}{2l_B} \sqrt{\frac{T_B}{\mu_B}}$.
Given for string $A$: $l_A = 2l_B$,$d_A = 2d_B \Rightarrow r_A = 2r_B$,$T_A = 2T_B$,and $\rho_A = 2\rho_B$.
Calculating $\mu_A$: $\mu_A = \rho_A \pi r_A^2 = (2\rho_B) \pi (2r_B)^2 = 8 \rho_B \pi r_B^2 = 8\mu_B$.
Now,the fundamental frequency of string $A$ is: $n_A = \frac{1}{2l_A} \sqrt{\frac{T_A}{\mu_A}} = \frac{1}{2(2l_B)} \sqrt{\frac{2T_B}{8\mu_B}} = \frac{1}{4} \left( \frac{1}{2l_B} \sqrt{\frac{T_B}{\mu_B}} \right) = \frac{1}{4} n_B$.
The frequency of the $p^{\text{th}}$ overtone of string $A$ is $f_p = (p+1)n_A$.
We want $f_p = n_B$,so $(p+1) \frac{n_B}{4} = n_B$.
This gives $p+1 = 4$,or $p = 3$.
Thus,the $3^{\text{rd}}$ overtone of string $A$ has the same frequency as the fundamental frequency of string $B$.

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the wire.
When the bob of weight $W$ is in air,the tension $T_1 = W$. Thus,$n_1 = \frac{1}{2L} \sqrt{\frac{W}{\mu}}$.
When the bob is immersed in water,the buoyant force $F_B$ acts on it,reducing the tension to $T_2 = W - F_B$. Thus,$n_2 = \frac{1}{2L} \sqrt{\frac{W - F_B}{\mu}}$.
Taking the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{W}{W - F_B}} \implies \frac{n_1^2}{n_2^2} = \frac{W}{W - F_B}$.
Relative density $\sigma = \frac{W}{F_B}$.
Rearranging the ratio: $\frac{n_1^2}{n_2^2} = \frac{W}{W - (W/\sigma)} = \frac{\sigma}{\sigma - 1}$.
Solving for $\sigma$: $n_1^2(\sigma - 1) = n_2^2 \sigma \implies \sigma(n_1^2 - n_2^2) = n_1^2$.
Therefore,$\sigma = \frac{n_1^2}{n_1^2 - n_2^2}$.

(C) The frequency of the fundamental mode of a stretched string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension and $m$ is the linear mass density.
Let the initial tension be $T$. The initial frequency is $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
When the tension is increased by $3 \ kg \ wt$,the new tension becomes $T' = T + 3$. The new frequency is $f' = \frac{1}{2l} \sqrt{\frac{T+3}{m}}$.
Given the ratio of frequencies is $f : f' = 2 : 3$,we have:
$\frac{f}{f'} = \sqrt{\frac{T}{T+3}} = \frac{2}{3}$
Squaring both sides:
$\frac{T}{T+3} = \frac{4}{9}$
$9T = 4(T + 3)$
$9T = 4T + 12$
$5T = 12$
$T = 2.4 \ kg \ wt$
Thus,the initial tension in the string is $2.4 \ kg \ wt$.

(B) For a sonometer,the frequency of vibration $n$ is proportional to the square root of the tension $T$,i.e.,$n \propto \sqrt{T}$.
When the weight is in air,the tension $T_1 = V \cdot d \cdot \rho_w \cdot g$,where $V$ is the volume of the weight,$d$ is its specific gravity,and $\rho_w$ is the density of water.
Thus,$n \propto \sqrt{V \cdot d \cdot \rho_w \cdot g}$.
When the weight is immersed in water,the effective weight (tension) becomes $T_2 = V \cdot (d - 1) \cdot \rho_w \cdot g$ due to the buoyant force.
The new frequency $n'$ is given by $n' = n - x$ (since beats are produced).
Thus,$n' \propto \sqrt{V \cdot (d - 1) \cdot \rho_w \cdot g}$.
Taking the ratio of the two frequencies:
$\frac{n}{n-x} = \sqrt{\frac{V \cdot d \cdot \rho_w \cdot g}{V \cdot (d - 1) \cdot \rho_w \cdot g}}$
$\frac{n}{n-x} = \sqrt{\frac{d}{d-1}}$.

(A) The fundamental resonant frequency for a pipe open at both ends of length $L$ is given by $n_1 = \frac{V}{2L}$,where $V$ is the speed of sound in air.
The fundamental resonant frequency for a pipe closed at one end of length $L$ is given by $n_2 = \frac{V}{4L}$.
Comparing the two equations:
$n_2 = \frac{V}{4L} = \frac{1}{2} \left( \frac{V}{2L} \right)$
$n_2 = \frac{n_1}{2}$
Therefore,$n_1 = 2 n_2$.

(A) Let the lengths of the three segments be $L_1$,$L_2$,and $L_3$ respectively. The total length is $L_1 + L_2 + L_3 = 110 \ cm$.
For a sonometer wire under constant tension,the fundamental frequency $n$ is inversely proportional to the length $L$ $(n \propto 1/L)$,so $n_1 L_1 = n_2 L_2 = n_3 L_3 = k$.
Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 2 : 3$,we have:
$n_1 = x, n_2 = 2x, n_3 = 3x$.
Substituting these into the relation:
$x L_1 = 2x L_2 = 3x L_3$.
This gives $L_2 = L_1/2$ and $L_3 = L_1/3$.
Substituting into the total length equation:
$L_1 + L_1/2 + L_1/3 = 110 \ cm$.
Multiplying by $6$ to clear denominators: $6 L_1 + 3 L_1 + 2 L_1 = 660 \ cm$.
$11 L_1 = 660 \ cm \Rightarrow L_1 = 60 \ cm$.
Then $L_2 = 30 \ cm$ and $L_3 = 20 \ cm$.
The first bridge is placed at $60 \ cm$ from end $A$. The second bridge is placed at $60 + 30 = 90 \ cm$ from end $A$.

(B) The frequency of the $p^{th}$ harmonic for a string fixed at both ends is given by $f_p = \frac{p}{2L} \sqrt{\frac{T}{m}}$.
Given $f_p = 320 \ Hz$ and the next higher frequency $f_{p+1} = 400 \ Hz$.
We have the ratio: $\frac{f_{p+1}}{f_p} = \frac{p+1}{p}$.
Substituting the values: $\frac{400}{320} = \frac{p+1}{p}$.
Simplifying the fraction: $\frac{5}{4} = \frac{p+1}{p}$.
Cross-multiplying gives: $5p = 4(p+1) \Rightarrow 5p = 4p + 4$.
Therefore,$p = 4$.

(C) For a string fixed at both ends,the frequency of the $n$-th harmonic is given by $f_n = n \cdot f_1$,where $f_1$ is the fundamental frequency.
The $n$-th harmonic corresponds to $(n-1)$-th overtone.
Therefore,the third overtone corresponds to the $4$-th harmonic $(n = 4)$.
In the $n$-th harmonic,the number of loops is $n$.
For the $4$-th harmonic,there are $4$ loops.
The number of nodes is $(n + 1) = 4 + 1 = 5$.
The number of antinodes is $n = 4$.
Thus,there are $5$ nodes and $4$ antinodes.

(D) The end correction $e$ for a tube of circular cross-section is given by the formula $e = 0.6r$ or $e = 0.3d$,where $r$ is the radius and $d$ is the diameter of the tube.
Since the end correction $e$ is directly proportional to the diameter $d$ of the tube $(e \propto d)$,increasing the diameter of the tube will result in a larger end correction.
Therefore,the end correction will be more if the tube is widened.

(D) The difference between two consecutive resonant frequencies of a stretched string is equal to the fundamental frequency $(f_0)$.
$f_0 = f_2 - f_1 = 450 \ Hz - 375 \ Hz = 75 \ Hz$.
The formula for the fundamental frequency is $f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T = 180 \ N$ and $\mu = 2 \times 10^{-3} \ kg/m$.
Substituting the values: $75 = \frac{1}{2L} \sqrt{\frac{180}{2 \times 10^{-3}}} = \frac{1}{2L} \sqrt{90000} = \frac{300}{2L} = \frac{150}{L}$.
Thus,$L = \frac{150}{75} = 2 \ m$.
The total mass of the string is $m = \mu \times L = (2 \times 10^{-3} \ kg/m) \times (2 \ m) = 4 \times 10^{-3} \ kg = 4 \ g$.

(D) Given: $T_A = T_B = T$,$r_A = 2r_B$.
The frequency of the $n$-th overtone for a string is given by $f_n = (n+1) \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
First overtone of $A$ $(n=1)$: $f_{A,1} = 2 \times \frac{1}{2L_A} \sqrt{\frac{T}{\pi r_A^2 \rho}} = \frac{1}{L_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
Second overtone of $B$ $(n=2)$: $f_{B,2} = 3 \times \frac{1}{2L_B} \sqrt{\frac{T}{\pi r_B^2 \rho}} = \frac{3}{2L_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Equating the two frequencies: $\frac{1}{L_A r_A} = \frac{3}{2L_B r_B}$.
Substituting $r_A = 2r_B$: $\frac{1}{L_A (2r_B)} = \frac{3}{2L_B r_B}$.
Simplifying: $\frac{1}{2L_A} = \frac{3}{2L_B} \implies \frac{L_B}{L_A} = \frac{3}{1}$.

(D) The number of half-lives $n$ in time $t$ is given by $n = \frac{t}{T_{1/2}}$.
Given $t = 3 \text{ hours} = 180 \text{ minutes}$ and $T_{1/2} = 60 \text{ minutes}$.
So,$n = \frac{180}{60} = 3$.
The fraction of substance remaining is $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
The fraction of substance decayed is $1 - \frac{N}{N_0} = 1 - \frac{1}{8} = \frac{7}{8}$.
Converting to percentage: $\frac{7}{8} \times 100 \% = 87.5 \%$.

(B) $1$. In the first step,${ }_z X^A \rightarrow{ }_{z+1} Y^A$,the atomic number increases by $1$ while the mass number remains the same. This corresponds to the emission of a $\beta^-$ particle $(-1\beta^0)$.
$2$. In the second step,${ }_{z+1} Y^A \rightarrow{ }_{z-1} K^{A-4}$,the atomic number decreases by $2$ and the mass number decreases by $4$. This corresponds to the emission of an $\alpha$ particle $(2\text{He}^4)$.
$3$. In the third step,${ }_{z-1} K^{A-4} \rightarrow{ }_{z-1} K^{A-4}$,there is no change in atomic number or mass number,which corresponds to the emission of a $\gamma$ ray $(0\gamma^0)$.
$4$. Therefore,the sequence of emissions is $\beta, \alpha, \gamma$.

(D) The number of nuclei remaining after time $t$ is given by $N = N_0 e^{-\lambda t}$.
For substance $A$,$N_A = N_0 e^{-5 \lambda t} \dots (i)$.
For substance $B$,$N_B = N_0 e^{-\lambda t} \dots (ii)$.
Dividing equation $(i)$ by equation $(ii)$,we get:
$\frac{N_A}{N_B} = \frac{N_0 e^{-5 \lambda t}}{N_0 e^{-\lambda t}} = e^{-5 \lambda t + \lambda t} = e^{-4 \lambda t}$.
Given that the ratio $\frac{N_A}{N_B} = (1/e)^2 = e^{-2}$.
Equating the exponents: $-4 \lambda t = -2$.
Therefore,$t = \frac{2}{4 \lambda} = \frac{1}{2 \lambda}$.

(B) The equivalent power $P_{eq}$ of two thin lenses with powers $P_1$ and $P_2$ separated by a distance $d$ is given by the formula: $P_{eq} = P_1 + P_2 - d P_1 P_2$.
Given $P_1 + P_2 = 9 \text{ D}$ and $d = 20 \text{ cm} = 0.2 \text{ m}$.
Substituting the given values into the formula:
$9 - (0.2) P_1 P_2 = \frac{27}{5} = 5.4$.
Rearranging the terms to solve for $P_1 P_2$:
$0.2 P_1 P_2 = 9 - 5.4 = 3.6$.
$P_1 P_2 = \frac{3.6}{0.2} = 18$.
We have a system of two equations: $P_1 + P_2 = 9$ and $P_1 P_2 = 18$.
These are the roots of the quadratic equation $x^2 - 9x + 18 = 0$.
Factoring the quadratic: $(x - 3)(x - 6) = 0$.
Thus,the powers of the two lenses are $3 \text{ D}$ and $6 \text{ D}$.

(B) The critical angle $i_c$ is given by $i_c = \sin^{-1}(1/n)$.
According to Cauchy's equation,the refractive index $n$ is inversely proportional to the wavelength $\lambda$ $(n \propto 1/\lambda)$.
Violet,indigo,and blue light have smaller wavelengths than green light,so the refractive index $n$ for these colors is higher than that for green light.
Since $i_c = \sin^{-1}(1/n)$,a higher refractive index $n$ results in a smaller critical angle $i_c$.
Therefore,the critical angle for violet,indigo,and blue light is smaller than the angle of incidence (which is equal to the critical angle for green light).
Consequently,these colors will undergo total internal reflection.
Conversely,red,orange,and yellow light have longer wavelengths than green light,resulting in a lower refractive index and a larger critical angle,so they will emerge out into the air.

(C) For total internal reflection to occur,two conditions must be satisfied:
$1$. The light ray must travel from a denser medium to a rarer medium.
$2$. The angle of incidence $(i)$ must be greater than the critical angle $(i_c)$.

(D) The critical angle $i_{c}$ is given by the formula $\sin i_{c} = \frac{1}{n}$,where $n$ is the refractive index of the medium.
From this relation,it is clear that $i_{c}$ is minimum when the refractive index $n$ is maximum.
According to Cauchy's dispersion formula,the refractive index $n$ is inversely proportional to the wavelength $\lambda$ $(n \propto \frac{1}{\lambda^2})$.
Since blue light has the shortest wavelength among the given options,it has the highest refractive index in glass.
Therefore,the critical angle $i_{c}$ is minimum for blue light.

(A) The critical angle $c$ for total internal reflection $(TIR)$ when light travels from a medium of refractive index $n_1$ to a medium of lower refractive index $n_2$ is given by $\sin c = \frac{n_2}{n_1}$,where $n_1 > n_2$.
The smaller the ratio $\frac{n_2}{n_1}$,the smaller $\sin c$,and hence the smaller the critical angle $c$.
From the options,the valid scenarios for $TIR$ are only those in which the first medium has a higher index than the second.
For glass to air $(n_{\text{glass}} \approx 1.5, n_{\text{air}} \approx 1.0)$: $\sin c = \frac{1.0}{1.5} = \frac{2}{3} \approx 0.667$,which gives $c \approx 41.8^{\circ}$.
For glass to water $(n_{\text{glass}} \approx 1.5, n_{\text{water}} \approx 1.33)$: $\sin c = \frac{1.33}{1.5} \approx 0.887$,which gives $c \approx 62.5^{\circ}$.
Clearly,the critical angle is smaller for glass-to-air.
Hence,the critical angle is minimum when light travels from glass to air.

(D) The power of a combination of lenses in contact is the algebraic sum of the individual powers of the lenses.
$P = P_1 + P_2$
Given $P_1 = +2.50 \ D$ and $P_2 = -3.75 \ D$.
$P = 2.50 + (-3.75) = -1.25 \ D$.
The focal length $f$ is given by the reciprocal of the power $P$ in meters:
$f = \frac{1}{P} = \frac{1}{-1.25} \ m$.
$f = -0.8 \ m$.
Since $1 \ m = 100 \ cm$,we have:
$f = -0.8 \times 100 \ cm = -80 \ cm$.

(D) Given: Focal length $= f$,Magnification $m = -2$ (since the image is real and inverted).
Using the magnification formula $m = \frac{v}{u}$,we get $v = -2u$.
Applying the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{-2u} - \frac{1}{u} = \frac{1}{f}$
$\frac{-1 - 2}{2u} = \frac{1}{f}$
$\frac{-3}{2u} = \frac{1}{f}$
$2u = -3f$
$u = -1.5f$.
The magnitude of the object distance is $1.5f$.

(D) Given that the magnification $m = -n$ (since the image is real).
By definition of magnification,$m = \frac{v}{u} = -n$,which implies $v = -nu$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-nu} - \frac{1}{u}$.
$\frac{1}{f} = \frac{-1 - n}{nu} = -\frac{n+1}{nu}$.
Since $f$ is positive for a convex lens,we consider the magnitude of the object distance $u$ (where $u$ is negative,so let $u = -x$):
$\frac{1}{f} = \frac{n+1}{nx} \implies x = f(\frac{n+1}{n}) = f(1 + \frac{1}{n})$.
Thus,the distance of the object from the lens is $f(1 + \frac{1}{n})$.

(D) Using the Lens Maker's formula, the power $P$ of a lens in air is given by:
$P = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 2.5 \ D$
When the lens is placed in a liquid of refractive index $\mu_l$, the effective refractive index of the lens relative to the liquid is $\mu' = \frac{\mu}{\mu_l}$.
The new power $P'$ is given by:
$P' = (\mu' - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Taking the ratio of the two equations:
$\frac{P'}{P} = \frac{\mu' - 1}{\mu - 1}$
Given $\mu = 1.5$ and $\mu_l = 2$, we have $\mu' = \frac{1.5}{2} = 0.75$.
Substituting the values:
$\frac{P'}{2.5} = \frac{0.75 - 1}{1.5 - 1} = \frac{-0.25}{0.5} = -0.5$
$P' = -0.5 \times 2.5 = -1.25 \ D$

(D) Let $PO = QO = x$.
According to the sign convention,the object distance $u = -x$ and the image distance $v = +x$ (since the image is formed on the other side of the refracting surface).
The formula for refraction at a spherical surface is:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Here,$n_1 = 1$ (air) and $n_2 = 1.4$ (glass).
Substituting the values:
$\frac{1.4}{x} - \frac{1}{-x} = \frac{1.4 - 1}{R}$
$\frac{1.4}{x} + \frac{1}{x} = \frac{0.4}{R}$
$\frac{2.4}{x} = \frac{0.4}{R}$
$x = \frac{2.4}{0.4} R$
$x = 6R$
Therefore,the distance $PO = 6R$.

(B) Power $(P) = \frac{1}{f} \dots (i)$
Given,the ratio of magnitudes of power is $\frac{|P_{\text{concave}}|}{|P_{\text{convex}}|} = \frac{2}{3} \dots (ii)$
Let the focal length of the convex lens be $f_{\text{convex}} = f$ (positive) and the focal length of the concave lens be $f_{\text{concave}} = -f'$ (negative).
Since $P = \frac{1}{f}$,the ratio of powers is $\frac{1/f'}{1/f} = \frac{f}{f'} = \frac{2}{3}$,which implies $f' = \frac{3}{2}f$.
Thus,$f_{\text{convex}} = f$ and $f_{\text{concave}} = -\frac{3}{2}f$.
Using the formula for equivalent focal length of two lenses in contact: $\frac{1}{f_{eq}} = \frac{1}{f_{\text{convex}}} + \frac{1}{f_{\text{concave}}}$
$\frac{1}{30} = \frac{1}{f} - \frac{1}{\frac{3}{2}f} = \frac{1}{f} - \frac{2}{3f} = \frac{3-2}{3f} = \frac{1}{3f}$
$\frac{1}{30} = \frac{1}{3f} \Rightarrow 3f = 30 \Rightarrow f = 10 \ cm$.
Therefore,$f_{\text{convex}} = 10 \ cm$ and $f_{\text{concave}} = -\frac{3}{2}(10) = -15 \ cm$.

(A) The power of a lens is given by $P = \frac{1}{f}$ (where $f$ is in meters).
For a convex lens,$f_1 = +40 \ cm = +0.4 \ m$.
For a concave lens,$f_2 = -25 \ cm = -0.25 \ m$.
The power of the combination of thin lenses in contact is $P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2}$.
Substituting the values: $P = \frac{1}{0.4} + \frac{1}{-0.25}$.
$P = 2.5 - 4.0 = -1.5 \ D$.

(B) The power $P$ of a lens is given by $P = \frac{1}{f(m)}$,where $f$ is the focal length in meters.
For a convex lens,$f_1 = +40 \ cm = +0.4 \ m$. Thus,$P_1 = \frac{1}{0.4} = +2.5 \ D$.
For a concave lens,$f_2 = -25 \ cm = -0.25 \ m$. Thus,$P_2 = \frac{1}{-0.25} = -4.0 \ D$.
The power of the combination of lenses in contact is $P = P_1 + P_2$.
$P = 2.5 \ D + (-4.0 \ D) = -1.5 \ D$.

(C) For a concave mirror, the magnification $m$ for a real image is negative, so $m = -n$.
By definition of magnification, $m = -\frac{v}{u}$, where $v$ is the image distance and $u$ is the object distance.
Thus, $-n = -\frac{v}{u} \implies v = nu$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Since the mirror is concave, the focal length $f$ is negative, i.e., $-f$. The object distance $u$ is also negative, i.e., $-u$.
Substituting these values: $\frac{1}{-f} = \frac{1}{-nu} + \frac{1}{-u}$.
Multiplying by $-1$: $\frac{1}{f} = \frac{1}{nu} + \frac{1}{u} = \frac{1+n}{nu}$.
Rearranging for $u$: $u = \left(\frac{n+1}{n}\right) f$.

(B) The resolving power $(R.P.)$ of a telescope is given by the formula $R.P. = \frac{D}{1.22\lambda}$,where $D$ is the diameter of the objective lens (aperture) and $\lambda$ is the wavelength of light.
Since the wavelength $(\lambda)$ is fixed for a given observation,the resolving power is directly proportional to the aperture $(D)$ of the objective lens $(R.P. \propto D)$.
Therefore,to achieve greater resolution (the ability to distinguish between two closely spaced objects),telescopes are designed with large aperture objectives.

(D) The limit of resolution $(d)$ of a microscope is directly proportional to the wavelength $(\lambda)$ of the light used,given by the relation $d \propto \lambda$.
Therefore,we can write the ratio as: $\frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2}$.
Given: $d_1 = 0.1 \ mm$,$\lambda_1 = 6000 \ Å$,and $\lambda_2 = 4800 \ Å$.
Substituting the values: $\frac{0.1}{d_2} = \frac{6000}{4800}$.
Simplifying the fraction: $\frac{6000}{4800} = \frac{60}{48} = \frac{5}{4} = 1.25$.
Thus,$d_2 = \frac{0.1}{1.25} = 0.08 \ mm$.

(A) For an astronomical telescope in normal adjustment,the length of the telescope $L$ is given by the sum of the focal length of the objective $(f_0)$ and the focal length of the eyepiece $(f_e)$:
$L = f_0 + f_e$
Given:
$f_0 = 1.5 \ m$
$L = 1.56 \ m$
Substituting the values into the formula:
$1.56 = 1.5 + f_e$
$f_e = 1.56 - 1.5$
$f_e = 0.06 \ m$
Therefore,the focal length of the eyepiece is $0.06 \ m$.

(B) The resolving power $(P)$ of a compound microscope is given by the formula: $P = \frac{2 \mu \sin \theta}{1.22 \lambda}$.
Here,$\mu$ is the refractive index of the medium between the object and the objective lens,$\theta$ is the half-angle of the cone of light from the object,and $\lambda$ is the wavelength of light used.
To improve the resolving power,one must increase the numerator or decrease the denominator.
Therefore,increasing the refractive index $(\mu)$ of the medium between the object and the objective lens increases the resolving power.

(A) The resolving power $(R.P.)$ of a telescope is given by the formula $R.P. = \frac{D}{1.22 \lambda}$,where $D$ is the diameter of the objective lens (aperture) and $\lambda$ is the wavelength of light used.
Since the wavelength of light from celestial objects is fixed,the only way to increase the resolving power is by increasing the diameter $(D)$ of the objective lens.
$A$ larger aperture also allows more light to enter the telescope,making faint objects more visible.
Therefore,an astronomical telescope has a large aperture to have high resolution.

(A) The magnifying power $(M)$ of a simple microscope is given by the formula: $M = 1 + \frac{D}{f}$, where $D$ is the least distance of distinct vision and $f$ is the focal length of the convex lens.
According to Cauchy's dispersion formula, the refractive index $(\mu)$ of a material is higher for shorter wavelengths (blue light) and lower for longer wavelengths (red light).
Since the focal length $f$ of a lens is related to the refractive index by the lens maker's formula, $f$ increases as the wavelength of light increases.
Therefore, the focal length for red light $(f_{red})$ is greater than the focal length for blue light $(f_{blue})$.
As $M = 1 + \frac{D}{f}$, an increase in focal length $f$ results in a decrease in the magnifying power $M$.
Thus, when changing from blue light to red light, the magnifying power decreases.

(C) In a compound microscope,the objective lens is placed very close to the object to form a real,inverted,and magnified image.
To achieve high magnification and high resolving power,the objective lens must have a small focal length $(f_o)$.
Additionally,to collect as much light as possible from the small object and to improve resolution,the objective lens is designed with a large aperture.
Therefore,the objective lens of a compound microscope has a small focal length and a large aperture.

(A) Given: Prism angle $A = 30^{\circ}$,angle of incidence $i_1 = 60^{\circ}$,and angle of deviation $\delta = 30^{\circ}$.
For a prism,the deviation formula is $\delta = (i_1 + i_2) - A$.
Substituting the values: $30^{\circ} = (60^{\circ} + i_2) - 30^{\circ}$.
$30^{\circ} = 30^{\circ} + i_2$,which gives $i_2 = 0^{\circ}$.
Since the emergent angle $i_2 = 0^{\circ}$,the emergent ray is normal to the second face,implying the angle of refraction at the second face $r_2 = 0^{\circ}$.
Using the relation $r_1 + r_2 = A$,we get $r_1 + 0^{\circ} = 30^{\circ}$,so $r_1 = 30^{\circ}$.
Applying Snell's law at the first face: $\mu_1 \sin i_1 = \mu_2 \sin r_1$.
Assuming the surrounding medium is air $(\mu_1 = 1)$: $1 \cdot \sin 60^{\circ} = \mu_2 \sin 30^{\circ}$.
$\frac{\sqrt{3}}{2} = \mu_2 \cdot \frac{1}{2}$.
Therefore,$\mu_2 = \sqrt{3} \approx 1.732$.

(B) For a prism,the condition for minimum deviation is given by the formula $i = \frac{A + \delta_m}{2}$.
In an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the angle of incidence $i = 50^{\circ}$.
Substituting these values into the formula:
$50^{\circ} = \frac{60^{\circ} + \delta_m}{2}$
$100^{\circ} = 60^{\circ} + \delta_m$
$\delta_m = 100^{\circ} - 60^{\circ} = 40^{\circ}$.
Thus,the angle of minimum deviation is $40^{\circ}$.

(A) The angle of deviation $\delta$ for refraction is given by $\delta = i - r$,where $i$ is the angle of incidence and $r$ is the angle of refraction.
From Snell's Law,$\mu = \frac{\sin i}{\sin r}$,which implies $\sin r = \frac{\sin i}{\mu}$.
Since $\delta = i - r$,we have $r = i - \delta$.
Substituting $r$ in the expression for $\mu$:
$\mu = \frac{\sin i}{\sin(i - \delta)}$
Taking the reciprocal:
$\frac{1}{\mu} = \frac{\sin(i - \delta)}{\sin i}$
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$\frac{1}{\mu} = \frac{\sin i \cos \delta - \cos i \sin \delta}{\sin i}$
$\frac{1}{\mu} = \frac{\sin i \cos \delta}{\sin i} - \frac{\cos i \sin \delta}{\sin i}$
$\frac{1}{\mu} = \cos \delta - \frac{\sin \delta}{\tan i}$

(A) The general prism formula is given by $\mu = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$.
For a symmetric (equilateral) prism,the prism angle $A = 60^{\circ}$.
Substituting $A = 60^{\circ}$ into the formula:
$\mu = \frac{\sin \left(\frac{60^{\circ}+\delta_m}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)}$
$\mu = \frac{\sin \left(30^{\circ}+\frac{\delta_m}{2}\right)}{\sin 30^{\circ}}$
Since $\sin 30^{\circ} = \frac{1}{2}$,we have:
$\mu = \frac{\sin \left(30^{\circ}+\frac{\delta_m}{2}\right)}{1/2}$
$\mu = 2 \sin \left(30^{\circ}+\frac{\delta_m}{2}\right)$.

(D) For a thin prism,the angle of minimum deviation is given by $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the prism material relative to the surrounding medium and $A$ is the prism angle.
In air,the refractive index of glass relative to air is $_{a}\mu_{g} = \frac{3}{2}$. Thus,$\delta_1 = (_{a}\mu_{g} - 1)A = (\frac{3}{2} - 1)A = \frac{1}{2}A$.
When immersed in water,the refractive index of glass relative to water is $_{w}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{w}} = \frac{3/2}{4/3} = \frac{9}{8}$.
The new angle of minimum deviation is $\delta_2 = (_{w}\mu_{g} - 1)A = (\frac{9}{8} - 1)A = \frac{1}{8}A$.
Comparing $\delta_2$ with $\delta_1$: $\frac{\delta_2}{\delta_1} = \frac{\frac{1}{8}A}{\frac{1}{2}A} = \frac{2}{8} = \frac{1}{4}$.
Therefore,$\delta_2 = \frac{\delta_1}{4}$.

(C) The dispersive power $\omega$ is given by the formula $\omega = \frac{\delta_v - \delta_R}{\delta_y}$,where $\delta_y = \frac{\delta_v + \delta_R}{2}$ is the mean deviation.
For prism $A$:
$\delta_y = \frac{12^{\circ} + 10^{\circ}}{2} = 11^{\circ}$.
$\omega_A = \frac{12^{\circ} - 10^{\circ}}{11^{\circ}} = \frac{2}{11}$.
For prism $B$:
$\delta_y = \frac{10^{\circ} + 8^{\circ}}{2} = 9^{\circ}$.
$\omega_B = \frac{10^{\circ} - 8^{\circ}}{9^{\circ}} = \frac{2}{9}$.
The ratio of their dispersive powers is $\frac{\omega_A}{\omega_B} = \frac{2/11}{2/9} = \frac{9}{11}$.

(B) The magnification $(m)$ of a plane mirror is $+1$.
This is because a plane mirror creates an image that is the same size as the object $(h_i = h_o)$.
The image formed by a plane mirror is virtual and erect,which means the magnification must be positive.
Since the height of the image is equal to the height of the object,the ratio is $1$.
Therefore,the magnification is calculated as:
$m = \frac{h_i}{h_o} = 1$.

(B) Let $s_1$ be the distance of the source from the surface and $c$ be the speed of light in air. The time taken to reach the surface is $T_1 = \frac{s_1}{c}$.
Let $s_2$ be the thickness of the glass slab $(5 \ cm)$ and $v$ be the speed of light in glass. The time taken to travel through the slab is $T_2 = \frac{s_2}{v}$.
Given that $T_1 = T_2$,we have $\frac{s_1}{c} = \frac{s_2}{v}$.
Since the refractive index $\mu = \frac{c}{v}$,we can write $v = \frac{c}{\mu}$.
Substituting this into the equation: $\frac{s_1}{c} = \frac{s_2}{c/\mu} = \frac{s_2 \times \mu}{c}$.
Therefore,$s_1 = s_2 \times \mu$.
Given $s_2 = 5 \ cm$ and $\mu = 1.6$,we get $s_1 = 5 \times 1.6 = 8 \ cm$.

(C) Let the height of the water filled in the container be $x$.
The apparent depth of the bottom of the container as seen from the top is given by $d' = \frac{x}{n}$, where $n$ is the refractive index of water.
Given $n = \frac{4}{3}$, the apparent depth is $d' = \frac{x}{4/3} = \frac{3x}{4}$.
The container appears half-filled, which means the apparent depth of the bottom from the top surface of the water is equal to the height of the empty part of the container.
The height of the empty part is $30 - x$.
Therefore, we set $d' = 30 - x$.
$\frac{3x}{4} = 30 - x$
$3x = 120 - 4x$
$7x = 120$
$x = \frac{120}{7} \approx 17.14 \, cm$.
Wait, re-evaluating the condition "appears half-filled": If the total height is $30 \, cm$, half-filled means the apparent position of the bottom is at $15 \, cm$ from the top.
So, $(30 - x) + d' = 15$ is not correct. The condition "appears half-filled" means the apparent depth of the bottom is $15 \, cm$.
$d' = 15 \, cm$.
Since $d' = \frac{x}{n}$, we have $15 = \frac{x}{4/3}$.
$x = 15 \times \frac{4}{3} = 20 \, cm$.
Thus, the height of the water should be $20 \, cm$.

(A) Given that the velocity of light in diamond is $v_d = \frac{5}{12}c$ and in water is $v_w = \frac{3}{4}c$,where $c$ is the speed of light in air.
Refractive index of diamond with respect to water is given by ${}_w n_d = \frac{n_d}{n_w} = \frac{c/v_d}{c/v_w} = \frac{v_w}{v_d}$.
Substituting the values: ${}_w n_d = \frac{3/4 c}{5/12 c} = \frac{3}{4} \times \frac{12}{5} = \frac{9}{5}$.
Using Snell's Law: ${}_w n_d = \frac{\sin i}{\sin r}$.
Therefore,$\sin i = {}_w n_d \times \sin r = \frac{9}{5} \times \sin 30^{\circ}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,we get $\sin i = \frac{9}{5} \times \frac{1}{2} = \frac{9}{10}$.
Thus,$i = \sin^{-1}\left(\frac{9}{10}\right)$.

(A) The formula for apparent depth of a single layer is given by $\text{Apparent depth} = \frac{\text{Real depth} (R)}{\text{Refractive index} (\mu)}$.
For a combination of multiple immiscible liquids,the total apparent depth is the sum of the apparent depths of individual layers:
$\text{Total Apparent depth} = \frac{R_1}{\mu_1} + \frac{R_2}{\mu_2} + \frac{R_3}{\mu_3}$
Given:
$R_1 = 3 \ cm, \mu_1 = \frac{3}{2}$
$R_2 = 4 \ cm, \mu_2 = \frac{4}{3}$
$R_3 = 6 \ cm, \mu_3 = \frac{6}{5}$
Substituting these values:
$\text{Apparent depth} = \frac{3}{3/2} + \frac{4}{4/3} + \frac{6}{6/5}$
$= (3 \times \frac{2}{3}) + (4 \times \frac{3}{4}) + (6 \times \frac{5}{6})$
$= 2 + 3 + 5$
$= 10 \ cm$.

(C) Given: The number of waves in glass $(N_g)$ is equal to the number of waves in water $(N_w)$.
$N_g = N_w$
Since the number of waves $N = \frac{d}{\lambda}$,where $d$ is the thickness of the medium and $\lambda$ is the wavelength in that medium:
$\frac{d_g}{\lambda_g} = \frac{d_w}{\lambda_w}$
We know that $\lambda = \frac{\lambda_{\text{air}}}{\mu}$,where $\mu$ is the refractive index.
Substituting this into the equation:
$\frac{d_g \cdot \mu_g}{\lambda_{\text{air}}} = \frac{d_w \cdot \mu_w}{\lambda_{\text{air}}}$
$\mu_g \cdot d_g = \mu_w \cdot d_w$
Given $d_g = 5 \ cm$,$d_w = 6 \ cm$,and $\mu_g = 1.56$:
$1.56 \times 5 = \mu_w \times 6$
$\mu_w = \frac{1.56 \times 5}{6} = \frac{7.8}{6} = 1.30$
Thus,the refractive index of water is $1.30$.

(B) The normal shift $x$ produced by a glass slab of thickness $t$ and refractive index $\mu$ is given by the formula:
$x = t \left(1 - \frac{1}{\mu}\right)$
Rearranging the formula to solve for $t$:
$x = t \left(\frac{\mu - 1}{\mu}\right)$
$t = \frac{x \cdot \mu}{\mu - 1}$
Therefore,the thickness of the glass slab is $t = \frac{\mu x}{\mu - 1}$.

(C) The apparent depth of a vessel containing multiple layers of non-mixing liquids is given by the sum of the apparent depths of each individual layer.
The formula for apparent depth is $d_{app} = \sum \frac{d_i}{n_i}$,where $d_i$ is the real depth and $n_i$ is the refractive index of the $i$-th layer.
Given:
Layer $1$: $d_1 = 40 \ cm$,$n_1 = 1.6$
Layer $2$: $d_2 = 30 \ cm$,$n_2 = 1.5$
Calculating the apparent depth:
$d_{app} = \frac{40}{1.6} + \frac{30}{1.5}$
$d_{app} = 25 \ cm + 20 \ cm$
$d_{app} = 45 \ cm$
Therefore,the apparent depth of the vessel is $45 \ cm$.

(A) The formula for the apparent shift $(h)$ due to a glass slab is given by:
$h = t \left(1 - \frac{1}{n}\right)$
where $t$ is the real thickness of the slab and $n$ is the refractive index.
Given:
$t = 4.8 \text{ cm}$
$n = 1.5$
Substituting the values:
$h = 4.8 \left(1 - \frac{1}{1.5}\right)$
$h = 4.8 \left(1 - \frac{2}{3}\right)$
$h = 4.8 \left(\frac{1}{3}\right)$
$h = 1.6 \text{ cm}$
Therefore, the ink dot appears to be raised by $1.6 \text{ cm}$.

(B) According to Snell's law,the refractive index $\mu$ is given by $\mu = \frac{\sin i}{\sin r}$.
Given that the angle of incidence $i$ is twice the angle of refraction $r$,so $i = 2r$ or $r = \frac{i}{2}$.
Substituting this into Snell's law:
$\mu = \frac{\sin i}{\sin(i/2)}$.
Using the trigonometric identity $\sin i = 2 \sin(i/2) \cos(i/2)$,we get:
$\mu = \frac{2 \sin(i/2) \cos(i/2)}{\sin(i/2)}$.
$\mu = 2 \cos(i/2)$.
Rearranging for $i$:
$\frac{\mu}{2} = \cos(i/2)$.
$i/2 = \cos^{-1}\left(\frac{\mu}{2}\right)$.
Therefore,$i = 2 \cos^{-1}\left(\frac{\mu}{2}\right)$.

(A) The optical path length is defined as the product of the refractive index $(\mu)$ and the geometric distance $(d)$ travelled by the light in that medium.
Given that the optical paths are equal:
$\mu_{g} x_{g} = \mu_{m} x_{m}$
Where $\mu_{g} = 1.6$ is the refractive index of flint glass, $x_{g} = 3 \ cm$ is the distance in flint glass, $\mu_{m} = 1.25$ is the refractive index of the other medium, and $x_{m} = x$ is the distance in the other medium.
Substituting the values:
$1.6 \times 3 = 1.25 \times x$
$4.8 = 1.25 \times x$
$x = \frac{4.8}{1.25} = 3.84 \ cm$
Therefore, the value of '$x$' is $3.84 \ cm$.

(B) Let the total depth of the liquid in the tank be $H$. The object $P$ is at a height $h$ above the mirror,so its depth from the liquid surface is $(H-h)$.
The apparent depth of the object $P$ as seen by the observer $O$ is $d_1 = \frac{H-h}{\mu}$.
The image of the object $P$ formed by the plane mirror is at a distance $h$ below the mirror. The total depth of this image from the liquid surface is $(H+h)$.
The apparent depth of the image as seen by the observer $O$ is $d_2 = \frac{H+h}{\mu}$.
The apparent distance between the object $P$ and its image is the difference between their apparent depths:
$\text{Apparent distance} = d_2 - d_1 = \frac{H+h}{\mu} - \frac{H-h}{\mu} = \frac{H+h-H+h}{\mu} = \frac{2h}{\mu}$.

(B) For small angles,Snell's law is given by $n = \frac{\sin i}{\sin r} \approx \frac{i}{r}$.
Given that the velocity is reduced by $25 \%$,the new velocity $v = c - 0.25c = 0.75c = \frac{3}{4}c$.
The refractive index $n$ is defined as $n = \frac{c}{v} = \frac{c}{0.75c} = \frac{1}{0.75} = \frac{4}{3}$.
Equating the two expressions for $n$: $\frac{i}{r} = \frac{4}{3} \implies r = \frac{3}{4}i$.
The angle of deviation $\delta$ is given by $\delta = i - r$.
Substituting the value of $r$: $\delta = i - \frac{3}{4}i = \frac{i}{4}$.

(B) The Zener diode is connected in parallel with resistor $R_2$. Therefore,the voltage across $R_2$ is equal to the Zener breakdown voltage,$V_Z = 8 \text{ V}$.
The current flowing through resistor $R_2$ is:
$I_{R_2} = \frac{V_Z}{R_2} = \frac{8 \text{ V}}{1600 \text{ }\Omega} = 5 \times 10^{-3} \text{ A} = 5 \text{ mA} \quad \dots(i)$
The total voltage supplied is $20 \text{ V}$. The voltage drop across resistor $R_1$ is:
$V_{R_1} = V_{\text{source}} - V_Z = 20 \text{ V} - 8 \text{ V} = 12 \text{ V}$
The total current flowing through the circuit (which passes through $R_1$) is:
$I_{R_1} = \frac{V_{R_1}}{R_1} = \frac{12 \text{ V}}{400 \text{ }\Omega} = 3 \times 10^{-2} \text{ A} = 30 \text{ mA} \quad \dots(ii)$
Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $I_Z$ is:
$I_Z = I_{R_1} - I_{R_2}$
$I_Z = 30 \text{ mA} - 5 \text{ mA} = 25 \text{ mA}$

(D) When a $p-n$ junction is formed,electrons from the $n$-region diffuse into the $p$-region,and holes from the $p$-region diffuse into the $n$-region.
This diffusion leaves behind immobile ionized donor atoms (positive charges) in the $n$-side and immobile ionized acceptor atoms (negative charges) in the $p$-side near the junction.
This region,which is devoid of mobile charge carriers,is called the depletion region.
The accumulation of these fixed positive and negative charges creates an electric field that opposes further diffusion,resulting in a potential barrier.

(A) In the given circuit,the $p$-terminal of the diode is connected to the negative terminal of the $4 \ V$ battery,and the $n$-terminal is connected towards the $1 \ V$ battery.
Specifically,the potential at the $p$-side is $-4 \ V$ and the potential at the $n$-side is $-1 \ V$.
Since the potential at the $p$-side is lower than the potential at the $n$-side $(-4 \ V < -1 \ V)$,the diode is in a reverse-biased condition.
An ideal diode in reverse bias acts as an open circuit,meaning it offers infinite resistance.
Therefore,no current will flow through the circuit.
The magnitude of the current is $0 \ A$.

(B) $p-n$ junction diode allows current to flow easily when it is forward biased.
When the terminals are reversed,the diode becomes reverse biased.
In a reverse biased state,the depletion region widens,offering very high resistance to the flow of charge carriers.
Consequently,the current drops to almost zero.
Therefore,the device $X$ is a $p-n$ junction diode.

(B) The Zener diode is in parallel with the load resistor of $1 \text{ k}\Omega$. Since the Zener breakdown voltage is $15 \text{ V}$, the voltage across the load resistor is $V_L = 15 \text{ V}$.
The current through the load resistor is $I_L = \frac{V_L}{R_L} = \frac{15 \text{ V}}{1000 \Omega} = 15 \times 10^{-3} \text{ A} = 15 \text{ mA}$.
The voltage drop across the series resistor $R_s = 250 \Omega$ is $V_s = V_{in} - V_L = 20 \text{ V} - 15 \text{ V} = 5 \text{ V}$.
The total current supplied by the source is $I = \frac{V_s}{R_s} = \frac{5 \text{ V}}{250 \Omega} = 0.02 \text{ A} = 20 \text{ mA}$.
Applying Kirchhoff's Current Law at the junction, the current through the Zener diode $I_z$ is $I_z = I - I_L = 20 \text{ mA} - 15 \text{ mA} = 5 \text{ mA}$.

(C) The correct option is $C$.
When a $p-n$ junction is formed,there is a high concentration of electrons in the $n$-region and a high concentration of holes in the $p$-region.
Due to this concentration gradient,electrons diffuse from the $n$-side to the $p$-side,and holes diffuse from the $p$-side to the $n$-side.
As these charge carriers cross the junction,they recombine near the junction interface.
This recombination leaves behind immobile ionized impurity atoms (positive ions on the $n$-side and negative ions on the $p$-side),which create an electric field that opposes further diffusion.
This region,depleted of mobile charge carriers,is known as the depletion layer.

(B) The zener diode is in parallel with the load resistor $R_L = 1200 \Omega$. Since the zener breakdown voltage is $15 \text{ V}$,the voltage across the load resistor is $V_L = 15 \text{ V}$.
The current through the load resistor is $I_L = \frac{V_L}{R_L} = \frac{15 \text{ V}}{1200 \Omega} = 0.0125 \text{ A} = 12.5 \text{ mA}$.
The voltage across the series resistor $R_S = 300 \Omega$ is $V_S = V_{in} - V_L = 24 \text{ V} - 15 \text{ V} = 9 \text{ V}$.
The total current supplied by the source is $I = \frac{V_S}{R_S} = \frac{9 \text{ V}}{300 \Omega} = 0.03 \text{ A} = 30 \text{ mA}$.
Applying Kirchhoff's Current Law at the junction,the current through the zener diode $I_Z$ is given by $I_Z = I - I_L$.
$I_Z = 30 \text{ mA} - 12.5 \text{ mA} = 17.5 \text{ mA}$.