MHT CET 2020 Physics Question Paper with Answer and Solution

(B) The gravitational force is given by $F = \frac{G M m}{r^2}$.
Acceleration due to gravity is $g = \frac{F}{m} = \frac{G M}{r^2}$.
Therefore,the ratio $\frac{G}{g}$ can be expressed as:
$\frac{G}{g} = \frac{G}{\frac{G M}{r^2}} = \frac{r^2}{M}$.
The unit of $r$ is $m$ (meters) and the unit of $M$ is $kg$ (kilograms).
Thus,the $SI$ unit of $\frac{G}{g}$ is $\frac{m^2}{kg}$.

(A) Given the equation $x = \pi R \left( \frac{P^2 - Q^2}{2} \right)$.
Here,$P, Q,$ and $R$ are lengths,so their dimensions are $[L]$.
The term $(P^2 - Q^2)$ has dimensions $[L^2] - [L^2] = [L^2]$.
The constant $\pi$ and the factor $1/2$ are dimensionless.
Therefore,the dimension of $x$ is $[L] \times [L^2] = [L^3]$.
Physical quantities with dimensions of $[L^3]$ represent volume.

(C) The relationship between phase difference $\Delta \phi$ and time interval $\Delta t$ is given by:
$\Delta \phi = \omega \Delta t = (2\pi f) \Delta t$
Given frequency $f = 50 \,Hz$ and time interval $\Delta t = 0.01 \,s$.
Substituting the values:
$\Delta \phi = 2 \times \pi \times 50 \times 0.01$
$\Delta \phi = 100 \pi \times 0.01$
$\Delta \phi = \pi \,rad$.

(C) Given: Mass of string $m = 0.1 \, kg$, Tension $T = 1.6 \, N$, Length $\ell = 1 \, m$.
Mass per unit length $\mu = \frac{m}{\ell} = \frac{0.1}{1} = 0.1 \, kg/m$.
The velocity $v$ of the transverse wave in the string is given by the formula $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values, $v = \sqrt{\frac{1.6}{0.1}} = \sqrt{16} = 4 \, m/s$.
The time taken $t$ to reach the other end is $t = \frac{\ell}{v} = \frac{1}{4} = 0.25 \, s$.

(B) The speed of a transverse wave on a stretched string is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
According to Hooke's Law,$T = kx$,where $k$ is the spring constant and $x$ is the extension.
Since the mass of the string remains constant,$\mu = \frac{m}{L}$. When the string is extended,its length $L$ increases. However,for small extensions,the change in $\mu$ is negligible compared to the change in tension $T$.
Thus,$V \propto \sqrt{T} \propto \sqrt{x}$.
Given the initial extension is $x_1 = x$ and the final extension is $x_2 = x + 0.5x = 1.5x$.
The ratio of the speeds is $\frac{V_2}{V_1} = \sqrt{\frac{x_2}{x_1}} = \sqrt{\frac{1.5x}{x}} = \sqrt{1.5}$.
Calculating the value,$\sqrt{1.5} \approx 1.2247$.
Therefore,the new speed $V_2 \approx 1.22 V$.

(B) Given: Wave speed $v = 960 \,m/s$.
Number of waves $n = 900$.
Time $t = 0.5 \,min = 30 \,s$.
The frequency $f$ is defined as the number of waves passing a point per unit time:
$f = \frac{n}{t} = \frac{900}{30} = 30 \,Hz$.
The relationship between wave speed, frequency, and wavelength is given by $v = f \lambda$.
Therefore, the wavelength $\lambda = \frac{v}{f} = \frac{960}{30} = 32 \,m$.

(B) The speed of a transverse wave in a stretched wire is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{M}{L}$.
Also,mass $M = \text{Volume} \times \text{Density} = (A \times L) \times \rho$,where $A$ is the area of cross-section.
Therefore,$\mu = \frac{A \times L \times \rho}{L} = A\rho$.
Substituting this into the wave speed formula: $V = \sqrt{\frac{T}{A\rho}}$.
Squaring both sides: $V^{2} = \frac{T}{A\rho}$.
Rearranging for the area of cross-section $A$: $A = \frac{T}{V^{2}\rho}$.

(C) The speed of a transverse wave in a stretched wire is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
According to Hooke's law,the tension $T$ in the wire is proportional to the extension $x$,i.e.,$T = kx$.
Initially,$T_1 = kx$,so $V_1 = \sqrt{\frac{kx}{\mu}} = V$.
When the extension is increased to $4x$,the new tension is $T_2 = k(4x) = 4kx$.
The new speed of sound $V_2$ is given by $V_2 = \sqrt{\frac{T_2}{\mu}} = \sqrt{\frac{4kx}{\mu}}$.
Therefore,$V_2 = 2 \sqrt{\frac{kx}{\mu}} = 2V$.

(B) The speed of a transverse wave on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{M}{L}$.
Since mass $M = \text{Volume} \times \text{density} = (A \times L) \times \rho$,where $A$ is the cross-sectional area.
Therefore,$\mu = \frac{A \times L \times \rho}{L} = A \rho$.
Substituting this into the wave speed formula: $v = \sqrt{\frac{T}{A \rho}}$.
Squaring both sides: $v^{2} = \frac{T}{A \rho}$.
Rearranging for the area $A$: $A = \frac{T}{v^{2} \rho}$.

(B) The general equation for a wave is given by $Y = A \sin(\omega t)$.
For the first wave,$Y_1 = 0.25 \sin(316t)$,comparing with the general equation,we get angular frequency $\omega_1 = 316 \text{ rad/s}$.
Since $\omega = 2\pi f$,the frequency $f_1 = \frac{\omega_1}{2\pi} = \frac{316}{2\pi} \text{ Hz}$.
For the second wave,$Y_2 = 0.25 \sin(310t)$,comparing with the general equation,we get angular frequency $\omega_2 = 310 \text{ rad/s}$.
The frequency $f_2 = \frac{\omega_2}{2\pi} = \frac{310}{2\pi} \text{ Hz}$.
The number of beats produced per second is the difference in frequencies: $f_{beat} = |f_1 - f_2|$.
$f_{beat} = \frac{316}{2\pi} - \frac{310}{2\pi} = \frac{6}{2\pi} = \frac{3}{\pi} \text{ Hz}$.

(B) The frequency $n$ of a vibrating wire is given by $n = \frac{1}{2 \ell} \sqrt{\frac{T}{m}}$.
Since the wires are identical,$\ell$ and $m$ are constant,so $n \propto \sqrt{T}$.
Let the initial frequency be $n_1$ and the initial tension be $T_1$. After increasing the tension by $2 \%$,the new tension $T_2 = T_1 + 0.02 T_1 = 1.02 T_1$.
The new frequency $n_2$ is given by $\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{1.02} = 1.01$.
Thus,$n_2 = 1.01 n_1$.
The beat frequency is given as $n_2 - n_1 = 5 \ Hz$.
Substituting $n_2$,we get $1.01 n_1 - n_1 = 5$.
$0.01 n_1 = 5$.
$n_1 = \frac{5}{0.01} = 500 \ Hz$.

(B) Let the frequency of tuning fork $C$ be $f_C$.
Given that the frequency of $A$ is $1.5 \%$ more than $f_C$:
$f_A = f_C + 0.015 f_C = 1.015 f_C$.
Given that the frequency of $B$ is $2.5 \%$ less than $f_C$:
$f_B = f_C - 0.025 f_C = 0.975 f_C$.
When $A$ and $B$ are sounded together, the beat frequency is $12 \text{ Hz}$:
$|f_A - f_B| = 12$.
Substituting the expressions:
$1.015 f_C - 0.975 f_C = 12$.
$0.040 f_C = 12$.
$f_C = \frac{12}{0.040} = 300 \text{ Hz}$.

(B) The frequency of the sound heard by the wall (which acts as a stationary observer) is given by $f' = f \left( \frac{v}{v - v_s} \right)$, where $v = 335 \,m/s$ and $v_s = 5 \,m/s$.
$f' = 165 \left( \frac{335}{335 - 5} \right) = 165 \left( \frac{335}{330} \right) = 165 \times \frac{335}{330} = \frac{335}{2} = 167.5 \,Hz$.
Now, the wall acts as a source reflecting this sound back to the bus (which acts as a moving observer). The frequency $f''$ heard by the passengers is $f'' = f' \left( \frac{v + v_o}{v} \right)$, where $v_o = 5 \,m/s$.
$f'' = 167.5 \left( \frac{335 + 5}{335} \right) = 167.5 \left( \frac{340}{335} \right) = 167.5 \times \frac{340}{335} = 167.5 \times 1.0149 \approx 170 \,Hz$.
The number of beats per second is the difference between the reflected frequency and the original frequency: $f_{beats} = f'' - f = 170 \,Hz - 165 \,Hz = 5 \,Hz$.

(A) Let the frequency of tuning fork $A$ be $f_A$ and the frequency of tuning fork $B$ be $f_B = 480 \ Hz$.
The beat frequency is given by $|f_A - f_B| = 5 \ Hz$.
This implies $f_A = 480 \pm 5$,so $f_A$ is either $485 \ Hz$ or $475 \ Hz$.
When wax is added to tuning fork $A$,its frequency $f_A$ decreases.
After adding wax,the new beat frequency is $|f_A' - 480| = 2 \ Hz$,where $f_A' < f_A$.
If $f_A = 475 \ Hz$,adding wax would decrease the frequency further (e.g.,to $473 \ Hz$),making the beat frequency $|473 - 480| = 7 \ Hz$,which is not $2 \ Hz$.
If $f_A = 485 \ Hz$,adding wax decreases the frequency towards $480 \ Hz$,making the beat frequency $|482 - 480| = 2 \ Hz$.
Thus,the initial frequency of tuning fork $A$ was $485 \ Hz$.

(D) The Doppler effect formula for an observer moving towards a stationary source is given by $n^{\prime} = n \left( \frac{v + v_{o}}{v} \right)$,where $n^{\prime}$ is the observed frequency,$n$ is the source frequency,$v$ is the speed of sound,and $v_{o}$ is the velocity of the observer.
Given that the observed frequency is double the source frequency,we have $n^{\prime} = 2n$.
Substituting this into the formula: $2n = n \left( \frac{v + v_{o}}{v} \right)$.
Dividing both sides by $n$: $2 = \frac{v + v_{o}}{v}$.
Multiplying by $v$: $2v = v + v_{o}$.
Solving for $v_{o}$: $v_{o} = v$.
Therefore,the observer must move towards the source with a velocity equal to the speed of sound.

(B) Let $v$ be the speed of sound and $V$ be the speed of the train/observer. The natural frequency is $n$.
In the first case,the source (train) moves away from the stationary observer:
$n' = n \left( \frac{v}{v + V} \right)$
Given $\frac{n}{n'} = 1.2$,so $\frac{v + V}{v} = 1.2 \implies 1 + \frac{V}{v} = 1.2 \implies \frac{V}{v} = 0.2$.
In the second case,the source is at rest and the observer moves away from it:
$n'' = n \left( \frac{v - V}{v} \right) = n \left( 1 - \frac{V}{v} \right)$
The ratio of the natural frequency to the apparent frequency is $\frac{n}{n''} = \frac{1}{1 - \frac{V}{v}}$.
Substituting $\frac{V}{v} = 0.2$:
$\frac{n}{n''} = \frac{1}{1 - 0.2} = \frac{1}{0.8} = 1.25$.
Thus,the ratio is $1.25: 1$.

(D) Let the original frequency of the source be $F_{o}$.
According to the Doppler effect,when the observer moves towards a stationary source,the apparent frequency $F_{1}$ is given by:
$F_{1} = F_{o} \left[ \frac{V + V_{1}}{V} \right]$ ...$(1)$
When the observer moves away from the stationary source,the apparent frequency $F_{2}$ is given by:
$F_{2} = F_{o} \left[ \frac{V - V_{1}}{V} \right]$ ...$(2)$
Dividing equation $(1)$ by equation $(2)$,we get:
$\frac{F_{1}}{F_{2}} = \frac{V + V_{1}}{V - V_{1}}$
Given that $\frac{F_{1}}{F_{2}} = 2$,we substitute this into the equation:
$2 = \frac{V + V_{1}}{V - V_{1}}$
$2(V - V_{1}) = V + V_{1}$
$2V - 2V_{1} = V + V_{1}$
$V = 3V_{1}$
Therefore,$\frac{V}{V_{1}} = 3$.

(A) When an obstacle moves towards a stationary source with velocity $v$,the frequency of the reflected sound wave $f_{r}$ is given by the Doppler effect formula for a moving reflector: $f_{r} = f \left(\frac{c+v}{c-v}\right)$.
Since the speed of sound $c$ remains constant in the same medium after reflection,we use the relation $c = f \lambda$.
For the reflected wave,$c = f_{r} \lambda_{r}$,which implies $\lambda_{r} = \frac{c}{f_{r}}$.
Substituting the expression for $f_{r}$:
$\lambda_{r} = \frac{c}{f \left(\frac{c+v}{c-v}\right)} = \frac{c}{f} \left(\frac{c-v}{c+v}\right)$.
Since $\frac{c}{f} = \lambda$,we get $\lambda_{r} = \left(\frac{c-v}{c+v}\right) \lambda$.

(D) Given: Velocity of source $V_s = 30 \,m/s$,frequency of source $n_0 = 256 \,Hz$,and speed of sound $V = 330 \,m/s$.
When the source is approaching the stationary observer,the observed frequency $n_1$ is given by $n_1 = n_0 \left( \frac{V}{V - V_s} \right)$.
When the source is moving away after crossing the observer,the observed frequency $n_2$ is given by $n_2 = n_0 \left( \frac{V}{V + V_s} \right)$.
The ratio of the frequencies is $\frac{n_1}{n_2} = \frac{n_0 \left( \frac{V}{V - V_s} \right)}{n_0 \left( \frac{V}{V + V_s} \right)} = \frac{V + V_s}{V - V_s}$.
Substituting the values: $\frac{n_1}{n_2} = \frac{330 + 30}{330 - 30} = \frac{360}{300} = \frac{6}{5}$.

(C) The Doppler effect depends on the relative velocity between the source and the observer along the line joining them.
In this scenario,the engine is moving on a circular path,and the observer is at the centre.
The velocity vector of the engine is always tangent to the circular path.
The line joining the observer (at the centre) and the source (on the circumference) is the radius of the circle.
Since the tangent to a circle is always perpendicular to the radius at the point of contact,the velocity of the source is always perpendicular to the line joining the source and the observer.
Therefore,the component of the source's velocity along the line of sight is $v_s \cos(90^{\circ}) = 0$.
Since there is no relative velocity along the line joining the source and the observer,there is no Doppler shift.
Thus,the frequency heard by the observer remains equal to the source frequency,which is $250 \,Hz$.

(B) Let the amplitude of each wave be $A$. The intensity $I$ is proportional to the square of the amplitude,so $I \propto A^2$. Let $I_0 = kA^2$ be the intensity of each wave.
Case $1$: When the waves arrive in phase,the phase difference $\phi = 0$. The resultant amplitude is $A_R = A + A = 2A$. The resultant intensity is $I_1 = k(2A)^2 = 4kA^2 = 4I_0$.
Case $2$: When the waves arrive $90^{\circ}$ ($\pi/2$ radians) out of phase,the resultant amplitude is $A_R = \sqrt{A^2 + A^2 + 2AA \cos(90^{\circ})} = \sqrt{2A^2} = A\sqrt{2}$. The resultant intensity is $I_2 = k(A\sqrt{2})^2 = 2kA^2 = 2I_0$.
The ratio of the resultant intensities is $\frac{I_1}{I_2} = \frac{4I_0}{2I_0} = \frac{2}{1}$.

(C) The given progressive waves are $Y_{1} = \sin 2\pi(\frac{t}{0.4} - \frac{x}{4})$ and $Y_{2} = \sin 2\pi(\frac{t}{0.4} + \frac{x}{4})$.
Comparing these with the standard wave equation $Y = A \sin 2\pi(\frac{t}{T} \pm \frac{x}{\lambda})$,we get the wavelength $\lambda = 4 \ m$ and time period $T = 0.4 \ s$.
When two waves of equal amplitude and frequency traveling in opposite directions superpose,they form a standing wave given by $Y = Y_{1} + Y_{2} = 2A \cos(\frac{2\pi x}{\lambda}) \sin(\frac{2\pi t}{T})$.
Here,the amplitude of the standing wave at any position $x$ is given by $A_{res} = |2A \cos(\frac{2\pi x}{\lambda})|$.
Given $A = 1$,$\lambda = 4 \ m$,and $x = 0.5 \ m$,we substitute these values:
$A_{res} = 2 \times 1 \times |\cos(\frac{2\pi \times 0.5}{4})|$
$A_{res} = 2 \cos(\frac{\pi}{4})$
Since $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,we get $A_{res} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \ m$.

(A) The frequency of the $m^{\text{th}}$ harmonic in an open organ pipe is given by $f_m = \frac{m v}{2 L}$.
For the fundamental frequency $(m=1)$, $f_1 = \frac{v}{2 L}$.
The frequency of the $n^{\text{th}}$ harmonic in a closed organ pipe is given by $f_n^{\prime} = \frac{n v}{4 L}$, where $n$ must be an odd integer.
The frequency of the $3^{\text{rd}}$ harmonic in a closed organ pipe is $f_3^{\prime} = \frac{3 v}{4 L}$.
According to the problem, $f_3^{\prime} - f_1 = 50 \,Hz$.
Substituting the expressions: $\frac{3 v}{4 L} - \frac{v}{2 L} = 50$.
$\frac{3 v - 2 v}{4 L} = 50 \Rightarrow \frac{v}{4 L} = 50 \Rightarrow \frac{v}{L} = 200$.
The fundamental frequency of the open organ pipe is $f_1 = \frac{v}{2 L} = \frac{200}{2} = 100 \,Hz$.

(B) For a pipe open at both ends, the resonant frequencies are given by the formula: $f_n = \frac{n v}{2L}$, where $n = 1, 2, 3, \dots$ is the harmonic number, $v$ is the speed of sound, and $L$ is the length of the pipe.
Given: $v = 340 \,m/s$ and $L = 1 \,m$.
Substituting these values, we get: $f_n = \frac{n \times 340}{2 \times 1} = n \times 170 \,Hz$.
This means the pipe can resonate at frequencies that are integer multiples of $170 \,Hz$ (i.e., $170 \,Hz, 340 \,Hz, 510 \,Hz, \dots$).
Comparing this with the given options, $85 \,Hz$ is not a multiple of $170 \,Hz$ and therefore the air column cannot resonate at this frequency.

(A) The frequency of the first overtone of a closed organ pipe of length $\ell$ is given by $f = \frac{3V}{4\ell}$.
The frequency of the first overtone of an open organ pipe of length $\ell^{\prime}$ is given by $f^{\prime} = \frac{2V}{2\ell^{\prime}} = \frac{V}{\ell^{\prime}}$.
Given that the frequencies are identical,we set $f = f^{\prime}$:
$\frac{3V}{4\ell} = \frac{V}{\ell^{\prime}}$.
Rearranging the terms to find the ratio of lengths $\frac{\ell}{\ell^{\prime}}$:
$\frac{\ell}{\ell^{\prime}} = \frac{3}{4}$.
Thus,the ratio of their lengths is $3: 4$.

(B) For an open organ pipe,the fundamental frequency is given by $n = \frac{V}{2\ell}$,where $V$ is the speed of sound and $\ell$ is the length of the pipe.
For the first pipe: $\ell_{1} = \frac{V}{2n_{1}}$.
For the second pipe: $\ell_{2} = \frac{V}{2n_{2}}$.
When joined in series,the total length of the new pipe is $\ell = \ell_{1} + \ell_{2}$.
The fundamental frequency of the new pipe is $n = \frac{V}{2\ell}$.
Substituting the values of $\ell_{1}$ and $\ell_{2}$ into the equation for $\ell$:
$\frac{V}{2n} = \frac{V}{2n_{1}} + \frac{V}{2n_{2}}$.
Dividing both sides by $\frac{V}{2}$,we get: $\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}}$.
Solving for $n$: $\frac{1}{n} = \frac{n_{1} + n_{2}}{n_{1}n_{2}}$,which gives $n = \frac{n_{1}n_{2}}{n_{1} + n_{2}}$.

(B) In an organ pipe closed at one end,the standing wave pattern always has a node at the closed end and an antinode at the open end.
For the fundamental mode,there is $1$ node and $1$ antinode.
For the first overtone,there are $2$ nodes and $2$ antinodes.
Since the question specifies $2$ nodes and $2$ antinodes,this corresponds to the first overtone.

(C) In the fundamental mode of a closed pipe,the length of the pipe $L$ corresponds to one-fourth of the wavelength,i.e.,$L = \frac{\lambda}{4}$.
Given that the time taken for the sound wave to travel the length $L$ is $t$,we have $t = \frac{L}{v}$,where $v$ is the speed of sound.
Since $L = \frac{\lambda}{4}$,we can write $t = \frac{\lambda}{4v}$.
The time period $T$ of the wave is given by $T = \frac{\lambda}{v}$.
Substituting $\lambda = vT$ into the expression for $t$,we get $t = \frac{vT}{4v} = \frac{T}{4}$.
Therefore,the time period $T = 4t$.
The frequency $f$ is the reciprocal of the time period,so $f = \frac{1}{T} = \frac{1}{4t}$.

(C) Given: Length of the tube $L = 0.8 \ m$,Frequency $f = 500 \ Hz$,Speed of sound $v = 340 \ m/s$.
First,calculate the wavelength $\lambda$ of the sound wave:
$\lambda = \frac{v}{f} = \frac{340}{500} = 0.68 \ m$.
The resonance condition for a tube closed at one end is $L_n = \frac{(2n-1)\lambda}{4}$,where $n = 1, 2, 3, \dots$.
We need to find the number of values of $n$ such that $L_n \le 0.8 \ m$.
For $n=1$: $L_1 = \frac{\lambda}{4} = \frac{0.68}{4} = 0.17 \ m$.
For $n=2$: $L_2 = \frac{3\lambda}{4} = 3 \times 0.17 = 0.51 \ m$.
For $n=3$: $L_3 = \frac{5\lambda}{4} = 5 \times 0.17 = 0.85 \ m$.
Since $L_3 = 0.85 \ m > 0.8 \ m$,the third resonance cannot be formed within the tube length.
Therefore,only $2$ resonances will be heard.

(C) $1$. The fundamental frequency of a pipe closed at one end is given by $f_p = \frac{v}{4L}$,where $v = 320 \ m/s$ and $L = 0.8 \ m$.
$f_p = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 \ Hz$.
$2$. The string is vibrating in its second harmonic. The frequency of the $n$-th harmonic for a string fixed at both ends is $f_s = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $n=2$,$l = 0.5 \ m$,$T = 50 \ N$,and $\mu = \frac{m}{l}$.
$3$. Since the string resonates with the pipe,$f_s = f_p = 100 \ Hz$.
$100 = \frac{2}{2 \times 0.5} \sqrt{\frac{50}{m/0.5}} = 2 \sqrt{\frac{25}{m}} = 2 \times 5 \sqrt{\frac{1}{m}} = \frac{10}{\sqrt{m}}$.
$4$. Squaring both sides: $100 = \frac{100}{m}$,which gives $m = 1 \ kg = 1000 \ g$.
Wait,re-evaluating the calculation: $100 = \frac{10}{\sqrt{m}} \implies \sqrt{m} = 0.1 \implies m = 0.01 \ kg = 10 \ g$.

(A) For a closed organ pipe of length $L_c$,the frequency of the $n^{\text{th}}$ overtone is given by $f_c = \frac{(2n+1)v}{4L_c}$,where $n$ is the overtone number.
For the $3^{\text{rd}}$ overtone $(n=3)$,$f_c = \frac{(2(3)+1)v}{4L_c} = \frac{7v}{4L_c}$.
For an open organ pipe of length $L_o$,the frequency of the $n^{\text{th}}$ overtone is given by $f_o = \frac{(n+1)v}{2L_o}$.
For the $3^{\text{rd}}$ overtone $(n=3)$,$f_o = \frac{(3+1)v}{2L_o} = \frac{4v}{2L_o} = \frac{2v}{L_o}$.
Since the frequencies are in unison,$f_c = f_o$:
$\frac{7v}{4L_c} = \frac{2v}{L_o}$.
Rearranging for the ratio $\frac{L_c}{L_o}$:
$\frac{L_c}{L_o} = \frac{7}{4 \times 2} = \frac{7}{8}$.
Thus,the correct option is $A$.

(A) Let the length of the pipe be $\ell$ and the speed of sound be $v$.
For an open pipe, the fundamental frequency is $f_0 = \frac{v}{2\ell}$. The $2^{nd}$ overtone of an open pipe is $3f_0 = \frac{3v}{2\ell}$.
For a closed pipe, the fundamental frequency is $f_c = \frac{v}{4\ell}$. The $3^{rd}$ overtone of a closed pipe is $7f_c = \frac{7v}{4\ell}$.
According to the problem, the $3^{rd}$ overtone of the closed pipe is $150 \,Hz$ higher than the $2^{nd}$ overtone of the open pipe:
$\frac{7v}{4\ell} = \frac{3v}{2\ell} + 150$
$\frac{7v}{4\ell} - \frac{6v}{4\ell} = 150$
$\frac{v}{4\ell} = 150$
Since the fundamental frequency of the open pipe is $f_0 = \frac{v}{2\ell}$, we can write:
$f_0 = 2 \times \left(\frac{v}{4\ell}\right) = 2 \times 150 = 300 \,Hz$.

(D) For a closed organ pipe of length $\ell_c$,the fundamental frequency is $f_c = \frac{V}{4\ell_c}$. The first overtone is the third harmonic,so $f_{c,1} = 3f_c = \frac{3V}{4\ell_c}$.
For an open organ pipe of length $\ell_o$,the fundamental frequency is $f_o = \frac{V}{2\ell_o}$. The first overtone is the second harmonic,so $f_{o,1} = 2f_o = \frac{2V}{2\ell_o} = \frac{V}{\ell_o}$.
Given that the frequencies of the first overtones are identical,we have $\frac{3V}{4\ell_c} = \frac{V}{\ell_o}$.
Rearranging the terms to find the ratio $\frac{\ell_o}{\ell_c}$,we get $\frac{\ell_o}{\ell_c} = \frac{4}{3}$.

(B) In a closed pipe,the fundamental mode of vibration occurs when the length of the pipe $\ell$ is equal to one-fourth of the wavelength $\lambda$ of the sound wave.
Formula: $\ell = \frac{\lambda}{4}$
Given: $\lambda = 62 \,cm$
Calculation: $\ell = \frac{62}{4} = 15.5 \,cm$
Therefore,the correct option is $B$.

(D) The frequency of vibration of a stretched wire is given by $n = \frac{1}{2\ell} \sqrt{\frac{T}{m}}$.
Assuming the tension $T$ is provided by a mass $M$ hanging from the wire,$T = Mg$,so $n = \frac{1}{2\ell} \sqrt{\frac{Mg}{m}}$.
Let $\ell_1$ and $g_1$ be the length and acceleration due to gravity at the poles,and $\ell_2$ and $g_2$ be the length and acceleration due to gravity at the equator.
Since the frequency $n$ remains the same for the same tuning fork,we have $\frac{1}{2\ell_1} \sqrt{\frac{Mg_1}{m}} = \frac{1}{2\ell_2} \sqrt{\frac{Mg_2}{m}}$.
This simplifies to $\frac{\sqrt{g_1}}{\ell_1} = \frac{\sqrt{g_2}}{\ell_2}$,which implies $\frac{\ell_2}{\ell_1} = \sqrt{\frac{g_2}{g_1}}$.
Since the acceleration due to gravity is greater at the poles than at the equator $(g_1 > g_2)$,it follows that $\ell_1 > \ell_2$.
Therefore,to maintain resonance at the equator,the vibrating length of the wire must be decreased.

(B) The fundamental frequency of a closed pipe of length $L$ is given by $f_1 = \frac{v}{4L} = 400 \,Hz$.
From this,we get $v = 1600L$.
When $1/3$ of the pipe is filled with water,the length of the air column becomes $L' = L - \frac{L}{3} = \frac{2L}{3}$.
The new fundamental frequency of the air column is $f'_1 = \frac{v}{4L'} = \frac{v}{4(2L/3)} = \frac{3v}{8L}$.
Substituting $v = 1600L$,we get $f'_1 = \frac{3(1600L)}{8L} = 3 \times 200 = 600 \,Hz$.
In a closed pipe,the harmonics are odd multiples of the fundamental frequency $(f_1, 3f_1, 5f_1, \dots)$. However,the question asks for the $2^{\text{nd}}$ harmonic of the pipe in its new state. The $2^{\text{nd}}$ harmonic (first overtone) of a closed pipe is $3f'_1$.
Therefore,the frequency of the $2^{\text{nd}}$ harmonic is $3 \times 600 \,Hz = 1800 \,Hz$.

(B) The fundamental frequency of a stretched wire is $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
The fundamental frequency of a closed pipe of length $L$ is $n = \frac{v}{4L}$.
Given that both have the same initial frequency, we have $\frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{v}{4L}$.
When tension is increased by $16 \,N$, the new frequency $n'$ is $n' = \frac{1}{2L} \sqrt{\frac{T+16}{\mu}}$.
This new frequency resonates with the $2^{\text{nd}}$ harmonic of the closed pipe. The harmonics of a closed pipe are odd multiples of the fundamental frequency $(n, 3n, 5n, \dots)$. However, the question specifies the $2^{\text{nd}}$ harmonic, which implies the $3n$ frequency (as the $1^{\text{st}}$ is $n$, $2^{\text{nd}}$ is $3n$).
Thus, $n' = 3n$.
$\frac{1}{2L} \sqrt{\frac{T+16}{\mu}} = 3 \left( \frac{1}{2L} \sqrt{\frac{T}{\mu}} \right)$.
Squaring both sides: $\frac{T+16}{T} = 3^2 = 9$.
$T+16 = 9T \implies 8T = 16 \implies T = 2 \,N$.

(B) Given: Length of tube $L = 0.8 \,m$, Frequency $f = 375 \,Hz$, Speed of sound $v = 330 \,m/s$.
For the fundamental frequency in a resonance tube (closed at one end), the length of the air column $\ell$ is given by $\ell = \frac{v}{4f}$.
Substituting the values: $\ell = \frac{330}{4 \times 375} = \frac{330}{1500} = 0.22 \,m$.
This $\ell$ represents the length of the air column from the top.
The water level from the bottom is the total length of the tube minus the length of the air column: $h = L - \ell$.
$h = 0.8 \,m - 0.22 \,m = 0.58 \,m$.

(D) The fundamental frequency of an open pipe is given by $n = \frac{v}{2 \ell}$.
For a closed pipe, the harmonics are odd multiples of the fundamental frequency $n_0 = \frac{v}{4 \ell}$.
The $2^{nd}$ harmonic of a closed pipe is actually the $1^{st}$ overtone, which is the $3^{rd}$ harmonic, given by $3 n_0 = \frac{3v}{4 \ell}$.
According to the problem, the frequency of the $2^{nd}$ harmonic of the closed pipe is $200 \,Hz$ higher than $n$:
$\frac{3v}{4 \ell} - n = 200$
Since $n = \frac{v}{2 \ell}$, we have $\frac{v}{4 \ell} = \frac{n}{2}$.
Substituting this into the equation: $3(\frac{n}{2}) - n = 200$.
$1.5n - n = 200 \implies 0.5n = 200$.
$n = 400 \,Hz$.

(C) The given equation of the stationary wave is $y = 10 \sin \left( \frac{\pi x}{4} \right) \cos (20 \pi t)$.
Comparing this with the standard equation of a stationary wave,$y = 2A \sin (kx) \cos (\omega t)$,we get the propagation constant $k = \frac{\pi}{4} \ cm^{-1}$.
We know that the propagation constant $k$ is related to the wavelength $\lambda$ by the formula $k = \frac{2\pi}{\lambda}$.
Substituting the value of $k$: $\frac{\pi}{4} = \frac{2\pi}{\lambda}$.
Solving for $\lambda$,we get $\lambda = 8 \ cm$.
The distance between two consecutive nodes in a stationary wave is equal to half of the wavelength,which is $\frac{\lambda}{2}$.
Therefore,the distance is $\frac{8 \ cm}{2} = 4 \ cm$.

(C) stationary wave is formed by the superposition of two identical progressive waves traveling in opposite directions with the same frequency and amplitude.
In a stationary wave,the distance between two consecutive nodes or two consecutive antinodes is $\frac{\lambda}{2}$.
The wavelength of the stationary wave is defined as the distance between two consecutive points that are in the same phase,which is equal to the wavelength of the constituent progressive waves.
Therefore,the wavelength of the stationary wave is $\lambda$.

(D) In Melde's experiment,the frequency of the vibrating string is given by $f = \frac{P}{2L} \sqrt{\frac{T}{\mu}}$,where $P$ is the number of loops,$T$ is the tension,$L$ is the length,and $\mu$ is the linear mass density.
Since the frequency $f$ and the length $L$ remain constant,we have $P \propto \frac{1}{\sqrt{T}}$,which implies $T P^{2} = \text{constant}$.
Let the initial tension be $T_1$ and the final tension be $T_2 = T_1 - 0.009 \ kg-wt$.
Given $P_1 = 4$ and $P_2 = 5$.
Using the relation $T_1 P_1^{2} = T_2 P_2^{2}$:
$T_1 (4)^{2} = (T_1 - 0.009) (5)^{2}$
$16 T_1 = 25 T_1 - 0.009 \times 25$
$25 T_1 - 16 T_1 = 0.225$
$9 T_1 = 0.225$
$T_1 = \frac{0.225}{9} = 0.025 \ kg-wt$.

(C) In a stationary wave, the distance between two consecutive nodes is $\lambda / 2$.
For $4$ nodes, there are $3$ such segments of length $\lambda / 2$ between them.
Given the total length of the string is $L = 120 \,cm$.
Therefore, $3 \times (\lambda / 2) = 120 \,cm$.
$3 \lambda / 2 = 120 \,cm$.
$\lambda = (120 \times 2) / 3 \,cm$.
$\lambda = 240 / 3 \,cm$.
$\lambda = 80 \,cm$.

(A) In a stationary wave,the medium particles oscillate with different amplitudes at different positions.
Nodes are points where the amplitude of vibration is zero,meaning the displacement is always zero.
Antinodes are points where the amplitude of vibration is maximum.
Therefore,the displacement at a node is zero and at an antinode is maximum.

(B) Given:
Velocity of waves,$v = 50 \,m/s$
Frequency of waves,$f = 200 \,Hz$
First,we calculate the wavelength $\lambda$ of the progressive waves:
$v = f \lambda$
$\lambda = \frac{v}{f} = \frac{50}{200} = 0.25 \,m$
When two identical progressive waves travel in opposite directions,they form a standing wave.
In a standing wave,the distance between two consecutive antinodes is equal to half the wavelength $(\frac{\lambda}{2})$.
Distance $= \frac{\lambda}{2} = \frac{0.25 \,m}{2} = 0.125 \,m$
Therefore,the correct option is $B$.

(D) The fundamental frequency of a pipe open at one end is given by $f_p = \frac{v}{4L_p}$, where $v = 320 \,m/s$ and $L_p = 0.8 \,m$.
$f_p = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 \,Hz$.
For a string of length $L_s = 0.5 \,m$ vibrating in its $1^{\text{st}}$ overtone, the frequency is $f_s = 2 \times \left( \frac{1}{2L_s} \sqrt{\frac{T}{\mu}} \right) = \frac{1}{L_s} \sqrt{\frac{T}{\mu}}$, where $\mu = \frac{M}{L_s}$ is the linear mass density.
Since the string resonates with the pipe, $f_s = f_p = 100 \,Hz$.
$100 = \frac{1}{0.5} \sqrt{\frac{50}{M/0.5}} = 2 \sqrt{\frac{50 \times 0.5}{M}} = 2 \sqrt{\frac{25}{M}} = \frac{2 \times 5}{\sqrt{M}} = \frac{10}{\sqrt{M}}$.
$\sqrt{M} = \frac{10}{100} = 0.1$.
$M = (0.1)^2 = 0.01 \,kg = 10 \,g$.

(C) The fundamental frequency of a vibrating string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \text{Area} \times \text{Density} = (\pi R^2) \rho$.
Substituting $\mu$ into the frequency formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\pi R^2 \rho}} = \frac{1}{2LR} \sqrt{\frac{T}{\pi \rho}}$.
Given that the material is the same,$\rho$ is constant. Since $T$ is also constant,we have $f \propto \frac{1}{LR}$.
For the first string: $L_1 = L$,$R_1 = 2r$. Thus,$f_1 \propto \frac{1}{L \cdot 2r} = \frac{1}{2Lr}$.
For the second string: $L_2 = 2L$,$R_2 = r$. Thus,$f_2 \propto \frac{1}{2L \cdot r} = \frac{1}{2Lr}$.
The ratio of their frequencies is $\frac{f_1}{f_2} = \frac{1/2Lr}{1/2Lr} = 1:1$.

(A) The frequency of a sonometer wire is given by the formula $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $p$ is the number of loops (antinodes),$L$ is the length,$T$ is the tension $(T = Mg)$,and $\mu$ is the linear mass density.
Since the frequency $n$,length $L$,and linear mass density $\mu$ remain constant,we have $p \propto \sqrt{T} \propto \sqrt{M}$.
For the first case: $p_1 = 5$ and $M_1 = 9 \ kg$.
For the second case: $p_2 = 3$ and $M_2 = m$.
Using the ratio: $\frac{p_1}{p_2} = \sqrt{\frac{M_1}{M_2}}$.
Substituting the values: $\frac{5}{3} = \sqrt{\frac{9}{m}}$.
Squaring both sides: $\frac{25}{9} = \frac{9}{m}$.
Solving for $m$: $m = \frac{9 \times 9}{25} = \frac{81}{25} = 3.24 \ kg$.
Wait,re-evaluating the relationship: $p_1 \sqrt{M_1} = p_2 \sqrt{M_2}$.
$5 \sqrt{9} = 3 \sqrt{m} \implies 5 \times 3 = 3 \sqrt{m} \implies 5 = \sqrt{m} \implies m = 25 \ kg$.

(A) The frequency of the $p$-th overtone for a string fixed at both ends is given by $f = (p+1) \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Since the material is the same,density $\rho$ is constant. The tension $T$ is also constant.
For the first wire,the first overtone $(p=1)$ is $f_1 = 2 \cdot \frac{1}{2 \ell_1} \sqrt{\frac{T}{\pi r_1^2 \rho}} = \frac{1}{\ell_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.
For the second wire,the second overtone $(p=2)$ is $f_2 = 3 \cdot \frac{1}{2 \ell_2} \sqrt{\frac{T}{\pi r_2^2 \rho}} = \frac{3}{2 \ell_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_1 = f_2$,we have $\frac{1}{\ell_1 r_1} = \frac{3}{2 \ell_2 r_2}$.
Given $r_1 = 2 r_2$,substituting this into the equation: $\frac{1}{\ell_1 (2 r_2)} = \frac{3}{2 \ell_2 r_2}$.
Simplifying,$\frac{1}{2 \ell_1} = \frac{3}{2 \ell_2}$,which gives $\frac{\ell_1}{\ell_2} = \frac{1}{3}$.

(C) For the $5^{\text{th}}$ harmonic, a string of length $L$ fixed at both ends vibrates in $5$ loops.
The relationship between length $L$ and wavelength $\lambda$ is given by $L = \frac{5\lambda}{2}$.
Comparing the given equation $Y = 3 \sin(0.4x) \cos(200\pi t)$ with the standard wave equation $Y = A \sin(kx) \cos(\omega t)$, we find the wave number $k = 0.4 \text{ rad/cm}$.
Since $k = \frac{2\pi}{\lambda}$, we have $\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.4} = 5\pi \text{ cm}$.
Substituting the value of $\lambda$ into the length equation: $L = \frac{5 \times 5\pi}{2} = \frac{25\pi}{2} = 12.5\pi \text{ cm}$.

(B) Given: Slit width $a = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m}$.
Distance of screen $D = 3 \text{ m}$.
Distance of first minima from central maximum $x = 5.5 \text{ mm} = 5.5 \times 10^{-3} \text{ m}$.
For a single slit diffraction, the condition for the first minima is given by $a \sin \theta = \lambda$.
For small angles, $\sin \theta \approx \tan \theta = \frac{x}{D}$.
Therefore, $a \left( \frac{x}{D} \right) = \lambda$.
Substituting the values: $\lambda = \frac{ax}{D} = \frac{(0.3 \times 10^{-3} \text{ m}) \times (5.5 \times 10^{-3} \text{ m})}{3 \text{ m}}$.
$\lambda = 0.1 \times 10^{-3} \times 5.5 \times 10^{-3} \text{ m} = 0.55 \times 10^{-6} \text{ m} = 5.5 \times 10^{-7} \text{ m}$.
Converting to $\text{ Å}$: $\lambda = 5500 \times 10^{-10} \text{ m} = 5500 \text{ Å}$.

(D) For a single slit diffraction,the condition for the $n^{th}$ minima is given by $a \sin \theta = n \lambda$.
Here,$a = 10^{-3} \ mm = 10^{-6} \ m$,$\lambda = 5 \times 10^{-7} \ m$,and for the first minima,$n = 1$.
Substituting the values,we get $\sin \theta = \frac{n \lambda}{a} = \frac{1 \times 5 \times 10^{-7} \ m}{10^{-6} \ m}$.
$\sin \theta = \frac{5 \times 10^{-7}}{10 \times 10^{-7}} = 0.5 = \frac{1}{2}$.
Therefore,the angle of diffraction is $\theta = \sin^{-1}\left(\frac{1}{2}\right)$.

(C) The fringe width $Z$ in an interference experiment is given by the formula $Z = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the source and the screen (or eyepiece),and $d$ is the distance between the two virtual sources.
Since $\lambda$ and $D$ are constant,the fringe width is inversely proportional to the distance between the sources,i.e.,$Z \propto \frac{1}{d}$.
Therefore,we can write the ratio as $\frac{Z_{A}}{Z_{B}} = \frac{d_{B}}{d_{A}}$.
Thus,the correct option is $C$.

(C) The linear width of the principal maxima in a single-slit diffraction pattern is given by the formula $\beta = \frac{2 \lambda D}{d}$.
According to the problem,the linear width of the principal maxima is equal to the width of the slit,so $\beta = d$.
Substituting this into the formula,we get $d = \frac{2 \lambda D}{d}$.
Rearranging the equation to solve for $D$,we get $D = \frac{d^{2}}{2 \lambda}$.

(C) The angular width of the central maximum in a single slit diffraction experiment is given by the formula $\theta = \frac{2 \lambda}{d}$,where $\lambda$ is the wavelength of the light used and $d$ is the width of the slit.
Since the angular width depends only on the wavelength $\lambda$ and the slit width $d$,it is independent of the distance $D$ between the slit and the screen.
Therefore,when the distance $D$ is doubled,the angular separation between the fringes remains the same.

(A) The intensity $I$ of a wave is proportional to the square of its amplitude $a$,i.e.,$I \propto a^{2}$.
This implies that the amplitude $a$ is proportional to the square root of the intensity,i.e.,$a \propto \sqrt{I}$.
Let the amplitudes of the two coherent sources be $a_{1}$ and $a_{2}$,where $a_{1} \propto \sqrt{I_{1}}$ and $a_{2} \propto \sqrt{I_{2}}$.
In an interference pattern,the maximum intensity $(I_{\max})$ occurs at points of constructive interference,where the phase difference is an even multiple of $\pi$ $(2n\pi)$.
At these points,the resultant amplitude is the sum of the individual amplitudes: $A_{\max} = a_{1} + a_{2}$.
Since $I_{\max} \propto A_{\max}^{2}$,we have $I_{\max} \propto (a_{1} + a_{2})^{2}$.
Substituting the expressions for $a_{1}$ and $a_{2}$,we get $I_{\max} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}$.

(D) The intensity of the resultant wave in interference is given by $I = 4I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference and $I_0$ is the intensity of each individual source.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x = 0$,the phase difference $\phi = 0$. Thus,$I_1 = 4I_0 \cos^2(0) = 4I_0$.
For path difference $\Delta x = \lambda/2$,the phase difference $\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{2} = \pi$. Thus,$I_2 = 4I_0 \cos^2(\pi/2) = 4I_0 \cdot 0 = 0$.
The ratio of intensities is $\frac{I_1}{I_2} = \frac{4I_0}{0} = \infty$.
Therefore,the ratio is $\infty:1$.

(C) In an interference pattern, the intensity $I$ is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For destructive interference, the phase difference $\phi = (2n+1)\pi$, so $\cos \phi = -1$.
The minimum intensity is $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
If the amplitudes are different, then $I_1 \neq I_2$, which implies $\sqrt{I_1} \neq \sqrt{I_2}$.
Therefore, $I_{min} \neq 0$.
This means that in the region of destructive interference, there is some residual intensity of light instead of complete darkness.

(A) Given: Slit separation $d = 1 \,mm = 10^{-3} \,m$,Wavelength $\lambda = 5000 \text{ Å} = 5 \times 10^{-7} \,m$,Change in fringe width $\Delta \beta = \beta_{2} - \beta_{1} = 12.5 \times 10^{-5} \,m$.
Fringe width formula is $\beta = \frac{\lambda D}{d}$.
The change in fringe width is given by $\Delta \beta = \frac{\lambda}{d} \Delta D$,where $\Delta D = D_{2} - D_{1}$ is the shift in the screen position.
Rearranging for $\Delta D$: $\Delta D = \frac{d \cdot \Delta \beta}{\lambda}$.
Substituting the values: $\Delta D = \frac{10^{-3} \times 12.5 \times 10^{-5}}{5 \times 10^{-7}}$.
$\Delta D = \frac{12.5 \times 10^{-8}}{5 \times 10^{-7}} = 2.5 \times 10^{-1} \,m = 0.25 \,m = 25 \,cm$.
Therefore,the screen should be moved away or towards the slit by $25 \,cm$.

(D) For destructive interference to occur,the two light waves must arrive at a point with a phase difference that results in the cancellation of their amplitudes.
This occurs when the waves are in anti-phase,meaning one wave is at a crest while the other is at a trough.
The condition for destructive interference is given by the phase difference $\Delta \phi = (2n + 1)\pi$,where $n = 0, 1, 2, \ldots$.
Substituting values for $n$,we get $\Delta \phi = \pi, 3\pi, 5\pi, \ldots$.

(D) According to Brewster's Law,the refractive index $\mu$ is given by $\mu = \tan i_p$,where $i_p$ is the polarizing angle.
Given $\mu = 1.73$ and $\tan 60^{\circ} = 1.73$,we have $\tan i_p = \tan 60^{\circ}$,which implies $i_p = 60^{\circ}$.
When light is incident at the polarizing angle,the reflected ray and the refracted ray are perpendicular to each other,meaning $i_p + r = 90^{\circ}$.
Substituting the value of $i_p$,we get $60^{\circ} + r = 90^{\circ}$.
Therefore,the angle of refraction $r = 90^{\circ} - 60^{\circ} = 30^{\circ}$.

(C) When unpolarised light passes through the first polaroid, it becomes linearly polarised with intensity $I_0 = I_{in} / 2$.
When this polarised light hits the second polaroid, the intensity transmitted is given by Malus' Law: $I = I_0 \cos^2 \theta$.
Since the polaroids are crossed, the angle between their transmission axes is $\theta = 90^{\circ}$.
Substituting this value, we get $I = I_0 \cos^2(90^{\circ}) = I_0 \times 0 = 0$.
Therefore, the light is completely blocked by the second polaroid.

(C) According to Brewster's Law,the refractive index $\mu$ of a transparent medium is given by $\mu = \tan \theta$,where $\theta$ is the polarising angle.
We also know that the refractive index $\mu$ is defined as the ratio of the speed of light in vacuum $(c)$ to the speed of light in the medium $(V)$:
$\mu = \frac{c}{V}$
Equating the two expressions for $\mu$:
$\tan \theta = \frac{c}{V}$
Taking the reciprocal on both sides:
$\cot \theta = \frac{V}{c}$
Therefore,the relation between $\theta$ and $V$ is:
$\theta = \cot^{-1}\left(\frac{V}{c}\right)$

(A) Brewster's law states that the tangent of the polarizing angle (Brewster's angle, $i_p$) is equal to the refractive index $(\mu)$ of the medium, given by the formula: $\tan(i_p) = \mu$.
Since the refractive index $(\mu)$ of a transparent medium depends on the wavelength $(\lambda)$ of light (due to dispersion), the refractive index is different for different colours of light.
Consequently, the Brewster's angle $(i_p = \arctan(\mu))$ will also be different for lights of different colours.
Therefore, the correct statement is that Brewster's angle is different for lights of different colours.

(C) Let the width of the incident wavefront be $w_i$ and the width of the refracted wavefront be $w_r$.
From the geometry of wavefronts,the width of a wavefront is related to the beam width by $w = L \cos \theta$,where $L$ is the length of the wavefront segment on the interface.
For the incident wavefront,$w_i = L \cos i$,where $i = 60^{\circ}$.
For the refracted wavefront,$w_r = L \cos r$,where $r = 45^{\circ}$.
The ratio of the width of the incident wavefront to the refracted wavefront is $\frac{w_i}{w_r} = \frac{\cos i}{\cos r}$.
Substituting the values: $\frac{w_i}{w_r} = \frac{\cos 60^{\circ}}{\cos 45^{\circ}} = \frac{1/2}{1/\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

(C) The number of waves $N$ in a medium of thickness $d$ and refractive index $\mu$ is given by $N = \frac{d}{\lambda_m}$,where $\lambda_m = \frac{\lambda_0}{\mu}$ is the wavelength in the medium.
Thus,$N = \frac{d \cdot \mu}{\lambda_0}$.
Since the number of waves is the same for both media,we have:
$\frac{d_g \cdot \mu_g}{\lambda_0} = \frac{d_w \cdot \mu_w}{\lambda_0}$
$d_g \cdot \mu_g = d_w \cdot \mu_w$
Given $d_g = 4 \,cm$,$\mu_g = \frac{5}{3}$,and $\mu_w = \frac{4}{3}$.
Substituting the values: $4 \times \frac{5}{3} = x \times \frac{4}{3}$
$\frac{20}{3} = \frac{4x}{3}$
$4x = 20$
$x = 5 \,cm$.

(D) The intensity of light in Young's double slit experiment is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x = \lambda$,the phase difference is $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.
Given intensity $K = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(1)^2 = 4I_0$.
For path difference $\Delta x = \frac{\lambda}{4}$,the phase difference is $\phi' = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $I'$ is $I' = 4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $I' = 4I_0 \cdot (\frac{1}{\sqrt{2}})^2 = 4I_0 \cdot \frac{1}{2} = 2I_0$.
Since $K = 4I_0$,then $2I_0 = \frac{K}{2}$.
Therefore,the intensity is $\frac{K}{2}$.

(B) The position of the $n^{\text{th}}$ maximum in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$,where $n$ is the order of the maximum,$\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the slit separation.
For the $6^{\text{th}}$ maximum with wavelength $\lambda_1$,the distance is $d_1 = \frac{6 \lambda_1 D}{d}$.
For the $4^{\text{th}}$ maximum with wavelength $\lambda_2$,the distance is $d_2 = \frac{4 \lambda_2 D}{d}$.
Taking the ratio of $d_1$ to $d_2$:
$\frac{d_1}{d_2} = \frac{6 \lambda_1 D / d}{4 \lambda_2 D / d} = \frac{6 \lambda_1}{4 \lambda_2} = \frac{3}{2} \frac{\lambda_1}{\lambda_2}$.

(B) The fringe width $\beta$ in a medium of refractive index $\mu$ is given by $\beta = \frac{\lambda_w D}{d}$.
Here, the wavelength in water is $\lambda_w = \frac{\lambda_{air}}{\mu} = \frac{6300 \times 10^{-10} \,m}{1.33}$.
Given: $D = 1.33 \,m$, $d = 1 \,mm = 10^{-3} \,m$, and $\mu = 1.33$.
Substituting these values into the formula:
$\beta = \frac{(6300 \times 10^{-10} / 1.33) \times 1.33}{10^{-3}} \,m$.
$\beta = \frac{6300 \times 10^{-10}}{10^{-3}} \,m$.
$\beta = 6300 \times 10^{-7} \,m = 6.3 \times 10^{-4} \,m$.

(C) The resultant intensity $I_R$ in Young's double slit experiment is given by $I_R = 4I_0 \cos^2(\frac{\phi}{2})$,where $I_0$ is the intensity of each individual slit and $\phi$ is the phase difference.
Phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Case $1$: When $\Delta x = \lambda$,$\phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi$. The intensity $I = 4I_0 \cos^2(\frac{2\pi}{2}) = 4I_0 \cos^2(\pi) = 4I_0(-1)^2 = 4I_0$.
Case $2$: When $\Delta x = \frac{\lambda}{4}$,$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The new intensity $I' = 4I_0 \cos^2(\frac{\pi/2}{2}) = 4I_0 \cos^2(\frac{\pi}{4}) = 4I_0 (\frac{1}{\sqrt{2}})^2 = 4I_0 \times \frac{1}{2} = 2I_0$.
Since $I = 4I_0$,we have $I_0 = \frac{I}{4}$.
Substituting $I_0$ into the expression for $I'$,we get $I' = 2(\frac{I}{4}) = \frac{I}{2}$.

(B) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the separation between the slits.
To make the fringes more closely spaced,the fringe width $\beta$ must be decreased.
From the formula,$\beta \propto \lambda$ and $\beta \propto \frac{1}{d}$.
Since the wavelength of blue light is shorter than that of green light $(\lambda_{blue} < \lambda_{green})$,using blue light will decrease the fringe width $\beta$.
Therefore,the correct option is $B$.

(A) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
When the experiment is performed in water,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$,where $\mu$ is the refractive index of water $(\mu > 1)$.
Since $\mu > 1$,the new wavelength $\lambda'$ is smaller than the original wavelength $\lambda$ in air.
Consequently,the new fringe width $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.
Since $\mu > 1$,the fringe width $\beta'$ decreases.

(C) In Young's Double Slit Experiment,the fringe-width $(z)$ is given by the formula:
$z = \frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen (or eye-piece),and $d$ is the distance between the two slits.
Since $\lambda$ and $d$ are kept constant,we have $z \propto D$.
This represents a linear relationship passing through the origin,which corresponds to a straight line graph.
Looking at the given options,graph $(C)$ represents a direct linear relationship $(z \propto D)$.
Therefore,the correct graph is $(C)$.

(D) Given: Slit separation $d = 3 \,mm = 3 \times 10^{-3} \,m$,distance to screen $D = 2 \,m$,wavelengths $\lambda_1 = 480 \,nm = 480 \times 10^{-9} \,m$ and $\lambda_2 = 600 \,nm = 600 \times 10^{-9} \,m$.
The position of the $n^{th}$ order bright fringe is given by $y_n = n \frac{\lambda D}{d}$.
For the $5^{th}$ order bright fringe of the first wavelength: $y_5^{(1)} = 5 \frac{\lambda_1 D}{d}$.
For the $5^{th}$ order bright fringe of the second wavelength: $y_5^{(2)} = 5 \frac{\lambda_2 D}{d}$.
The separation between these fringes is $\Delta y = y_5^{(2)} - y_5^{(1)} = \frac{5D}{d} (\lambda_2 - \lambda_1)$.
Substituting the values: $\Delta y = \frac{5 \times 2}{3 \times 10^{-3}} \times (600 - 480) \times 10^{-9} \,m$.
$\Delta y = \frac{10}{3 \times 10^{-3}} \times 120 \times 10^{-9} \,m = \frac{1200}{3} \times 10^{-6} \,m = 400 \times 10^{-6} \,m = 4 \times 10^{-4} \,m$.

(C) Given the ratio of intensities of two waves is $\frac{I_1}{I_2} = \frac{9}{4}$.
Since intensity $I \propto a^2$,the ratio of amplitudes is $\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
The maximum intensity is $I_{max} = (a_1 + a_2)^2$ and the minimum intensity is $I_{min} = (a_1 - a_2)^2$.
The ratio of maximum to minimum intensity is $\frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2$.
Substituting the values,$\frac{I_{max}}{I_{min}} = \left( \frac{3 + 2}{3 - 2} \right)^2 = \left( \frac{5}{1} \right)^2 = \frac{25}{1}$.
Thus,the ratio is $25: 1$.

(A) The intensity at any point on the screen is given by $I = 4I_0 \cos^2 \left( \frac{\phi}{2} \right)$,where $\phi$ is the phase difference.
The phase difference $\phi$ is related to path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
For path difference $\Delta x_1 = 0$,the phase difference $\phi_1 = 0$. Thus,$I_1 = 4I_0 \cos^2(0) = 4I_0$.
For path difference $\Delta x_2 = \frac{\lambda}{4}$,the phase difference $\phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity $I_2 = 4I_0 \cos^2 \left( \frac{\pi/2}{2} \right) = 4I_0 \cos^2 \left( \frac{\pi}{4} \right) = 4I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = 4I_0 \times \frac{1}{2} = 2I_0$.
The ratio of intensities is $\frac{I_1}{I_2} = \frac{4I_0}{2I_0} = \frac{2}{1}$ or $2:1$.