MHT CET 2025 Physics Question Paper with Answer and Solution

(A) The given wave equation is $y = A \sin (\omega t - k x)$,where $A = 60 \ \mu m = 60 \times 10^{-6} \ m$,$\omega = 1200 \ rad/s$,and $k = 6 \ m^{-1}$.
The maximum particle velocity $(v_p)$ is given by $v_p = A \omega$.
$v_p = (60 \times 10^{-6} \ m) \times (1200 \ rad/s) = 72000 \times 10^{-6} \ m/s = 7.2 \times 10^{-2} \ m/s$.
The wave velocity $(v_w)$ is given by $v_w = \frac{\omega}{k}$.
$v_w = \frac{1200}{6} = 200 \ m/s$.
The ratio of maximum particle velocity to wave velocity is $\frac{v_p}{v_w} = \frac{A \omega}{\omega/k} = A k$.
$\frac{v_p}{v_w} = (60 \times 10^{-6} \ m) \times (6 \ m^{-1}) = 360 \times 10^{-6} = 3.6 \times 10^{-4}$.

(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula: $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Given $\Delta \phi = 45^{\circ} = \frac{\pi}{4} \text{ radians}$ and $\Delta x = 4.0 \text{ m}$.
Substituting these values: $\frac{\pi}{4} = \frac{2\pi}{\lambda} \times 4.0$.
Solving for wavelength $\lambda$: $\lambda = 8 \times 4.0 = 32 \text{ m}$.
The wave velocity $v$ is given by $v = f \lambda$,where $f = 300 \text{ Hz}$.
$v = 300 \text{ Hz} \times 32 \text{ m} = 9600 \text{ m/s}$.
Converting to $\text{km/s}$: $v = \frac{9600}{1000} \text{ km/s} = 9.6 \text{ km/s}$.

(D) For a closed pipe of length $L$,the fundamental frequency is given by $f_c = \frac{v}{4L}$.
For an open pipe of length $L' = XL$,the frequencies are given by $f_n = \frac{n v}{2L'}$,where $n = 1, 2, 3, \dots$.
The first overtone is $n=2$ and the second overtone is $n=3$.
Therefore,the second overtone frequency of the open pipe is $f_o = \frac{3v}{2(XL)}$.
According to the problem,$f_c = f_o$,so $\frac{v}{4L} = \frac{3v}{2XL}$.
Canceling $v$ and $L$ from both sides,we get $\frac{1}{4} = \frac{3}{2X}$.
Solving for $X$,we get $2X = 12$,which implies $X = 6$.

(C) Given: Linear mass density $\mu = 0.1 \ kg/m$,Length $L = 0.9 \ m$,Tension $T = 40 \ N$,Amplitude $A = 0.3 \ cm = 0.003 \ m$,Number of segments $n = 3$.
Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.1}} = \sqrt{400} = 20 \ m/s$.
The frequency of the $n^{th}$ harmonic is $f_n = \frac{n v}{2L} = \frac{3 \times 20}{2 \times 0.9} = \frac{60}{1.8} = \frac{600}{18} = \frac{100}{3} \ Hz$.
Angular frequency $\omega = 2 \pi f_n = 2 \pi \times \frac{100}{3} = \frac{200 \pi}{3} \ rad/s$.
Maximum particle velocity $v_{max} = A \omega = 0.003 \times \frac{200 \pi}{3} = \frac{3}{1000} \times \frac{200 \pi}{3} = \frac{\pi}{5} \ m/s$.

(B) For a closed pipe of length $L_c$,the frequency of the $n^{th}$ overtone is given by $f_c = \frac{(2n+1)v}{4L_c}$,where $n$ is the overtone number. For the fourth overtone $(n=4)$,$f_c = \frac{(2(4)+1)v}{4L_c} = \frac{9v}{4L_c}$.
For an open pipe of length $L_o$,the frequency of the $m^{th}$ overtone is given by $f_o = \frac{(m+1)v}{2L_o}$,where $m$ is the overtone number. For the fifth overtone $(m=5)$,$f_o = \frac{(5+1)v}{2L_o} = \frac{6v}{2L_o} = \frac{3v}{L_o}$.
Given that the frequencies are in unison,$f_c = f_o$,so $\frac{9v}{4L_c} = \frac{3v}{L_o}$.
Simplifying the equation: $\frac{9}{4L_c} = \frac{3}{L_o} \implies \frac{L_c}{L_o} = \frac{9}{4 \times 3} = \frac{9}{12} = \frac{3}{4}$.
Thus,the ratio of the length of the closed pipe to the open pipe is $3: 4$.

(C) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.
Since $\mu = \text{Area} \times \text{Density} = (\pi r^2) \rho = \pi (d/2)^2 \rho = \frac{\pi d^2 \rho}{4}$,we can write $n = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi d^2 \rho}{4}}} = \frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}$.
For wire $A$: $n_A = \frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}$.
For wire $B$: $n_B = \frac{1}{(2L)(3d)} \sqrt{\frac{2T}{\pi (2\rho)}} = \frac{1}{6Ld} \sqrt{\frac{T}{\pi \rho}}$.
Taking the ratio $n_B : n_A = \frac{\frac{1}{6Ld} \sqrt{\frac{T}{\pi \rho}}}{\frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}} = \frac{1}{6}$.
Thus,the ratio $n_B : n_A$ is $1 : 6$.

(D) Let the initial tension be $T$. The frequency of the stretched wire is $f_w = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
The fundamental frequency of a closed pipe of length $L$ is $f_c = \frac{v}{4L}$.
Given $f_w = f_c$,so $\frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{v}{4L} \implies \sqrt{\frac{T}{\mu}} = \frac{v}{2}$.
When tension is increased by $8 \ N$,the new frequency is $f_w' = \frac{1}{2L} \sqrt{\frac{T+8}{\mu}}$.
The first overtone of the closed pipe is $3f_c = \frac{3v}{4L}$.
Given $f_w' = 3f_c$,so $\frac{1}{2L} \sqrt{\frac{T+8}{\mu}} = \frac{3v}{4L} \implies \sqrt{\frac{T+8}{\mu}} = \frac{3v}{2}$.
Dividing the two equations: $\sqrt{\frac{T+8}{T}} = 3$.
Squaring both sides: $\frac{T+8}{T} = 9 \implies T+8 = 9T \implies 8T = 8 \implies T = 1 \ N$.

(A) The frequency of the $n^{\text{th}}$ harmonic for a stretched string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Thus,$f = \frac{n}{2L r} \sqrt{\frac{T}{\pi \rho}}$.
For the $1^{\text{st}}$ wire,the $1^{\text{st}}$ overtone is the $2^{\text{nd}}$ harmonic $(n_1 = 2)$.
So,$f_1 = \frac{2}{2L_1 r_1} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{L_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.
For the $2^{\text{nd}}$ wire,the $2^{\text{nd}}$ overtone is the $3^{\text{rd}}$ harmonic $(n_2 = 3)$.
So,$f_2 = \frac{3}{2L_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_1 = f_2$ and $r_1 = 2r_2$:
$\frac{1}{L_1 (2r_2)} = \frac{3}{2L_2 r_2}$.
$\frac{1}{2L_1} = \frac{3}{2L_2} \implies \frac{L_1}{L_2} = \frac{1}{3}$.
Therefore,the ratio is $1: 3$.

(D) The speed of sound $v = 330 \ m/s$ and frequency $f = 330 \ Hz$. The wavelength $\lambda = v/f = 330/330 = 1 \ m$.
When one end of an open pipe is dipped in water,it acts as a pipe closed at one end.
The length of the air column is $L' = L - x$,where $L = 1.5 \ m$ is the total length and $x$ is the length immersed.
For a pipe closed at one end,the frequencies of harmonics are given by $f_n = (2n-1) \frac{v}{4L'}$,where $n=1$ is the fundamental,$n=2$ is the $1^{\text{st}}$ overtone,and $n=3$ is the $2^{\text{nd}}$ overtone.
For the $2^{\text{nd}}$ overtone,$n=3$,so $f_3 = 5 \frac{v}{4L'}$.
Given $f_3 = 330 \ Hz$,we have $330 = 5 \times \frac{330}{4L'}$.
This simplifies to $1 = \frac{5}{4L'}$,which gives $4L' = 5$,or $L' = 1.25 \ m$.
Since $L' = L - x$,we have $1.25 = 1.5 - x$.
Therefore,$x = 1.5 - 1.25 = 0.25 \ m$.

(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $T$ and $\mu$ are constant,$f \propto \frac{1}{L}$.
Let the initial length be $L_1 = L$ and the initial fundamental frequency be $f_1 = 50 \text{ Hz}$.
The new length is $L_2 = L + 0.25L = 1.25L = \frac{5}{4}L$.
The new fundamental frequency $f_2$ is given by $\frac{f_2}{f_1} = \frac{L_1}{L_2} = \frac{L}{1.25L} = \frac{1}{1.25} = 0.8$.
So,$f_2 = 0.8 \times 50 \text{ Hz} = 40 \text{ Hz}$.
The frequency of the second harmonic is $f'_n = n \times f_n$. For the second harmonic $(n=2)$,$f'_2 = 2 \times f_2 = 2 \times 40 \text{ Hz} = 80 \text{ Hz}$.
The initial frequency of the second harmonic was $f'_1 = 2 \times f_1 = 2 \times 50 \text{ Hz} = 100 \text{ Hz}$.
The change in frequency is $\Delta f = f'_2 - f'_1 = 80 - 100 = -20 \text{ Hz}$.
The percentage change is $\frac{\Delta f}{f'_1} \times 100 = \frac{-20}{100} \times 100 = -20 \%$.
Thus,the frequency of the second harmonic decreases by $20 \%$.

(A) For a sonometer wire,the frequency $f$ is inversely proportional to the length $l$ of the wire,i.e.,$f \propto 1/l$ or $f \cdot l = k$ (constant).
Let the frequencies of the two tuning forks be $f_1$ and $f_2$.
Given $f_1 \cdot 23 = f_2 \cdot 24 = k$.
This implies $f_1 = k/23$ and $f_2 = k/24$.
Since $f_1 > f_2$,the beat frequency is $f_1 - f_2 = 4$.
Substituting the values: $k/23 - k/24 = 4$.
Taking $k$ as common: $k(24 - 23) / (23 \cdot 24) = 4$.
$k / 552 = 4 \implies k = 2208$.
Now,$f_1 = 2208 / 23 = 96 \ Hz$.
And $f_2 = 2208 / 24 = 92 \ Hz$.
Thus,the frequencies are $96 \ Hz$ and $92 \ Hz$.

(C) The fundamental frequency $n$ of a sonometer wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since $\mu = \pi r^2 \rho$ (where $r$ is the radius and $\rho$ is the density),the frequency can be written as: $n = \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{2Lr} \sqrt{\frac{T}{\pi \rho}}$.
Given that the diameter $D = 2r$,we have $n \propto \frac{1}{LD} \sqrt{T}$.
Let the initial frequency be $n_1 = n$. The new frequency $n_2$ is given by the ratios of the changes:
$n_2 = n_1 \times \left( \frac{L_1}{L_2} \right) \times \left( \frac{D_1}{D_2} \right) \times \sqrt{\frac{T_2}{T_1}}$.
Given $L_2 = 3L_1$,$D_2 = 2D_1$,and $T_2 = 3T_1$:
$n_2 = n \times \left( \frac{1}{3} \right) \times \left( \frac{1}{2} \right) \times \sqrt{3} = n \times \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{2 \times 3} n = \frac{\sqrt{3}}{2 \times \sqrt{3} \times \sqrt{3}} n = \frac{n}{2 \sqrt{3}}$.
Thus,the new frequency is $\frac{n}{2 \sqrt{3}}$.

(A) The frequency of a vibrating stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the vibrating length,$T$ is the tension,and $\mu$ is the mass per unit length.
Since the frequency $f$ of the tuning fork is constant,we have $L \propto \sqrt{T}$.
The tension $T$ in the wire is provided by a suspended mass $M$,so $T = Mg$,where $g$ is the acceleration due to gravity.
Thus,$L \propto \sqrt{g}$.
At the equator,the value of $g$ is less than at the poles $(g_{equator} < g_{poles})$.
Since $L$ is directly proportional to $\sqrt{g}$,the vibrating length $L$ must be decreased at the equator to maintain the same frequency $f$.

(B) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $f \propto \frac{1}{L}$.
Let the frequency of the tuning fork be $n$.
For length $L_1 = 40 \ cm$,the frequency of the wire is $f_1 = \frac{k}{40}$. Since it produces $5$ beats,$f_1 = n \pm 5$.
For length $L_2 = 39 \ cm$,the frequency of the wire is $f_2 = \frac{k}{39}$. Since it produces $5$ beats,$f_2 = n \pm 5$.
Since $L_2 < L_1$,$f_2 > f_1$. Thus,$f_1 = n - 5$ and $f_2 = n + 5$.
So,$\frac{k}{40} = n - 5$ and $\frac{k}{39} = n + 5$.
From the first equation,$k = 40(n - 5)$.
From the second equation,$k = 39(n + 5)$.
Equating the two: $40n - 200 = 39n + 195$.
$40n - 39n = 195 + 200$.
$n = 395 \ Hz$.

(C) In a Sonometer experiment,the string vibrates in harmonics which are integer multiples of the fundamental frequency $(f_0)$.
Given the fundamental frequency $f_0 = 288 \ Hz$.
The frequencies of the harmonics are given by $f_n = n \times f_0$,where $n = 1, 2, 3, ...$.
Calculating the first few harmonics:
$n=1: f_1 = 1 \times 288 = 288 \ Hz$
$n=2: f_2 = 2 \times 288 = 576 \ Hz$
$n=3: f_3 = 3 \times 288 = 864 \ Hz$
Comparing these with the given options,$844 \ Hz$ is not an integer multiple of $288 \ Hz$. Therefore,harmonics will not be produced at $844 \ Hz$.

(C) For an air column of length $L$ closed at one end,the frequency of the $n^{\text{th}}$ overtone is given by $f_{c} = (2n + 1) \frac{v}{4L}$. For the $5^{\text{th}}$ overtone,$n = 5$,so $f_{c} = (2(5) + 1) \frac{v}{4L} = 11 \frac{v}{4L}$.
For an air column of length $L$ open at both ends,the frequency of the $n^{\text{th}}$ overtone is given by $f_{o} = (n + 1) \frac{v}{2L}$. For the $5^{\text{th}}$ overtone,$n = 5$,so $f_{o} = (5 + 1) \frac{v}{2L} = 6 \frac{v}{2L} = 12 \frac{v}{4L}$.
The ratio of the frequencies is $\frac{f_{c}}{f_{o}} = \frac{11v/4L}{12v/4L} = \frac{11}{12}$.

(C) The fundamental frequency of a sonometer wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since $\mu = \text{Area} \times \text{Density} = (\pi r^2) \rho = \pi (\frac{d}{2})^2 \rho = \frac{\pi d^2 \rho}{4}$,we can write $n \propto \frac{1}{L d} \sqrt{\frac{T}{\rho}}$.
Given the initial conditions: $n_1 = n$,$L_1 = L$,$T_1 = T$,$d_1 = d$.
New conditions: $L_2 = 3L$,$T_2 = 3T$,$d_2 = 2d$.
The new frequency $n_2$ is given by: $n_2 = n_1 \times (\frac{L_1}{L_2}) \times (\frac{d_1}{d_2}) \times \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $n_2 = n \times (\frac{L}{3L}) \times (\frac{d}{2d}) \times \sqrt{\frac{3T}{T}}$.
$n_2 = n \times \frac{1}{3} \times \frac{1}{2} \times \sqrt{3} = \frac{\sqrt{3}}{6} n = \frac{\sqrt{3}}{2 \times 3} n = \frac{n}{2 \sqrt{3}}$.

(C) Let the length of both pipes be $L$. The speed of sound is $v$.
For an open pipe,the frequencies are $f_{open, n} = n(v/2L)$. The first overtone $(n=2)$ is $f_{open, 1st} = 2(v/2L) = v/L$.
For a closed pipe,the frequencies are $f_{closed, n} = (2n-1)(v/4L)$. The first overtone $(n=2)$ is $f_{closed, 1st} = 3(v/4L)$.
Given that the beat frequency is $4$ Hz: $|v/L - 3v/4L| = 4 \implies v/4L = 4 \implies v/L = 16$.
Now,the new length of the open pipe is $L' = L/2$ and the new length of the closed pipe is $L'' = 2L$.
The fundamental frequency of the new open pipe is $f'_{open} = v/(2L') = v/(2(L/2)) = v/L = 16$ Hz.
The fundamental frequency of the new closed pipe is $f''_{closed} = v/(4L'') = v/(4(2L)) = v/8L = 16/8 = 2$ Hz.
The number of beats produced is $|f'_{open} - f''_{closed}| = |16 - 2| = 14$ Hz.

(C) Let the frequency of fork $A$ be $f_A$ and fork $B$ be $f_B$. Given $|f_A - f_B| = 5 \ Hz$.
For a closed pipe of length $L_c = 20 \ cm = 0.2 \ m$,the fundamental frequency is $f_A = \frac{v}{4L_c} = \frac{v}{4 \times 0.2} = \frac{v}{0.8}$.
For an open pipe of length $L_o = 40.5 \ cm = 0.405 \ m$,the fundamental frequency is $f_B = \frac{v}{2L_o} = \frac{v}{2 \times 0.405} = \frac{v}{0.81}$.
Since $f_A > f_B$,we have $f_A - f_B = 5$.
$\frac{v}{0.8} - \frac{v}{0.81} = 5 \implies v \left( \frac{0.81 - 0.8}{0.8 \times 0.81} \right) = 5$.
$v \left( \frac{0.01}{0.648} \right) = 5 \implies v = \frac{5 \times 0.648}{0.01} = 500 \times 0.648 = 324 \ m/s$.
Now,$f_A = \frac{324}{0.8} = 405 \ Hz$ and $f_B = \frac{324}{0.81} = 400 \ Hz$.

(D) For an open pipe of length $L$,the fundamental frequency is $f_{open, 1} = \frac{v}{2L}$. The first overtone is the second harmonic,given by $f_{open, 2} = 2 \times f_{open, 1} = 2 \times \frac{v}{2L} = \frac{v}{L}$.
For a closed pipe of length $L$,the fundamental frequency is $f_{closed, 1} = \frac{v}{4L}$. The first overtone is the third harmonic,given by $f_{closed, 2} = 3 \times f_{closed, 1} = 3 \times \frac{v}{4L} = \frac{3v}{4L}$.
The ratio of the frequency of the first overtone of the open pipe to that of the closed pipe is $\frac{f_{open, 2}}{f_{closed, 2}} = \frac{v/L}{3v/4L} = \frac{1}{1} \times \frac{4}{3} = \frac{4}{3}$.

(A) For an organ pipe closed at one end,the frequencies of the harmonics are given by $f_n = (2n - 1)f_0$,where $n = 1, 2, 3, \dots$ and $f_0 = 1500 \ Hz$.
The fundamental frequency is $f_1 = 1500 \ Hz$.
The overtones are the frequencies higher than the fundamental frequency,given by $f_n = (2n - 1)f_0$ for $n > 1$.
We need $f_n \leq 19500 \ Hz$.
$(2n - 1) \times 1500 \leq 19500$.
$2n - 1 \leq \frac{19500}{1500} = 13$.
$2n \leq 14$,so $n \leq 7$.
The possible values for $n$ are $1, 2, 3, 4, 5, 6, 7$.
Since $n=1$ is the fundamental frequency,the overtones correspond to $n = 2, 3, 4, 5, 6, 7$.
There are $7 - 1 = 6$ overtones.

(B) Let $v_h$ be the speed at the highest point and $v_l$ be the speed at the lowest point.
At the highest point,the tension $T_h$ is given by $T_h = \frac{mv_h^2}{L} - mg$.
At the lowest point,the tension $T_l$ is given by $T_l = \frac{mv_l^2}{L} + mg$.
Using the conservation of energy between the highest and lowest points: $\frac{1}{2}mv_l^2 = \frac{1}{2}mv_h^2 + mg(2L)$,which simplifies to $v_l^2 = v_h^2 + 4gL$.
Substituting $v_l^2$ into the expression for $T_l$: $T_l = \frac{m(v_h^2 + 4gL)}{L} + mg = \frac{mv_h^2}{L} + 4mg + mg = \frac{mv_h^2}{L} + 5mg$.
Given the ratio $\frac{T_l}{T_h} = 3$,we have $\frac{\frac{mv_h^2}{L} + 5mg}{\frac{mv_h^2}{L} - mg} = 3$.
Let $x = \frac{v_h^2}{L}$. Then $\frac{x + 5g}{x - g} = 3 \implies x + 5g = 3x - 3g \implies 2x = 8g \implies x = 4g$.
Since $x = \frac{v_h^2}{L}$,we have $\frac{v_h^2}{L} = 4g$,which gives $v_h^2 = 4gL$.
Therefore,$v_h = 2\sqrt{gL}$.

(B) Given: Tension $T_{max} = 3.7 \ kg \ wt = 3.7 \times 10 \ N = 37 \ N$. Mass $m = 500 \ g = 0.5 \ kg$. Radius $r = 4 \ m$. $g = 10 \ m/s^2$.
In a vertical circular motion,the tension is maximum at the lowest point of the path.
The formula for tension at the lowest point is $T = mg + \frac{mv^2}{r}$.
Substituting the values: $37 = (0.5 \times 10) + \frac{0.5 \times v^2}{4}$.
$37 = 5 + \frac{0.5 \times v^2}{4}$.
$32 = \frac{0.5 \times v^2}{4}$.
$128 = 0.5 \times v^2$.
$v^2 = 256$.
$v = 16 \ m/s$.
Since $v = r\omega$,we have $\omega = \frac{v}{r} = \frac{16}{4} = 4 \ rad/s$.

(B) Let $m$ be the mass of the bob and $l$ be the length of the string. The tension in the string at any angle $\phi$ is given by $T = mg \cos \phi + \frac{mv^2}{l}$.
At the extreme position,$\phi = \theta$ and $v = 0$,so the minimum tension is $T_{min} = mg \cos \theta$.
At the lowest point,$\phi = 0$. Using conservation of energy,$mg(l - l \cos \theta) = \frac{1}{2}mv^2$,so $mv^2 = 2mgl(1 - \cos \theta)$.
The maximum tension is $T_{max} = mg + \frac{mv^2}{l} = mg + 2mg(1 - \cos \theta) = mg(3 - 2 \cos \theta)$.
Given $T_{max} = 4 T_{min}$,we have $mg(3 - 2 \cos \theta) = 4mg \cos \theta$.
$3 - 2 \cos \theta = 4 \cos \theta \implies 6 \cos \theta = 3 \implies \cos \theta = 0.5$.
Therefore,$\theta = \cos^{-1}(0.5)$.

(B) For a particle to complete a vertical circular motion,the minimum velocity at the top point $A$ is $v_A = \sqrt{g\ell}$.
Using the law of conservation of energy between point $A$ and point $C$:
$\frac{1}{2}mv_A^2 + mg(2\ell) = \frac{1}{2}mv_C^2 + mg\ell$
$\frac{1}{2}m(g\ell) + 2mg\ell = \frac{1}{2}mv_C^2 + mg\ell$
$\frac{1}{2}g\ell + mg\ell = \frac{1}{2}mv_C^2$
$\frac{3}{2}g\ell = \frac{1}{2}v_C^2 \implies v_C^2 = 3g\ell$.
The centripetal acceleration at point $A$ is $a_A = \frac{v_A^2}{\ell} = \frac{g\ell}{\ell} = g$.
The centripetal acceleration at point $C$ is $a_C = \frac{v_C^2}{\ell} = \frac{3g\ell}{\ell} = 3g$.
The increase in centripetal acceleration is $\Delta a = a_C - a_A = 3g - g = 2g$.

(D) Given: Mass $= m$ $kg$,Radius $r = 49$ $cm = 0.49$ $m$,Frequency $f = 30$ $rpm = 0.5$ $rev/s$.
Angular velocity $\omega = 2\pi f = 2 \times \pi \times 0.5 = \pi$ $rad/s$.
At the lowermost point,the tension $T$ is given by $T = mg + mr\omega^2$.
Substituting the values: $T = m(10) + m(0.49)(\pi^2)$.
Using $\pi^2 = 10$: $T = 10m + m(0.49)(10) = 10m + 4.9m = 14.9m$ $N$.
Rounding to the nearest integer,$T \approx 15m$ $N$.

(C) Let the initial velocity of the stone be $u$. The initial kinetic energy is $E = \frac{1}{2} m u^2$.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,while the horizontal component remains constant at $v_x = u \cos \theta$.
Therefore,the kinetic energy at the highest point is $E' = \frac{1}{2} m v_x^2$.
Substituting $v_x = u \cos \theta$,we get $E' = \frac{1}{2} m (u \cos \theta)^2$.
$E' = \frac{1}{2} m u^2 \cos^2 \theta$.
Since $E = \frac{1}{2} m u^2$,we have $E' = E \cos^2 \theta$.

(C) Let the length of the pendulum be $L = 1 \,m$ and the initial speed at the lowest point be $v_0 = 4 \,m/s$.
At the lowest point, the potential energy is taken as $0$. The total energy is $E = \frac{1}{2}mv_0^2$.
When the string makes an angle $\theta = 60^{\circ}$ with the vertical, the height of the bob is $h = L(1 - \cos \theta)$.
Substituting the values, $h = 1(1 - \cos 60^{\circ}) = 1(1 - 0.5) = 0.5 \,m$.
By the law of conservation of energy, the total energy at the lowest point equals the total energy at the angle $\theta$:
$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgh$
$v^2 = v_0^2 - 2gh$
$v^2 = (4)^2 - 2(10)(0.5)$
$v^2 = 16 - 10 = 6$
$v = \sqrt{6} \,m/s$.

(A) Given: Mass $m = 1 \ kg$,Force $\vec{F} = t \hat{i} + 2t^2 \hat{j}$.
Using Newton's second law,$\vec{F} = m \vec{a}$,so $\vec{a} = \frac{\vec{F}}{m} = t \hat{i} + 2t^2 \hat{j}$.
Velocity $\vec{v} = \int \vec{a} \ dt = \int (t \hat{i} + 2t^2 \hat{j}) \ dt = \frac{t^2}{2} \hat{i} + \frac{2t^3}{3} \hat{j}$ (assuming initial velocity is zero).
Power $P = \vec{F} \cdot \vec{v} = (t \hat{i} + 2t^2 \hat{j}) \cdot (\frac{t^2}{2} \hat{i} + \frac{2t^3}{3} \hat{j})$.
$P = \frac{t^3}{2} + \frac{4t^5}{3}$.
At $t = 3 \ s$:
$P = \frac{3^3}{2} + \frac{4(3^5)}{3} = \frac{27}{2} + 4(3^4) = 13.5 + 4(81) = 13.5 + 324 = 337.5 \ W$.

(D) The displacement vector $\vec{d}$ is given by $\vec{d} = \vec{r}_Q - \vec{r}_P$.
Given $\vec{r}_P = (3 \hat{i} + 4 \hat{j} + 5 \hat{k}) \ m$ and $\vec{r}_Q = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \ m$.
So,$\vec{d} = (2-3) \hat{i} + (3-4) \hat{j} + (4-5) \hat{k} = (-1 \hat{i} - 1 \hat{j} - 1 \hat{k}) \ m$.
The work done $W$ by a constant force $\vec{F}$ is given by the dot product $W = \vec{F} \cdot \vec{d}$.
$W = (3 \hat{i} + 4 \hat{j} + 5 \hat{k}) \cdot (-1 \hat{i} - 1 \hat{j} - 1 \hat{k})$.
$W = (3 \times -1) + (4 \times -1) + (5 \times -1) = -3 - 4 - 5 = -12 \ J$.

(B) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{r}$.
Substituting the given values: $W = (3\hat{i} - 2\hat{j} - \hat{k}) \cdot (2\hat{i} - 3\hat{j} - 3\hat{k})$.
$W = (3 \times 2) + (-2 \times -3) + (-1 \times -3) = 6 + 6 + 3 = 15 \text{ J}$.
Power $P$ is defined as the rate of work done: $P = \frac{W}{t}$.
Given $t = 2 \text{ s}$,$P = \frac{15}{2} = 7.5 \text{ W}$.
Thus,the work done is $15 \text{ J}$ and the power is $7.5 \text{ W}$.

(C) $1$. The upper branch consists of an $AND$ gate followed by a $NOT$ gate ($NAND$ gate). The inputs to the $AND$ gate are $A$ and $B$,so its output is $(A \cdot B)$. After the $NOT$ gate,the output becomes $(\overline{A \cdot B})$.
$2$. The lower branch consists of a $NOT$ gate on input $A$ (giving $\overline{A}$) followed by an $AND$ gate with input $B$. Thus,the output of this branch is $(\overline{A} \cdot B)$.
$3$. These two outputs are fed into an $OR$ gate. Therefore,the final output $Y$ is the sum of these two expressions: $Y = (\overline{A \cdot B}) + (\overline{A} \cdot B)$.
$4$. Comparing this with the given options,option $C$ matches our derived expression.

(C) The given circuit consists of a $NOR$ gate followed by an $AND$ gate.
Inputs $B$ and $C$ are fed into a $NOR$ gate,which produces an output $X = \overline{B+C}$.
This output $X$ and input $A$ are then fed into an $AND$ gate to produce the final output $Y$.
Therefore,the Boolean expression for the output is $Y = A \cdot X = A \cdot (\overline{B+C})$.
For the output $Y$ to be $HIGH$ $(Y=1)$,both $A$ must be $1$ and $(\overline{B+C})$ must be $1$.
$(\overline{B+C}) = 1$ only when $B+C = 0$,which means $B=0$ and $C=0$.
Thus,$Y=1$ when $A=1, B=0, C=0$.

(D) The given logic circuit consists of a $NOR$ gate followed by an $AND$ gate.
Let the output of the $NOR$ gate be $X$. Then $X = \overline{A + B}$.
The final output $Y$ is the output of the $AND$ gate,so $Y = X \cdot C = (\overline{A + B}) \cdot C$.
We want the output $Y = 0$.
Let's check the options:
$A) A=1, B=1, C=0 \implies Y = (\overline{1+1}) \cdot 0 = 0 \cdot 0 = 0$.
$B) A=0, B=1, C=0 \implies Y = (\overline{0+1}) \cdot 0 = 0 \cdot 0 = 0$.
$C) A=1, B=0, C=1 \implies Y = (\overline{1+0}) \cdot 1 = 0 \cdot 1 = 0$.
$D) A=0, B=0, C=1 \implies Y = (\overline{0+0}) \cdot 1 = 1 \cdot 1 = 1$.
Since the question asks for the input set that does $NOT$ result in an output of $0$,the correct answer is $D$.

(C) The logic gate that produces a '$HIGH$' or '$1$' output only when an odd number of inputs are '$HIGH$' or '$1$' is the Exclusive-$OR$ (Ex-$OR$) gate.
For a two-input Ex-$OR$ gate,the output $Y$ is given by $Y = A \oplus B = A\bar{B} + \bar{A}B$.
The truth table for a two-input Ex-$OR$ gate is:
If $A=0, B=0$,then $Y=0$.
If $A=0, B=1$,then $Y=1$.
If $A=1, B=0$,then $Y=1$.
If $A=1, B=1$,then $Y=0$.
As observed,the output is '$1$' only when the number of '$HIGH$' inputs is odd (i.e.,$1$ input is '$HIGH$').

(D) The given circuit consists of an $OR$ gate followed by an $AND$ gate.
Let the output of the $OR$ gate be $X$.
The inputs to the $OR$ gate are $B$ and $C$,so $X = B + C$.
The inputs to the $AND$ gate are $A$ and $X$.
Therefore,the final output is $Y = A \cdot X = A \cdot (B + C)$.
For the output $Y = 1$,both $A$ must be $1$ and $(B + C)$ must be $1$.
Checking the options:
$A) 0, 0, 0 \implies Y = 0 \cdot (0 + 0) = 0$
$B) 0, 1, 0 \implies Y = 0 \cdot (1 + 0) = 0$
$C) 1, 0, 0 \implies Y = 1 \cdot (0 + 0) = 0$
$D) 1, 0, 1 \implies Y = 1 \cdot (0 + 1) = 1 \cdot 1 = 1$
Thus,the set of inputs $A=1, B=0, C=1$ gives the output $Y=1$.

(C) The truth table for the given conditions is:
$A=0, B=0 \implies C=1$
$A=0, B=1 \implies C=1$
Let us check the truth tables for the given options:
$1$. $OR$ gate: $0+0=0, 0+1=1, 1+0=1, 1+1=1$. (Does not match)
$2$. $AND$ gate: $0 \cdot 0=0, 0 \cdot 1=0, 1 \cdot 0=0, 1 \cdot 1=1$. (Does not match)
$3$. $NAND$ gate: $\overline{0 \cdot 0}=1, \overline{0 \cdot 1}=1, \overline{1 \cdot 0}=1, \overline{1 \cdot 1}=0$. (Matches the given conditions)
$4$. $NOR$ gate: $\overline{0+0}=1, \overline{0+1}=0, \overline{1+0}=0, \overline{1+1}=0$. (Does not match)
Thus,the logic gate is a $NAND$ gate.

(A) Let's analyze each logic gate with the given inputs:
$(A)$ This is a $NAND$ gate with inputs $1$ and $1$. The output of a $NAND$ gate is $Y = \overline{A \cdot B}$. For $A=1, B=1$,$Y = \overline{1 \cdot 1} = \overline{1} = 0$.
$(B)$ This is a $NOR$ gate with inputs $0$ and $0$. The output of a $NOR$ gate is $Y = \overline{A + B}$. For $A=0, B=0$,$Y = \overline{0 + 0} = \overline{0} = 1$.
$(C)$ This is an $AND$ gate with inputs $1$ and $0$. The output of an $AND$ gate is $Y = A \cdot B$. For $A=1, B=0$,$Y = 1 \cdot 0 = 0$.
$(D)$ This is an $XOR$ gate with inputs $0$ and $1$. The output of an $XOR$ gate is $Y = A \oplus B$. For $A=0, B=1$,$Y = 0 \oplus 1 = 1$.
Comparing the outputs,gates $(A)$ and $(C)$ give an output of '$0$'.
Therefore,the correct option is $(A)$ and $(C)$.

(B) $1$. An $AND$ gate takes two inputs $A$ and $B$ and produces an output $X = A \cdot B$.
$2$. This output $X$ is then passed through a $NOT$ gate.
$3$. $A$ $NOT$ gate inverts the input,so the final output $Y = \overline{X}$.
$4$. Substituting the value of $X$,we get $Y = \overline{A \cdot B}$.
$5$. This combination of an $AND$ gate followed by a $NOT$ gate is known as a $NAND$ gate.

(C) In a centre-tap full-wave rectifier,the secondary winding of the transformer is divided into two equal halves by the centre tap.
If the total secondary voltage across the entire secondary winding is $V_s$,then the voltage across each half of the secondary winding is $\frac{V_s}{2}$.
During the positive half-cycle,one diode conducts and the voltage across the load is the voltage across the top half of the secondary,which is $\frac{V_s}{2}$.
During the negative half-cycle,the other diode conducts and the voltage across the load is the voltage across the bottom half of the secondary,which is also $\frac{V_s}{2}$.
Therefore,the output voltage with respect to each diode is $\frac{1}{2} V_s$.

(C) In a semiconductor,as the temperature $(T)$ increases,the number of charge carriers (electrons and holes) increases exponentially due to the breaking of covalent bonds.
This increase in the number of charge carriers dominates over the effect of increased lattice vibrations (which would otherwise increase resistivity).
Consequently,the resistivity $(\varrho)$ of a semiconductor decreases rapidly with an increase in temperature.
The relationship is given by $\varrho = \varrho_0 e^{E_g / 2k_BT}$,where $E_g$ is the band gap energy and $k_B$ is the Boltzmann constant.
Graph $(C)$ correctly represents this exponential decrease of resistivity with temperature.

(B) Given: Intrinsic carrier concentration $n_i = 10^{16} \ m^{-3}$.
Phosphorus is a pentavalent impurity,so it acts as a donor. The concentration of donor atoms added is $N_D = 10^{21} \ m^{-3}$.
Since $N_D \gg n_i$,the electron concentration in the n-type semiconductor is approximately equal to the donor concentration: $n_e \approx N_D = 10^{21} \ m^{-3}$.
According to the law of mass action for semiconductors,$n_e \cdot n_h = n_i^2$,where $n_h$ is the hole concentration.
Substituting the values: $10^{21} \cdot n_h = (10^{16})^2$.
$10^{21} \cdot n_h = 10^{32}$.
$n_h = \frac{10^{32}}{10^{21}} = 10^{11} \ m^{-3}$.
Therefore,the new hole concentration is $10^{11} \ m^{-3}$.

(C) In an $n$-type semiconductor,pentavalent impurity atoms (like $P, As, Sb$) are added to the intrinsic semiconductor (like $Si$ or $Ge$).
These impurity atoms provide extra electrons for conduction.
The energy levels corresponding to these donor electrons are called donor energy levels $(E_D)$.
These donor energy levels are located in the forbidden energy gap,just below the conduction band edge $(E_C)$.
Because they are very close to the conduction band,the electrons can easily gain thermal energy at room temperature and jump into the conduction band to contribute to electrical conductivity.

(C) In an $n$-type semiconductor,the intrinsic semiconductor (like $Si$ or $Ge$) is doped with pentavalent impurity atoms (such as $P, As, Sb$).
These pentavalent atoms provide extra electrons to the conduction band,making electrons the majority charge carriers.
Conversely,holes are generated only due to thermal excitation,making them the minority charge carriers.
Therefore,the statement that 'Holes are minority carriers and pentavalent atoms are dopants' is correct.

(C) In a semiconductor,the number of charge carriers (electrons and holes) increases exponentially with an increase in temperature because more covalent bonds are broken,releasing more charge carriers.
As the number of charge carriers increases,the electrical conductivity of the semiconductor increases,which leads to a decrease in its electrical resistance.
According to Ohm's law,$I = V/R$. Since the voltage $V$ remains constant and the resistance $R$ decreases,the current $I$ in the circuit will increase.

(C) In an $n$-type semiconductor,the semiconductor is doped with pentavalent impurity atoms (like $P$,$As$,$Sb$).
These pentavalent atoms provide extra electrons to the conduction band,making electrons the majority charge carriers.
Conversely,holes are generated only due to thermal excitation,making them the minority charge carriers.
Therefore,electrons are majority carriers and pentavalent atoms are dopants,while holes are minority carriers.
Comparing this with the given options,option $C$ is correct because it states that holes are minority carriers and pentavalent atoms are dopants.

(D) solar cell is a device that converts solar energy into electrical energy. The solar radiation spectrum has a maximum intensity around $1.5 \ eV$. To efficiently absorb solar radiation,the semiconductor material used in a solar cell must have a band gap that matches this energy range. Materials with a band gap between $1.0 \ eV$ and $1.8 \ eV$ are ideal for this purpose,as they can absorb a significant portion of the solar spectrum. Silicon,with a band gap of approximately $1.1 \ eV$,is the most commonly used material for solar cells.

(B) The wavelength of light emitted by an $LED$ depends on the band gap energy $(E_g)$ of the semiconductor material used.
Aluminium gallium arsenide $(AlGaAs)$ is a semiconductor material with a band gap energy that corresponds to the infra-red region of the electromagnetic spectrum.
Therefore,LEDs manufactured using $AlGaAs$ emit infra-red radiation.

(A) The linear width of the central maximum in a Fraunhofer diffraction pattern is given by the formula $w = \frac{2 \lambda D}{d}$.
According to the problem,the linear width of the central maximum is equal to two times the width of the slit,so $w = 2d$.
Equating the two expressions: $\frac{2 \lambda D}{d} = 2d$.
Dividing both sides by $2$,we get $\frac{\lambda D}{d} = d$.
Solving for $D$,we find $D = \frac{d^2}{\lambda}$.

(C) For Fraunhofer diffraction,the position of the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$. For small angles,$\sin \theta \approx \theta = \frac{y_n}{D}$,where $y_n$ is the distance from the central maximum.
Thus,$y_n = \frac{n \lambda D}{a}$.
The distance between the first minima on either side of the central maximum is $2y_1 = \frac{2 \lambda D}{a}$.
Given: $a = 0.2 \ mm = 2 \times 10^{-4} \ m$,$D = 2 \ m$,and $2y_1 = 1 \ cm = 10^{-2} \ m$.
Substituting the values: $10^{-2} = \frac{2 \times \lambda \times 2}{2 \times 10^{-4}}$.
$10^{-2} = \frac{4 \lambda}{2 \times 10^{-4}} = 2 \times 10^4 \lambda$.
$\lambda = \frac{10^{-2}}{2 \times 10^4} = 0.5 \times 10^{-6} \ m = 5000 \times 10^{-10} \ m = 5000 \ Å$.

(D) In a single slit diffraction pattern,the central maximum is the brightest and widest. As we move away from the center,the intensity of the secondary maxima decreases rapidly. The width of the central maximum is $2\lambda D/a$,while the width of the secondary maxima is $\lambda D/a$. Therefore,the fringes have unequal width and unequal intensity. Option $D$ states that the fringes have equal width and equal intensity,which is incorrect.

(B) According to the Rayleigh criterion for a circular aperture,the angular resolution $\theta$ is given by $\theta = 1.22 \lambda / D$,where $\lambda$ is the wavelength and $D$ is the diameter of the aperture.
Given: Wavelength $\lambda = 500 \ nm = 500 \times 10^{-9} \ m$,Radius $r = 0.25 \ cm = 2.5 \times 10^{-3} \ m$,so Diameter $D = 2r = 5.0 \times 10^{-3} \ m$.
The angular resolution is $\theta = (1.22 \times 500 \times 10^{-9}) / (5.0 \times 10^{-3}) = 1.22 \times 10^{-4} \ rad$.
The minimum separation $d$ at a distance $L = 25 \ cm = 0.25 \ m$ is $d = L \times \theta$.
$d = 0.25 \times 1.22 \times 10^{-4} = 0.305 \times 10^{-4} \ m = 30.5 \times 10^{-6} \ m = 30.5 \mu m$.
Thus,the minimum separation is nearly $30 \mu m$.

(C) For a single slit diffraction pattern,the condition for the $n^{th}$ secondary maximum is given by: $a \sin \theta = (n + \frac{1}{2}) \lambda$,where $a$ is the slit width,$\theta$ is the angle of diffraction,and $\lambda$ is the wavelength.
Since $\theta$ is very small,$\sin \theta \approx \tan \theta = \frac{y}{D}$,where $y$ is the distance from the center and $D$ is the distance to the screen.
For the second maximum,$n = 2$. Thus,$a \frac{y}{D} = (2 + \frac{1}{2}) \lambda = \frac{5}{2} \lambda$.
Given: $a = 0.65 \ mm = 0.65 \times 10^{-3} \ m$,$D = 1.3 \ m$,and $y = 2.6 \ mm = 2.6 \times 10^{-3} \ m$.
Substituting the values: $(0.65 \times 10^{-3}) \times \frac{2.6 \times 10^{-3}}{1.3} = \frac{5}{2} \lambda$.
$(0.65 \times 10^{-3}) \times (2 \times 10^{-3}) = 2.5 \lambda$.
$1.3 \times 10^{-6} = 2.5 \lambda$.
$\lambda = \frac{1.3 \times 10^{-6}}{2.5} = 0.52 \times 10^{-6} \ m = 5200 \ \times 10^{-10} \ m = 5200 \ Å$.

(B) For a single slit diffraction pattern,the condition for secondary maxima is given by $a \sin \theta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, \dots$ is the order of the secondary maximum.
For the second secondary maximum $(n = 2)$ with wavelength $\lambda_1 = 6384 Å$,the condition is $a \sin \theta = (2 + \frac{1}{2}) \lambda_1 = \frac{5}{2} \lambda_1$.
For the third secondary maximum $(n = 3)$ with wavelength $\lambda_0$,the condition is $a \sin \theta = (3 + \frac{1}{2}) \lambda_0 = \frac{7}{2} \lambda_0$.
Since the positions coincide,we equate the two expressions: $\frac{5}{2} \lambda_1 = \frac{7}{2} \lambda_0$.
This simplifies to $5 \lambda_1 = 7 \lambda_0$.
Substituting $\lambda_1 = 6384 Å$,we get $\lambda_0 = \frac{5 \times 6384}{7} = \frac{31920}{7} = 4560 Å$.

(D) Given: Slit width $a = 0.3 \ mm = 0.3 \times 10^{-3} \ m$,Screen distance $D = 1.5 \ m$,Wavelength $\lambda = 4500 \ Å = 4500 \times 10^{-10} \ m$.
For Fraunhofer diffraction,the position of the $n^{th}$ minimum is given by $a \sin \theta = n \lambda$. For small $\theta$,$\sin \theta \approx \theta = \frac{y}{D}$.
Thus,$a \left( \frac{y}{D} \right) = n \lambda \implies y_n = \frac{n \lambda D}{a}$.
The distance of the first minimum $(n=1)$ from the central maximum is $y_1 = \frac{\lambda D}{a}$.
The distance between the first minimum on either side of the central maximum is $2y_1 = \frac{2 \lambda D}{a}$.
Substituting the values: $2y_1 = \frac{2 \times 4500 \times 10^{-10} \times 1.5}{0.3 \times 10^{-3}}$.
$2y_1 = \frac{13500 \times 10^{-10}}{0.3 \times 10^{-3}} = 45000 \times 10^{-7} \ m = 4.5 \times 10^{-3} \ m = 4.5 \ mm$.

(C) The width of the central maximum in a single slit diffraction experiment is given by the formula: $y = \frac{2D\lambda}{a}$,where '$D$' is the distance of the screen from the slit,'$\lambda$' is the wavelength of light,and '$a$' is the width of the slit.
Initially,$y = \frac{2D\lambda}{a}$.
In the second case,the slit width is halved,so the new width is $a' = \frac{a}{2}$.
The new wavelength is $\lambda' = 1.5\lambda = \frac{3}{2}\lambda$.
The new width of the central maximum '$y'$' is given by: $y' = \frac{2D\lambda'}{a'} = \frac{2D(1.5\lambda)}{a/2}$.
Simplifying this,we get: $y' = \frac{2D(3/2\lambda)}{a/2} = \frac{3D\lambda}{a/2} = \frac{6D\lambda}{a}$.
Since $y = \frac{2D\lambda}{a}$,we can substitute this into the expression for $y'$:
$y' = 3 \times (\frac{2D\lambda}{a}) = 3y$.
Therefore,the new width of the central maximum is $3y$.

(B) The condition for the $n^{\text{th}}$ secondary maximum in a single slit diffraction pattern is given by: $a \sin \theta = (n + \frac{1}{2}) \lambda'$,where $n$ is the order of the secondary maximum.
For the $4^{\text{th}}$ secondary maximum of wavelength $\lambda_1$,the condition is: $a \sin \theta_1 = (4 + \frac{1}{2}) \lambda_1 = \frac{9}{2} \lambda_1$.
For the $3^{\text{rd}}$ secondary maximum of wavelength $\lambda$,the condition is: $a \sin \theta_2 = (3 + \frac{1}{2}) \lambda = \frac{7}{2} \lambda$.
Since the maxima coincide,$\theta_1 = \theta_2$,so $a \sin \theta_1 = a \sin \theta_2$.
Equating the two expressions: $\frac{9}{2} \lambda_1 = \frac{7}{2} \lambda$.
Solving for $\lambda_1$: $\lambda_1 = \frac{7}{9} \lambda$.

(B) Let the amplitudes of the two coherent sources be $a_1$ and $a_2$.
The intensity of the bright fringe is $I_{max} = (a_1 + a_2)^2$.
The intensity of the dark fringe is $I_{min} = (a_1 - a_2)^2$.
Given the ratio of intensities is $\frac{I_{max}}{I_{min}} = \frac{5}{1}$.
Thus,$\frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \frac{5}{1}$.
Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = \frac{\sqrt{5}}{1}$.
The resultant amplitude of the bright fringe is $A_{max} = a_1 + a_2$ and the resultant amplitude of the dark fringe is $A_{min} = a_1 - a_2$.
Therefore,the ratio of the resultant amplitudes is $\frac{A_{max}}{A_{min}} = \frac{a_1 + a_2}{a_1 - a_2} = \frac{\sqrt{5}}{1}$.

(B) Let the incident intensity be $I$.
At point $A$,$25\%$ of $I$ is reflected as ray $AB$. So,$I_1 = 0.25I = I/4$.
The intensity of the refracted ray is $0.75I$.
At point $C$,this ray is reflected. $25\%$ of the incident energy $(0.75I)$ is reflected. So,the intensity of the ray reaching $A^1$ is $0.25 \times 0.75I = 0.1875I = (3/16)I$.
At point $A^1$,this ray is refracted out as $A^1B^1$. Since $25\%$ is reflected at $A^1$,$75\%$ of the incident energy is transmitted.
So,$I_2 = 0.75 \times (3/16)I = (3/4) \times (3/16)I = (9/64)I$.
The amplitudes are $a_1 = \sqrt{I_1} = \sqrt{I/4} = (1/2)\sqrt{I}$ and $a_2 = \sqrt{I_2} = \sqrt{9I/64} = (3/8)\sqrt{I}$.
The ratio of amplitudes is $a_1/a_2 = (1/2) / (3/8) = 4/3$.
$I_{\text{max}} / I_{\text{min}} = (a_1 + a_2)^2 / (a_1 - a_2)^2 = ((4+3)/ (4-3))^2 = (7/1)^2 = 49/1$.

(C) The resultant intensity $I_R$ in an interference pattern is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference.
Given $I_1 = I$ and $I_2 = 9I$.
At point $P$,the phase difference $\phi_P = \pi / 2$. Thus,$I_P = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi / 2) = 10I + 6I(0) = 10I$.
At point $Q$,the phase difference $\phi_Q = \pi$. Thus,$I_Q = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi) = 10I + 6I(-1) = 10I - 6I = 4I$.
The difference between the resultant intensities at points $P$ and $Q$ is $|I_P - I_Q| = |10I - 4I| = 6I$.

(B) Let the intensities of the two coherent sources be $I_1$ and $I_2$. The ratio of intensities is given as $b = \frac{I_1}{I_2}$.
We know that $I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$.
The required ratio is $R = \frac{I_{\text{max}} + I_{\text{min}}}{I_{\text{max}} - I_{\text{min}}}$.
Substituting the expressions for $I_{\text{max}}$ and $I_{\text{min}}$:
$R = \frac{(\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2 - (\sqrt{I_1} - \sqrt{I_2})^2}$.
Expanding the terms:
$R = \frac{(I_1 + I_2 + 2\sqrt{I_1I_2}) + (I_1 + I_2 - 2\sqrt{I_1I_2})}{(I_1 + I_2 + 2\sqrt{I_1I_2}) - (I_1 + I_2 - 2\sqrt{I_1I_2})} = \frac{2(I_1 + I_2)}{4\sqrt{I_1I_2}} = \frac{I_1 + I_2}{2\sqrt{I_1I_2}}$.
Dividing numerator and denominator by $I_2$:
$R = \frac{\frac{I_1}{I_2} + 1}{2\sqrt{\frac{I_1}{I_2}}} = \frac{b + 1}{2\sqrt{b}}$.

(A) In the phenomenon of interference, the intensity of light is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$, where $I_1$ and $I_2$ are the intensities of the two waves.
Since $I \propto A^2$, where $A$ is the amplitude, the intensities at the points of destructive interference are given by $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (A_1 - A_2)^2$.
If the amplitudes $A_1$ and $A_2$ are different, then $(A_1 - A_2)^2 \neq 0$.
Therefore, the intensity at the region of destructive interference is not zero, meaning there is some residual intensity of light.

(A) Let the initial intensity of unpolarized light be $I_0 = 256 \ W/m^2$.
When unpolarized light passes through the first polaroid $P_1$,the intensity of the transmitted light becomes $I_1 = I_0 / 2 = 256 / 2 = 128 \ W/m^2$.
The pass axis of $P_2$ is at an angle $\theta_1 = 60^{\circ}$ with respect to $P_1$. According to Malus' Law,the intensity after $P_2$ is $I_2 = I_1 \cos^2(\theta_1) = 128 \times \cos^2(60^{\circ}) = 128 \times (0.5)^2 = 128 \times 0.25 = 32 \ W/m^2$.
The pass axis of $P_3$ is at an angle of $90^{\circ}$ with respect to $P_1$. The angle between the pass axes of $P_2$ and $P_3$ is $\theta_2 = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Applying Malus' Law again for $P_3$,the final intensity at point $O$ is $I_3 = I_2 \cos^2(\theta_2) = 32 \times \cos^2(30^{\circ}) = 32 \times (\sqrt{3}/2)^2 = 32 \times (3/4) = 24 \ W/m^2$.

(B) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the transmitted light is $I_1 = I_0 / 2$.
For subsequent polaroids,we use Malus' Law: $I_n = I_{n-1} \cos^2 \theta$,where $\theta = 30^{\circ}$.
Intensity after the second polaroid: $I_2 = I_1 \cos^2 30^{\circ} = (I_0 / 2) \times (\sqrt{3} / 2)^2 = (I_0 / 2) \times (3 / 4) = 3 I_0 / 8$.
Intensity after the third polaroid: $I_3 = I_2 \cos^2 30^{\circ} = (3 I_0 / 8) \times (3 / 4) = 9 I_0 / 32$.
Intensity after the fourth polaroid: $I_4 = I_3 \cos^2 30^{\circ} = (9 I_0 / 32) \times (3 / 4) = 27 I_0 / 128$.
Thus,the correct option is $B$.

(B) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the emergent light is $I_1 = \frac{I_0}{2}$.
According to Malus' Law,the intensity of light emerging from a polaroid is $I = I_{incident} \cos^2 \theta$,where $\theta$ is the angle between the pass axis of the polaroid and the plane of polarisation of the incident light.
For the second polaroid,the angle between its pass axis and the pass axis of the first polaroid is $\theta_1 = 30^{\circ}$. Thus,the intensity of light emerging from the second polaroid is $I_2 = I_1 \cos^2 30^{\circ} = \left(\frac{I_0}{2}\right) \left(\frac{\sqrt{3}}{2}\right)^2 = \left(\frac{I_0}{2}\right) \left(\frac{3}{4}\right) = \frac{3 I_0}{8}$.
For the third polaroid,the angle between its pass axis and the pass axis of the second polaroid is $\theta_2 = 90^{\circ} - 30^{\circ} = 60^{\circ}$. Thus,the intensity of light emerging from the third polaroid is $I_3 = I_2 \cos^2 60^{\circ} = \left(\frac{3 I_0}{8}\right) \left(\frac{1}{2}\right)^2 = \left(\frac{3 I_0}{8}\right) \left(\frac{1}{4}\right) = \frac{3 I_0}{32}$.

(B) $1$. When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the transmitted light becomes $I_1 = \frac{I_0}{2}$.
$2$. The first and second polaroids are crossed (angle between axes is $90^\circ$),so no light passes through the second polaroid initially.
$3$. $A$ third polaroid is placed between them at an angle $\theta$ with the first polaroid. The angle between the third and the second polaroid is $(90^\circ - \theta)$.
$4$. Intensity after the third polaroid: $I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta$.
$5$. Intensity after the second (last) polaroid using Malus' Law: $I_3 = I_2 \cos^2(90^\circ - \theta) = I_2 \sin^2 \theta$.
$6$. Substituting $I_2$: $I_3 = (\frac{I_0}{2} \cos^2 \theta) \sin^2 \theta = \frac{I_0}{2} (\sin \theta \cos \theta)^2 = \frac{I_0}{2} (\frac{\sin 2 \theta}{2})^2 = \frac{I_0}{8} \sin^2 2 \theta$.

(D) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the transmitted light becomes $I_1 = I_0 / 2$.
This light is now plane-polarised.
When this light passes through the second polaroid whose transmission axis makes an angle $\theta = 30^{\circ}$ with the first,the intensity of the transmitted light $I_2$ is given by Malus' Law: $I_2 = I_1 \cos^2 \theta$.
Substituting the values: $I_2 = (I_0 / 2) \cos^2 30^{\circ}$.
Since $\cos 30^{\circ} = \sqrt{3} / 2$,we have $\cos^2 30^{\circ} = 3 / 4$.
Therefore,$I_2 = (I_0 / 2) \times (3 / 4) = 3 I_0 / 8$.
Calculating the fraction: $3 / 8 = 0.375$,which is $37.5 \%$.

(D) According to Brewster's Law,the refractive index $\mu$ of a medium is given by $\mu = \tan(i_p)$,where $i_p$ is the polarising angle.
Also,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 $\mu$,we get: $\frac{C}{V} = \tan(i_p)$.
This can be written as $\frac{C}{V} = \frac{\sin(i_p)}{\cos(i_p)}$.
Rearranging the terms,we get $C \cos(i_p) = V \sin(i_p)$,which is equivalent to $V \sin(i_p) = C \cos(i_p)$.

(A) Brewster's law states that $\tan(i_B) = \mu_{21}$, where $\mu_{21} = \frac{\mu_2}{\mu_1}$ is the refractive index of the second medium with respect to the first.
For light propagating from air $(\mu_1 = 1)$ to glass $(\mu_2 = \mu)$, the Brewster's angle $\theta$ is given by $\tan(\theta) = \frac{\mu}{1} = \mu$.
For light propagating from glass $(\mu_1 = \mu)$ to air $(\mu_2 = 1)$, the Brewster's angle $i_B'$ is given by $\tan(i_B') = \frac{1}{\mu}$.
Since $\tan(\theta) = \mu$, we have $\frac{1}{\mu} = \frac{1}{\tan(\theta)} = \cot(\theta) = \tan(\frac{\pi}{2} - \theta)$.
Therefore, $i_B' = \frac{\pi}{2} - \theta$.

(C) According to Brewster's Law,the refractive index $\mu$ of a medium is given by $\mu = \tan \theta$,where $\theta$ is the polarising angle.
We also know that the refractive index $\mu$ is the ratio of the speed of light in air $(c)$ to the speed of light in the medium $(v)$,so $\mu = \frac{c}{v}$.
Equating the two expressions for $\mu$,we get $\tan \theta = \frac{c}{v}$.
Therefore,$\theta = \tan^{-1}\left(\frac{c}{v}\right)$.
However,looking at the options provided,there seems to be a reciprocal relationship intended. If we define the refractive index as $\mu = \frac{1}{\tan \theta}$ or if the question implies the angle of incidence relative to the surface,we re-evaluate: $\tan \theta = \frac{c}{v} \implies \cot \theta = \frac{v}{c}$.
Thus,$\theta = \cot^{-1}\left(\frac{v}{c}\right)$.

(A) According to Brewster's Law,the refractive index $\mu$ of the glass with respect to air is given by $\mu = \tan \theta$,where $\theta$ is the polarising angle.
We know that the refractive index $\mu$ is also defined as the ratio of the speed of light in air $(c_a)$ to the speed of light in glass $(c_g)$,which is $\mu = \frac{c_a}{c_g}$.
Since the frequency $f$ of light remains constant when it travels from one medium to another,the speed of light is related to wavelength by $c = f \lambda$.
Therefore,$\mu = \frac{f \lambda_a}{f \lambda_g} = \frac{\lambda_a}{\lambda_g}$.
Equating the two expressions for $\mu$,we get $\tan \theta = \frac{\lambda_a}{\lambda_g}$.
Rearranging this,we find $\lambda_a = \lambda_g \tan \theta$. However,looking at the options provided,there seems to be a discrepancy in the standard form. Re-evaluating the relationship: $\mu = \frac{\lambda_a}{\lambda_g} = \tan \theta$. Thus,$\lambda_g = \frac{\lambda_a}{\tan \theta} = \lambda_a \cot \theta$.

(B) Let the intensity of the incident light be $I_0$.
When light passes through the first polaroid,its intensity becomes $I_1 = I_0/2$ (assuming unpolarized light).
For the second polaroid,the angle between the transmission axis and the incident light is $\theta = 60^{\circ}$. According to Malus' Law,$I_2 = I_1 \cos^2(60^{\circ}) = (I_0/2) \times (1/2)^2 = I_0/8$.
For the third polaroid,the angle between its transmission axis and the second polaroid's axis is also $\theta = 60^{\circ}$.
Thus,$I_3 = I_2 \cos^2(60^{\circ}) = (I_0/8) \times (1/2)^2 = I_0/32$.
The fraction of the incident light intensity that passes through the system is $I_3/I_0 = 1/32$.

(B) Huygens' wave theory of light proposes that light travels as waves in a hypothetical medium called the luminiferous ether.
According to this theory,different colours of light correspond to different wavelengths (or frequencies) of these waves.
Option $A$ is correct as it aligns with the wave nature of light.
Option $B$ refers to Newton's Corpuscular theory,which suggests light consists of particles (corpuscles) of different sizes for different colours. This is $NOT$ part of Huygens' wave theory.
Option $C$ is a prediction of Huygens' theory,which correctly states that light travels slower in a denser medium compared to a rarer medium.
Option $D$ is correct because Huygens' principle successfully explains the laws of reflection and refraction using wave-fronts.
Therefore,the statement that is not correct according to Huygens' wave theory is $B$.

(D) The optical path length is defined as the product of the refractive index of the medium and the geometric distance traveled by the light.
In vacuum (or air),the optical path for a distance $t$ is $t_{opt} = 1 \times t = t$.
When a mica film of thickness $t$ and refractive index $n$ is introduced,the light travels a distance $t$ through the film.
The optical path through the film is $t'_{opt} = n \times t = nt$.
The change in optical path is $\Delta = t'_{opt} - t_{opt} = nt - t = (n-1)t$.
Since $n > 1$,the optical path increases by $(n-1)t$.

(C) The path difference at a point exactly in front of one of the slits is given by $\Delta x = \frac{d^2}{2D}$.
For the third minimum,the condition for path difference is $\Delta x = (2n - 1) \frac{\lambda}{2}$ where $n = 3$.
So,$\frac{d^2}{2D} = (2(3) - 1) \frac{\lambda}{2} = \frac{5\lambda}{2}$.
This implies $\frac{d^2}{D} = 5\lambda$.
Now,for the first minimum at the same point,let the new wavelength be $\lambda'$. The condition is $\Delta x = (2(1) - 1) \frac{\lambda'}{2} = \frac{\lambda'}{2}$.
Equating the path difference: $\frac{d^2}{2D} = \frac{\lambda'}{2}$,which gives $\frac{d^2}{D} = \lambda'$.
Substituting the value of $\frac{d^2}{D}$ from the first case: $\lambda' = 5\lambda$.
The change in wavelength is $\Delta \lambda = \lambda' - \lambda = 5\lambda - \lambda = 4\lambda$.

(C) The intensity $I$ at any point on the screen is given by $I = I_0 \cos^2(\frac{\phi}{2})$, where $I_0$ is the maximum intensity and $\phi$ is the phase difference.
Phase difference $\phi = \frac{2\pi}{\lambda} \times \Delta x$, where $\Delta x$ is the path difference.
For $\Delta x = \frac{\lambda}{4}$, the phase difference $\phi_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity $I_1 = I_0 \cos^2(\frac{\pi/2}{2}) = I_0 \cos^2(\frac{\pi}{4}) = I_0 (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{2}$.
Given $I_1 = \frac{K}{2}$, we have $\frac{I_0}{2} = \frac{K}{2}$, which implies $I_0 = K$.
For $\Delta x = \lambda$, the phase difference $\phi_2 = \frac{2\pi}{\lambda} \times \lambda = 2\pi$.
The intensity $I_2 = I_0 \cos^2(\frac{2\pi}{2}) = I_0 \cos^2(\pi) = I_0 (1)^2 = I_0$.
Since $I_0 = K$, the intensity $I_2 = K$.

(C) In Young's double slit experiment,the condition for destructive interference (dark fringe) is that the path difference $\Delta x$ must be an odd multiple of half the wavelength: $\Delta x = (2n-1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \ldots$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.
Substituting the expression for path difference into the phase difference formula:
$\Delta \phi = \frac{2\pi}{\lambda} \times (2n-1) \frac{\lambda}{2}$.
Simplifying the expression:
$\Delta \phi = (2n-1) \pi$ radians.
Thus,for the $n$th dark fringe,the phase difference is $(2n-1) \pi$.

(A) The number of fringes $N$ observed on a screen of fixed width $W$ is inversely proportional to the fringe width $\beta$.
The fringe width is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit separation.
Since the total width $W$ covered by $N$ fringes is $W = N \beta$,we have $W = N \frac{\lambda D}{d}$.
For a fixed screen width $W$,$N \lambda = \text{constant}$.
Therefore,$N_1 \lambda_1 = N_2 \lambda_2$.
Given $N_1 = 18$,$\lambda_1 = 600 \ nm$,and $\lambda_2 = 400 \ nm$.
Substituting the values: $18 \times 600 = N_2 \times 400$.
$N_2 = \frac{18 \times 600}{400} = \frac{18 \times 6}{4} = \frac{108}{4} = 27$.
Thus,the number of fringes observed is $27$.

(B) The lateral displacement (fringe shift) $y$ in Young's double slit experiment when a glass plate of thickness $t$ and refractive index $\mu$ is placed in front of one slit is given by the formula:
$y = \frac{D}{d} (\mu - 1) t$
Given:
Distance between screen and aperture $D = 1 \ m$
Slit separation $d = 2 \ mm = 2 \times 10^{-3} \ m$
Refractive index $\mu = 1.5$
Thickness $t = 0.04 \ mm = 0.04 \times 10^{-3} \ m = 4 \times 10^{-5} \ m$
Substituting the values:
$y = \frac{1}{2 \times 10^{-3}} (1.5 - 1) \times 4 \times 10^{-5}$
$y = \frac{1}{2 \times 10^{-3}} (0.5) \times 4 \times 10^{-5}$
$y = \frac{0.5 \times 4 \times 10^{-5}}{2 \times 10^{-3}}$
$y = \frac{2 \times 10^{-5}}{2 \times 10^{-3}} = 10^{-2} \ m$
$y = 1 \ cm$
Therefore,the lateral displacement of the fringes is $1 \ cm$.

(B) The fringe width in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$.
Given: $\lambda = 600 \ nm = 6 \times 10^{-7} \ m$,$d = 10^{-3} \ m$,and $\Delta\beta = 3 \times 10^{-5} \ m$.
The change in fringe width is $\Delta\beta = \frac{\lambda}{d} \Delta D$.
Substituting the values: $3 \times 10^{-5} = \frac{6 \times 10^{-7}}{10^{-3}} \Delta D$.
$3 \times 10^{-5} = 6 \times 10^{-4} \Delta D$.
$\Delta D = \frac{3 \times 10^{-5}}{6 \times 10^{-4}} = 0.5 \times 10^{-1} \ m = 0.05 \ m = 5 \ cm$.
If $\Delta\beta$ is positive (increase),the screen must be moved away by $5 \ cm$. If $\Delta\beta$ is negative (decrease),the screen must be moved towards the slits by $5 \ cm$. The question asks for the change in magnitude,so both moving away by $5 \ cm$ or towards by $5 \ cm$ result in a change of $3 \times 10^{-5} \ m$ in fringe width. Thus,both $(a)$ and $(b)$ are correct.

(B) In Young's double slit experiment,the path difference $\Delta x$ at a point exactly in front of one slit is given by the difference in distances from the two slits to that point.
Let the slits be at $y = d/2$ and $y = -d/2$. The point in front of one slit is at $y = d/2$.
The distance from the first slit is $0$.
The distance from the second slit is $\sqrt{D^2 + d^2}$.
Thus,the path difference is $\Delta x = \sqrt{D^2 + d^2} - D$.
Using the binomial approximation for $d \ll D$,$\Delta x \approx D(1 + \frac{d^2}{2D^2}) - D = \frac{d^2}{2D}$.
For a minimum (destructive interference),the path difference must be an odd multiple of half-wavelengths: $\Delta x = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$.
Equating the two expressions: $\frac{d^2}{2D} = (2n - 1) \frac{\lambda}{2}$.
Solving for $\lambda$,we get $\lambda = \frac{d^2}{D(2n - 1)}$.
For $n = 1, 2, 3, \dots$,the wavelengths are proportional to $\frac{1}{D}, \frac{1}{3D}, \frac{1}{5D}, \dots$.