MHT CET 2019 Physics Question Paper with Answer and Solution

(D) Given,height of the inclined plane,$h = 7 \ m$.
Acceleration due to gravity,$g = 10 \ m/s^2$.
By the law of conservation of energy,the potential energy lost by the solid sphere equals the total kinetic energy gained (translational + rotational).
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$ and the condition for rolling without slipping is $\omega = \frac{v}{R}$.
Substituting these values:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left(\frac{2}{5}mR^2\right) \left(\frac{v}{R}\right)^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$
$mgh = \frac{7}{10}mv^2$
$v^2 = \frac{10}{7}gh$
$v = \sqrt{\frac{10}{7} \times 10 \times 7} = \sqrt{100} = 10 \ m/s$.

(A) Initial linear velocity $= v$
Initial radius $= r$
Angular momentum $L = mvr$
When the string is halved,the new radius $r' = \frac{r}{2}$.
Since the angular momentum $L$ remains constant:
$mvr = mv'r'$
$mvr = mv' \left(\frac{r}{2}\right)$
$v = \frac{v'}{2}$
$v' = 2v$
Therefore,the new linear velocity is $2v$.

(A) The given relation is $v = r \omega$.
In a rotating rigid body,all particles rotate with the same angular velocity $\omega$ about the axis of rotation.
While the linear velocity $v$ of a particle depends on its distance $r$ from the axis (as $v = r \omega$),the angular velocity $\omega$ is a property of the entire rigid body's rotation.
Therefore,$\omega$ remains constant for all particles in the body regardless of their distance $r$ from the axis.
Thus,$\omega$ does not depend on $r$.

(B) Let the masses be at vertices $A, B,$ and $C$ of an equilateral triangle with side $L$. Let $O$ be the centroid of the triangle.
The distance from any vertex to the centroid $O$ is $R = \frac{L}{\sqrt{3}}$.
The moment of inertia of the system about the axis passing through $O$ and perpendicular to the plane of the triangle is:
$I = 3 \times (m R^2) = 3 \times m \times \left(\frac{L}{\sqrt{3}}\right)^2 = 3 \times m \times \frac{L^2}{3} = m L^2$.
For the system to rotate,the gravitational force between the masses provides the necessary centripetal force.
The gravitational force on one mass due to the other two is $F_{net} = 2 \times \left(\frac{G m^2}{L^2}\right) \cos 30^{\circ} = 2 \times \frac{G m^2}{L^2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{L^2}$.
This force acts as the centripetal force: $F_{net} = m \omega^2 R$.
$\frac{\sqrt{3} G m^2}{L^2} = m \omega^2 \left(\frac{L}{\sqrt{3}}\right)$.
$\omega^2 = \frac{\sqrt{3} G m}{L^2} \times \frac{\sqrt{3}}{L} = \frac{3 G m}{L^3}$.
Since $T = \frac{2 \pi}{\omega}$,we have $T^2 = \frac{4 \pi^2}{\omega^2} = \frac{4 \pi^2 L^3}{3 G m}$.
Thus,$T^2 \propto L^3$,which implies $T \propto L^{3/2}$.

(C) The moment of a couple is given by the product of the magnitude of one of the forces and the perpendicular distance between the lines of action of the two forces.
Let the angle between the rod and the direction of the force be $\theta = 30^{\circ}$.
The perpendicular distance between the two parallel forces $F$ acting at the ends of the rod of length $l$ is $d = l \sin \theta$.
The moment of the couple $\tau$ is given by:
$\tau = F \times d = F \times l \sin 30^{\circ}$
Given $\sin 30^{\circ} = 0.5 = \frac{1}{2}$.
Substituting the values:
$\tau = F \times l \times \frac{1}{2}$
$\tau = \frac{F l}{2}$
Rearranging to solve for $F$:
$F = \frac{2 \tau}{l}$

(A) Since there is no external torque about the pivot $C$,the angular momentum of the system is conserved.
Initial angular momentum $L_i = (2m)(v)(L) + (m)(2v)(2L) = 2mvL + 4mvL = 6mvL$.
Final moment of inertia $I_f = I_{\text{rod}} + I_{2m} + I_{m} = \frac{(8m)(6L)^2}{12} + (2m)(L)^2 + (m)(2L)^2$.
$I_f = \frac{8m \cdot 36L^2}{12} + 2mL^2 + 4mL^2 = 24mL^2 + 2mL^2 + 4mL^2 = 30mL^2$.
Using $L_i = I_f \omega$:
$6mvL = (30mL^2) \omega$.
$\omega = \frac{6mvL}{30mL^2} = \frac{v}{5L}$.

(C) Given,maximum wavelength,$\lambda_m = 289.8 \ nm = 289.8 \times 10^{-9} \ m = 2.898 \times 10^{-7} \ m$.
Stefan's constant,$\sigma = 5.67 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$.
Wien's constant,$b = 2898 \ \mu m \ K = 2898 \times 10^{-6} \ m \ K$.
According to Wien's displacement law,$\lambda_m = \frac{b}{T}$,so $T = \frac{b}{\lambda_m}$.
Substituting the values,$T = \frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} = \frac{2898 \times 10^{-6}}{2.898 \times 10^{-7}} = 10^4 \ K$.
According to Stefan-Boltzmann law,the intensity of radiation $I$ is given by $I = \sigma T^4$ (assuming the star acts as a black body,$e=1$).
Substituting the values,$I = (5.67 \times 10^{-8}) \times (10^4)^4 = 5.67 \times 10^{-8} \times 10^{16} = 5.67 \times 10^8 \ W \ m^{-2}$.

(D) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
Taking the derivative,the change in time period is $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$.
Since $\frac{dL}{L} = \alpha \Delta \theta$,we have $\frac{dT}{T} = \frac{1}{2} \alpha \Delta \theta$.
Here,$T = 0.5 \ s$,$\alpha = 9 \times 10^{-7} /^{\circ}C$,and $\Delta \theta = 30^{\circ}C - 20^{\circ}C = 10^{\circ}C$.
Substituting the values: $dT = T \times \frac{1}{2} \times \alpha \times \Delta \theta$.
$dT = 0.5 \times \frac{1}{2} \times (9 \times 10^{-7}) \times 10$.
$dT = 0.25 \times 9 \times 10^{-6} = 2.25 \times 10^{-6} \ s$.

(C) The coefficient of performance $\alpha$ of a refrigerator is defined as the ratio of heat extracted from the cold reservoir $(Q_2)$ to the work done $(W)$ on the system: $\alpha = \frac{Q_2}{W}$.
Since $W = Q_1 - Q_2$,where $Q_1$ is the heat released to the hot reservoir,we have $\alpha = \frac{Q_2}{Q_1 - Q_2}$.
Cross-multiplying gives: $\alpha(Q_1 - Q_2) = Q_2$.
Expanding the equation: $\alpha Q_1 - \alpha Q_2 = Q_2$.
Rearranging to solve for $Q_2$: $\alpha Q_1 = Q_2 + \alpha Q_2 = Q_2(1 + \alpha)$.
Therefore,$Q_2 = \frac{\alpha Q_1}{1 + \alpha}$.

(A) The dimension of torque is $[M^1 L^2 T^{-2}]$.
Torque is defined as the product of force and the perpendicular distance from the axis of rotation,given by $\tau = F \times r$.
The dimensional formula for force $(F)$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for distance $(r)$ is $[L^1]$.
Therefore,the dimensional formula for torque $(\tau)$ is $[M^1 L^1 T^{-2}] \times [L^1] = [M^1 L^2 T^{-2}]$.

(D) According to Stefan's law,the energy emitted by a body per unit area per unit time is proportional to the fourth power of its absolute temperature.
$E = \sigma T^4$
Where $E$ is the energy emitted per unit area per unit time,$T$ is the absolute temperature in Kelvin,and $\sigma$ is Stefan's constant.
Rearranging for $\sigma$: $\sigma = \frac{E}{T^4}$.
The unit of $E$ is $\frac{J}{m^2 s}$.
Therefore,the unit of $\sigma$ is $\frac{J}{m^2 s K^4}$.
The dimensions of energy are $[M L^2 T^{-2}]$,area is $[L^2]$,time is $[T]$,and temperature is $[K]$.
$\sigma = \frac{[M L^2 T^{-2}]}{[L^2] [T] [K^4]} = [M^1 L^0 T^{-3} K^{-4}]$.

(A) The dimension of force is $[F] = [M L T^{-2}]$.
The dimension of density is $[d] = [M L^{-3} T^0]$.
From the given relation,$F = \frac{y}{\sqrt{d}}$,we can write $y = F \sqrt{d}$.
Substituting the dimensions of $F$ and $d$:
$[y] = [M L T^{-2}] \times [M L^{-3}]^{1/2}$
$[y] = [M L T^{-2}] \times [M^{1/2} L^{-3/2}]$
$[y] = [M^{1 + 1/2} L^{1 - 3/2} T^{-2}]$
$[y] = [M^{3/2} L^{-1/2} T^{-2}]$.

(B) The gyromagnetic ratio $(\gamma)$ is defined as the ratio of the magnetic dipole moment $(M)$ to the angular momentum $(L)$ of a system.
$\gamma = \frac{M}{L}$
Since the magnetic moment $M = I \cdot A$ (where $I$ is current and $A$ is area),its dimensions are $[I^1 L^2]$.
Since angular momentum $L = mvr$,its dimensions are $[M^1 L^2 T^{-1}]$.
Therefore,the dimensions of the gyromagnetic ratio are:
$\text{Dimension} = \frac{[I^1 L^2]}{[M^1 L^2 T^{-1}]} = [M^{-1} L^0 T^1 I^1]$.

(A) The dimensional formula for torque is given by $\tau = r \times F$.
Since the dimensions of force $F$ are $[M^1L^1T^{-2}]$ and the dimensions of distance $r$ are $[L^1]$,the dimensions of torque are $[M^1L^1T^{-2}] \times [L^1] = [M^1L^2T^{-2}]$.
The moment of force is defined as the product of force and the perpendicular distance from the axis of rotation,which is identical to the definition of torque.
Therefore,the dimensions of the moment of force are also $[M^1L^2T^{-2}]$.
Thus,the dimensions of torque are the same as those of the moment of force.

(B) In damped $SHM$,the damping force $F_d$ is proportional to the velocity $v$ of the oscillator,given by $F_d = -bv$,where $b$ is the damping constant.
Therefore,the damping constant is $b = \frac{F_d}{v}$.
The $SI$ unit of force $F_d$ is $Newton$ $(N)$ or $kg \cdot m/s^2$.
The $SI$ unit of velocity $v$ is $m/s$.
Thus,the $SI$ unit of damping constant $b$ is $\frac{kg \cdot m/s^2}{m/s} = kg/s$.
Hence,the $SI$ unit of damping constant is $kg/s$.

(C) For a particle executing $SHM$,the displacement is $x = A \sin(\omega t + \phi)$.
Velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.
Acceleration is $a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) = -\omega^2 x$.
We know that angular frequency $\omega = \frac{2 \pi}{T}$.
Now,let us evaluate the expression in option $(c)$:
$\frac{a T}{v} = \frac{(-\omega^2 x) T}{A \omega \cos(\omega t + \phi)} = \frac{-\omega^2 (A \sin(\omega t + \phi)) T}{A \omega \cos(\omega t + \phi)} = -\omega T \tan(\omega t + \phi)$.
Wait,let's re-evaluate using the standard relation $a = -\omega^2 x$ and $v = \omega \sqrt{A^2 - x^2}$.
Actually,the ratio $\frac{a}{v}$ for $SHM$ is $\frac{-\omega^2 x}{v}$. Since $v = \omega \sqrt{A^2 - x^2}$,$\frac{a}{v} = \frac{-\omega^2 x}{\omega \sqrt{A^2 - x^2}} = -\omega \frac{x}{\sqrt{A^2 - x^2}}$. This is time-dependent.
However,checking the dimensions: $[a] = [L T^{-2}]$,$[T] = [T]$,$[v] = [L T^{-1}]$.
Dimension of $\frac{a T}{v} = \frac{[L T^{-2}] [T]}{[L T^{-1}]} = \frac{[L T^{-1}]}{[L T^{-1}]} = [M^0 L^0 T^0]$.
Since the expression $\frac{a T}{v}$ is dimensionless and represents a constant factor related to the angular frequency $(2\pi)$,it is the only quantity that remains constant in magnitude relative to the phase of motion.

(C) The dimensions of Planck's constant $h$ are given by $[h] = [M L^2 T^{-1}]$.
The dimensions of the moment of inertia $I$ are given by $[I] = [M L^2]$.
Taking the ratio of the dimensions of Planck's constant to the moment of inertia:
$\frac{[h]}{[I]} = \frac{[M L^2 T^{-1}]}{[M L^2]} = [T^{-1}]$.
The dimension $[T^{-1}]$ corresponds to the dimension of frequency.

(C) The diameter of the Earth $(D_e)$ is $2 \times 6371 \ km = 12742 \ km = 1.2742 \times 10^4 \ km$.
The order of magnitude of the diameter of the Earth is $10^4$.
The diameter of the orbit $(D_o)$ is $2 \times 149 \times 10^6 \ km = 298 \times 10^6 \ km = 2.98 \times 10^8 \ km$.
The order of magnitude of the diameter of the orbit is $10^8$.
The difference in the order of magnitude is $8 - 4 = 4$.
Therefore,the order of magnitude of the diameter of the orbit is greater than that of the Earth by $10^4$.

(B) The given equation of the $SHM$ wave is $y = 0.03 \sin \pi (2 t - 0.01 x) \ m$.
Expanding this,we get $y = 0.03 \sin (2 \pi t - 0.01 \pi x) \ m$.
Comparing this with the general wave equation $y = a \sin (\omega t - k x)$,we identify the wave number $k$ as $k = 0.01 \pi \ rad/m$.
The phase difference $\Delta \phi$ between two particles separated by a distance $\Delta x$ is given by the formula $\Delta \phi = k \Delta x$.
Given $\Delta x = 25 \ m$ and $k = 0.01 \pi \ rad/m$,we substitute these values into the formula:
$\Delta \phi = (0.01 \pi) \times 25 = 0.25 \pi = \frac{\pi}{4} \ rad$.
Thus,the phase difference is $\frac{\pi}{4} \ rad$.

(B) The frequency produced by a stretched wire is given by $f = \frac{p}{2l} \sqrt{\frac{T}{m}}$.
Here,$p$ is the number of loops,$l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Since $T$ and $m$ are constant for the same wire,we have $f \propto \frac{1}{l}$.
Given the ratio of frequencies $f_1 : f_2 : f_3 = 2 : 3 : 4$,the ratio of lengths is $l_1 : l_2 : l_3 = \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$.
Multiplying by the least common multiple $(12)$,we get $l_1 : l_2 : l_3 = 6 : 4 : 3$.
The sum of the ratio parts is $6 + 4 + 3 = 13$.
Given the total length $L = 260 \ cm$,the individual lengths are:
$l_1 = \frac{6}{13} \times 260 = 120 \ cm$
$l_2 = \frac{4}{13} \times 260 = 80 \ cm$
$l_3 = \frac{3}{13} \times 260 = 60 \ cm$.

(A) For a stretched string fixed at both ends,the length $l$ is given by $l = \frac{p \lambda}{2}$,where $p$ is the number of loops (or segments).
In a standing wave on a string fixed at both ends,the number of loops $p$ is equal to the number of anti-nodes.
The relationship between the number of nodes $m$ and the number of anti-nodes $p$ is given by $m = p + 1$.
Therefore,the number of loops is $p = m - 1$.
Substituting this value of $p$ into the length formula,we get $l = \frac{(m-1) \lambda}{2}$.

(D) Beats are the periodic variations in the intensity of sound heard when two sound waves of slightly different frequencies and comparable amplitudes interfere with each other.
For the formation of distinct beats,the two sound sources must have nearly equal frequencies so that the beat frequency $(f_{beat} = |f_1 - f_2|)$ is low enough to be perceived by the human ear.
Additionally,they should have nearly equal amplitudes to ensure that the interference results in clearly audible maximums and minimums of intensity.

(C) For an open organ pipe of length $L$,the frequency of the $P^{\text{th}}$ overtone is given by $f_{\text{open}} = (P+1) \frac{v}{2L}$.
For a closed organ pipe of the same length $L$,the frequency of the $P^{\text{th}}$ overtone is given by $f_{\text{closed}} = (2P+1) \frac{v}{4L}$.
Taking the ratio of the frequency of the $P^{\text{th}}$ overtone of the open pipe to that of the closed pipe:
Ratio $= \frac{(P+1) \frac{v}{2L}}{(2P+1) \frac{v}{4L}}$.
Ratio $= \frac{P+1}{2L} \times \frac{4L}{2P+1} = \frac{2(P+1)}{2P+1}$.

(D) For a closed pipe of length $l$,the fundamental frequency is given by $f_0 = \frac{v}{4l}$,where $v$ is the speed of sound.
The time $t$ taken by the sound wave to travel the length of the pipe $l$ is $t = \frac{l}{v}$,which implies $v = \frac{l}{t}$.
Substituting the value of $v$ into the frequency formula:
$f_0 = \frac{l/t}{4l} = \frac{1}{4t}$.
Thus,the frequency of vibration is $(4t)^{-1}$.

(D) For an open pipe,the fundamental frequency is given by $n = \frac{v}{2l}$,where $v$ is the speed of sound and $l$ is the length of the pipe.
For the two pipes,we have $l_1 = \frac{v}{2n_1}$ and $l_2 = \frac{v}{2n_2}$.
When the two pipes are joined,the new length becomes $L = l_1 + l_2$.
The new fundamental frequency $n'$ is given by $n' = \frac{v}{2L} = \frac{v}{2(l_1 + l_2)}$.
Substituting the values of $l_1$ and $l_2$:
$n' = \frac{v}{2(\frac{v}{2n_1} + \frac{v}{2n_2})} = \frac{v}{\frac{v}{n_1} + \frac{v}{n_2}} = \frac{1}{\frac{1}{n_1} + \frac{1}{n_2}} = \frac{n_1 n_2}{n_1 + n_2}$.

(A) The given equation is $Y = 0.04 \cos(\pi x) \sin(50 \pi t)$.
Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we can rewrite the equation as:
$Y = 0.02 \sin(50 \pi t + \pi x) + 0.02 \sin(50 \pi t - \pi x)$.
Comparing this with the standard wave equation $y = a \sin(\omega t \pm kx)$:
$1$. Amplitude $a = 0.02 \text{ m}$.
$2$. Angular frequency $\omega = 50 \pi \text{ rad/s}$.
$3$. Wave number $k = \pi \text{ m}^{-1}$.
Now,calculating the parameters:
- Time Period $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{50 \pi} = 0.04 \text{ s}$.
- Wavelength $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{\pi} = 2 \text{ m}$.
- Velocity $v = \frac{\omega}{k} = \frac{50 \pi}{\pi} = 50 \text{ m/s}$.
Comparing these results with the given options,the statement in option $A$ (Time Period $= 0.02 \text{ s}$) is incorrect,as the calculated time period is $0.04 \text{ s}$.

(C) The general equation of a transverse wave is given by $Y = A \sin(kx + \omega t)$.
Comparing this with the given equation $Y = 0.05 \sin(x + 15t)$,we get the angular frequency $\omega = 15 \ rad/s$ and the wave number $k = 1 \ m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{15}{1} = 15 \ m/s$.
The speed of a transverse wave on a string is also given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 10^{-3} \ kg/m$,we have $15 = \sqrt{\frac{T}{10^{-3}}}$.
Squaring both sides,$225 = \frac{T}{10^{-3}}$.
Therefore,$T = 225 \times 10^{-3} \ N = 0.225 \ N$.

(D) The frequency of vibration $v$ of a sonometer wire is given by the formula:
$v = \frac{1}{2L} \sqrt{\frac{T}{m}}$
where $T$ is the tension (weight $w$) and $m$ is the mass per unit length.
For the first case,the frequency is:
$v = \frac{1}{2L_1} \sqrt{\frac{w}{m}} \quad (i)$
For the second case,the weight is reduced to $\frac{w}{4}$ and the length becomes $L_2$:
$v = \frac{1}{2L_2} \sqrt{\frac{w/4}{m}} = \frac{1}{2L_2} \cdot \frac{1}{2} \sqrt{\frac{w}{m}} = \frac{1}{4L_2} \sqrt{\frac{w}{m}} \quad (ii)$
Since the tuning fork frequency $v$ remains constant,we equate $(i)$ and $(ii)$:
$\frac{1}{2L_1} \sqrt{\frac{w}{m}} = \frac{1}{4L_2} \sqrt{\frac{w}{m}}$
$\frac{1}{2L_1} = \frac{1}{4L_2}$
$4L_2 = 2L_1$
$\frac{L_1}{L_2} = \frac{4}{2} = 2$
Therefore,the ratio $\frac{L_1}{L_2}$ is $2:1$.

(C) The fundamental frequency of a stretched string is given by $v = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Since $l$ and $m$ are constant,$v \propto \sqrt{T}$.
Let the initial frequency be $v$ and initial tension be $T$.
When tension is increased by $69 \%$,the new tension $T' = T + 0.69T = 1.69T$.
The new frequency is $v' = v + 9$.
Taking the ratio: $\frac{v'}{v} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{1.69T}{T}} = \sqrt{1.69} = 1.3$.
Therefore,$v + 9 = 1.3v$.
$0.3v = 9$.
$v = \frac{9}{0.3} = 30 \ Hz$.

(D) The refractive index of medium '$x$' with respect to medium '$Y$' is given by the relation: $n_{xy} = \frac{1}{\sin \theta}$.
Also,the refractive index is defined as the ratio of the speed of light in the second medium to the speed of light in the first medium: $n_{xy} = \frac{V_{Y}}{V_{x}}$.
Equating the two expressions: $\frac{V_{Y}}{V_{x}} = \frac{1}{\sin \theta}$.
Therefore,the speed of light in medium '$Y$' is: $V_{Y} = \frac{V_{x}}{\sin \theta}$.

(D) The refractive index of a medium is inversely proportional to the speed of light in that medium,given by $\mu = \frac{c}{v}$.
When light travels from oil (medium $1$) to crystal (medium $2$),the ratio of the speeds is given by the inverse ratio of their refractive indices:
$\frac{v_{crystal}}{v_{oil}} = \frac{\mu_{oil}}{\mu_{crystal}}$
Given $\mu_{crystal} = 1.68$ and $\mu_{oil} = 1.2$.
Substituting the values:
$\frac{v_{crystal}}{v_{oil}} = \frac{1.2}{1.68} = \frac{120}{168} = \frac{10}{14} = \frac{5}{7}$.
Thus,the velocity changes by a factor of $\frac{5}{7}$.

(A) When light enters glass from vacuum,its wavelength decreases. This is because the speed of light in glass is less than that in a vacuum. The frequency of light remains constant when it travels from one medium to another. From the relation $v = f \lambda$,where $v$ is the speed,$f$ is the frequency,and $\lambda$ is the wavelength,we have $\lambda = \frac{v}{f}$. Since the speed $v$ decreases in glass compared to vacuum and the frequency $f$ remains constant,the wavelength $\lambda$ must decrease.

(B) For a thin prism,the angle of deviation $\delta$ is given by the formula: $\delta = (n - 1)A$,where $n$ is the refractive index of the material of the prism and $A$ is the refracting angle of the prism.
Given:
Refracting angle $A = 3^{\circ}$
Angle of deviation $\delta = 1^{\circ}$
Substituting these values into the formula:
$1^{\circ} = (n - 1) \cdot 3^{\circ}$
$n - 1 = \frac{1}{3}$
$n = 1 + \frac{1}{3} = \frac{4}{3}$
$n = 1.33$
Therefore,the refractive index of water is $1.33$.

(B) An $LED$ (Light Emitting Diode) is a $p-n$ junction diode that emits light when it is forward-biased.
Therefore,its $V-I$ characteristics are similar to those of a standard $p-n$ junction diode operating in the forward bias region.
In the forward bias region,the current increases exponentially with the applied voltage after the threshold voltage (knee voltage) is reached.
Graph $(b)$ represents this exponential increase in current with voltage in the first quadrant,which is the characteristic behavior of an $LED$ in forward bias.
Graph $(a)$ shows both forward and reverse bias,which is for a general diode.
Graph $(c)$ represents the characteristics of a solar cell.
Graph $(d)$ represents the characteristics of a photodiode.

(B) In a $p-n$ junction diode,when it is reverse biased,the applied voltage supports the barrier potential,which causes the width of the depletion region to increase.
In forward biasing,the width of the depletion region decreases because the forward voltage opposes the potential barrier.
Additionally,the width of the depletion region decreases with heavy doping because the increased concentration of charge carriers leads to a narrower space charge region.

(C) The diode is connected in forward bias because the $p$-side is at $+3 \text{ V}$ and the $n$-side is at $+1 \text{ V}$.
Since the diode is ideal, its resistance in forward bias is zero.
The potential difference across the resistor $R = 100 \ \Omega$ is $V = 3 \text{ V} - 1 \text{ V} = 2 \text{ V}$.
Using Ohm's law, the current $I$ is given by $I = \frac{V}{R} = \frac{2 \text{ V}}{100 \ \Omega} = 0.02 \text{ A}$.
Converting to milliamperes, $I = 0.02 \times 1000 \text{ mA} = 20 \text{ mA}$.

(C) In a transistor,the emitter region is heavily doped because its primary function is to inject a large number of charge carriers into the base.
The base region is very lightly doped and kept thin to minimize the recombination of charge carriers,allowing most of them to reach the collector.
The collector region is moderately doped compared to the emitter and base.
Therefore,the emitter is heavily doped and the base is lightly doped.

(B) The ratio of collector current $(I_C)$ to emitter current $(I_E)$ is known as the common-base current amplification factor,denoted by $\alpha$. Given: $\alpha = \frac{I_C}{I_E} = 0.98$.
The ratio of collector current $(I_C)$ to base current $(I_B)$ is known as the common-emitter current amplification factor,denoted by $\beta$.
The relationship between $\alpha$ and $\beta$ is given by the formula: $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the given value: $\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.
Therefore,the ratio of collector current to base current is $49$.

(B) The current gain $\beta_{dc}$ for a common-emitter transistor configuration is defined as the ratio of the collector current $(i_c)$ to the base current $(i_b)$.
Mathematically,$\beta_{dc} = \frac{i_c}{i_b} = \frac{\text{Collector current}}{\text{Base current}}$.

(A) Electric polarisation $(P)$ is defined as the dipole moment per unit volume.
$P = \frac{p}{V} = \frac{q \cdot d}{A \cdot d} = \frac{q}{A}$
Where $q$ is charge and $A$ is area.
The dimensional formula for charge $q$ is $[I^1 T^1]$.
The dimensional formula for area $A$ is $[L^2]$.
Therefore,the dimensional formula for $P$ is:
$[P] = \frac{[I^1 T^1]}{[L^2]} = [M^0 L^{-2} T^1 I^1]$.

(B) The electric field $E$ is defined as force per unit charge: $E = \frac{F}{q}$.
Dimension of force $[F] = [M L T^{-2}]$ and dimension of charge $[q] = [A T]$.
Therefore,the dimension of electric field $[E] = \frac{[M L T^{-2}]}{[A T]} = [M L T^{-3} A^{-1}]$.
The power of mass in the electric field is $1$.
The electric dipole moment $p$ is defined as the product of charge and separation distance: $p = q \times d$.
Dimension of charge $[q] = [A T]$ and dimension of distance $[d] = [L]$.
Therefore,the dimension of electric dipole moment $[p] = [A T L] = [M^0 L^1 T^1 A^1]$.
The power of mass in the electric dipole moment is $0$.
Thus,the powers of mass are $1$ and $0$ respectively.

(C) The self or mutual inductance of a coil is defined by the flux linkage per unit current,given by the relation:
$L \text{ or } M = \frac{\phi}{I}$
Where $\phi$ is the magnetic flux and $I$ is the current.
The dimensional formula for magnetic flux $\phi$ is $[M L^2 T^{-2} I^{-1}]$.
The dimensional formula for current $I$ is $[I^1]$.
Therefore,the dimension of inductance is:
$[L] = \frac{[\phi]}{[I]} = \frac{[M L^2 T^{-2} I^{-1}]}{[I^1]} = [M L^2 T^{-2} I^{-2}]$

(C) The width of the central maximum in a single-slit diffraction pattern is given by the formula: $w = \frac{2 \lambda D}{a}$.
According to the problem,the slit width $a$ is equal to the width of the central maximum $w$.
Therefore,we set $a = \frac{2 \lambda D}{a}$.
Rearranging the equation to solve for $D$,we get: $a^2 = 2 \lambda D$.
Thus,$D = \frac{a^2}{2 \lambda}$.

(D) The luminous border that surrounds the profile of a mountain just before the sun rises behind it is caused by the diffraction of light. Diffraction is the phenomenon of bending of light around the corners of an obstacle or an aperture. When the sun is slightly below the horizon,the light rays pass near the edges of the mountain. These rays bend around the edges due to diffraction,making the mountain's profile appear luminous.

(C) In the phenomenon of interference,the energy is redistributed in the medium.
At points of constructive interference,the intensity (and thus energy) is maximum,while at points of destructive interference,the intensity is minimum.
Since the total energy remains constant throughout the medium,the phenomenon of interference is based on the principle of conservation of energy.

(B) The position of the $n^{th}$ dark fringe in Young's double slit experiment is given by:
$x_n = \frac{D}{d} (2n - 1) \frac{\lambda}{2}$
Rearranging for wavelength $\lambda$:
$\lambda = \frac{2 x_n d}{D(2n - 1)} \quad \dots (i)$
Given that the fifth dark fringe $(n=5)$ is formed opposite to one of the slits,the distance of this fringe from the central axis is half the distance between the slits:
$x_5 = \frac{d}{2} \implies 2x_5 = d$
Substituting $n=5$ and $2x_5 = d$ into equation $(i)$:
$\lambda = \frac{(2x_5) d}{D(2 \times 5 - 1)}$
$\lambda = \frac{d \cdot d}{D(10 - 1)}$
$\lambda = \frac{d^2}{9D}$

(D) The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$,where $D = 1.2 \ m$ is the distance between the source and the eyepiece,and $d = 0.84 \ mm = 0.84 \times 10^{-3} \ m$ is the distance between the virtual sources.
When the eyepiece is moved transversely by a distance $y = 2.799 \ cm = 2.799 \times 10^{-2} \ m$,the number of fringes shifted is $n = 30$.
The relationship is $y = n \beta = n \frac{\lambda D}{d}$.
Substituting the values: $2.799 \times 10^{-2} = 30 \times \frac{\lambda \times 1.2}{0.84 \times 10^{-3}}$.
$\lambda = \frac{2.799 \times 10^{-2} \times 0.84 \times 10^{-3}}{30 \times 1.2}$.
$\lambda = \frac{2.35116 \times 10^{-5}}{36} = 0.06531 \times 10^{-5} \ m = 6531 \times 10^{-10} \ m$.
Therefore,$\lambda = 6531 \ \mathring{A}$.

(C) According to the Doppler effect for light,when a source of light moves towards an observer,the observed frequency $f'$ increases compared to the source frequency $f$.
This is given by the formula $f' = f \sqrt{\frac{c+v}{c-v}}$,where $v$ is the velocity of the source towards the observer and $c$ is the speed of light.
Since the frequency increases,the wavelength $\lambda$ decreases $(\lambda = c/f)$.
$A$ decrease in wavelength corresponds to a shift towards the blue end of the visible spectrum,which is known as a 'blue shift'.
Therefore,the yellow light will appear to shift towards the blue color.