TS EAMCET 2022 Physics Question Paper with Answer and Solution

(A) The zeroth law of thermodynamics states that if two systems are each in thermal equilibrium with a third system,then they are in thermal equilibrium with each other.
This law forms the basis for the measurement of temperature,as it allows us to define a property (temperature) that is the same for all systems in thermal equilibrium.
Therefore,the zeroth law provides the fundamental concept of temperature.
Since the reason correctly explains why the zeroth law leads to the concept of temperature,$(A)$ is true,$(R)$ is true,and $(R)$ is the correct explanation for $(A)$.

(D) Dimensional analysis requires that both sides of an equation have the same dimensions and that the argument of any exponential function must be dimensionless.
$(A) E = 2 E_0 \frac{L}{L_0}$
$LHS$: $[M L^2 T^{-2}]$
$RHS$: $[M L^2 T^{-2}] \times \frac{[L]}{[L]} = [M L^2 T^{-2}]$
Status: Dimensionally correct.
$(B) E = E_0 e^{-\frac{2 L}{L_0}}$
Exponent: $\frac{[L]}{[L]} = [1]$ (Dimensionless). Correct.
$LHS$: $[M L^2 T^{-2}]$
$RHS$: $[M L^2 T^{-2}] \times [1] = [M L^2 T^{-2}]$
Status: Dimensionally correct.
$(C) E = 2 L e^{-\frac{L}{E_0}}$
Exponent: $\frac{[L]}{[M L^2 T^{-2}]} = [M^{-1} L^{-1} T^2]$. Since the exponent is not dimensionless,the expression is invalid.
$LHS$: $[M L^2 T^{-2}]$
Pre-factor: $[L]$
Status: Dimensionally incorrect.
$(D) E = 2 \left( \frac{E_0}{L_0} \right) e^{-\frac{L}{L_0}}$
Exponent: $\frac{[L]}{[L]} = [1]$. Correct.
$LHS$: $[M L^2 T^{-2}]$
$RHS$: $\frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$
Status: Dimensionally incorrect.
Therefore,equations $(C)$ and $(D)$ are incorrect.

(B) According to Noether's theorem,conservation laws are deeply connected to the symmetries of nature.
There are exactly four fundamental forces in nature: the gravitational force,the electromagnetic force,the strong nuclear force,and the weak nuclear force.
Therefore,statement $B$ is correct.

$(A)$ The dimensions are matched as follows:
$(A)$ Entropy $(S)$: $S = \frac{\Delta Q}{T}$. The unit is $J/K$, which is the same as the unit of the Boltzmann constant $(k_B)$. Thus, $(A) \rightarrow (II)$.
$(B)$ Young's modulus $(Y)$: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\text{Force}}{\text{Area}}$. The unit is $N/m^2$ or $J/m^3$, which is the same as the unit of energy density $(U/V)$. Thus, $(B) \rightarrow (III)$.
$(C)$ Angular momentum $(L)$: $L = mvr$. The unit is $kg \cdot m^2/s$, which is the same as the unit of Planck's constant $(h)$. Thus, $(C) \rightarrow (IV)$.
$(D)$ Decay constant $(\lambda)$: $\lambda = \frac{1}{\text{time}}$. The unit is $s^{-1}$, which is the same as the unit of angular velocity $(\omega)$. Thus, $(D) \rightarrow (I)$.
Therefore, the correct matching is $(A-II, B-III, C-IV, D-I)$.

(B) Let the density $\rho$ be expressed as $\rho = k C^a G^b h^c$.
Dimensional formulas are:
$[\rho] = M L^{-3}$
$[C] = L T^{-1}$
$[G] = M^{-1} L^3 T^{-2}$
$[h] = M L^2 T^{-1}$
Substituting these into the equation:
$M^1 L^{-3} T^0 = (L T^{-1})^a (M^{-1} L^3 T^{-2})^b (M L^2 T^{-1})^c$
$M^1 L^{-3} T^0 = M^{-b+c} L^{a+3b+2c} T^{-a-2b-c}$
Comparing the exponents of $M, L,$ and $T$:
$1$) $-b + c = 1 \implies c = 1 + b$
$2$) $a + 3b + 2c = -3$
$3$) $-a - 2b - c = 0 \implies a = -2b - c$
Substitute $c = 1 + b$ into $(3)$: $a = -2b - (1 + b) = -3b - 1$
Substitute $a$ and $c$ into $(2)$: $(-3b - 1) + 3b + 2(1 + b) = -3$
$-3b - 1 + 3b + 2 + 2b = -3$
$2b + 1 = -3 \implies 2b = -4 \implies b = -2$
Now find $c$: $c = 1 + (-2) = -1$
Now find $a$: $a = -3(-2) - 1 = 6 - 1 = 5$
Thus,the dimension of density is $[\rho] = C^5 G^{-2} h^{-1}$.

(A) The distance between two successive minima is equal to the wavelength $\lambda$. Therefore,$\lambda = 2.7 \ m$.
Frequency $f$ is defined as the number of waves passing a point per unit time.
Given that $5$ crests pass in $15.0 \ s$,the frequency is $f = \frac{5}{15} = \frac{1}{3} \ s^{-1}$.
The speed of the wave $v$ is given by the relation $v = f \lambda$.
Substituting the values,$v = \frac{1}{3} \times 2.7 = 0.9 \ m \ s^{-1}$.

(A) Let the frequency of string $A$ be $f_A = f_0$ and the frequency of string $B$ be $f_B$.
Initially,the beat frequency is $\Delta f_1 = |f_0 - f_B| > 0$.
When the tension in string $A$ is increased,its frequency $f_A$ increases to $f_A'$.
The new beat frequency is $\Delta f_2 = |f_A' - f_B| < \Delta f_1$.
Since the beat frequency decreased after increasing the frequency of $A$,it implies that $f_A$ was approaching $f_B$.
Therefore,$f_B$ must be greater than $f_A$.
Thus,$\Delta f_1 = f_B - f_0$,which gives $f_B = f_0 + \Delta f_1$.

(B) The intensity $I$ of a wave is proportional to the square of its amplitude $A$,i.e.,$I \propto A^2$.
For two waves with amplitudes $A_1$ and $A_2$,the maximum intensity $I_{max}$ occurs at constructive interference,where $I_{max} \propto (A_1 + A_2)^2$.
The minimum intensity $I_{min}$ occurs at destructive interference,where $I_{min} \propto (A_1 - A_2)^2$.
Given the ratio of intensities $\frac{I_{max}}{I_{min}} = \frac{9}{4}$,we have:
$\frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{9}{4}$
Taking the square root on both sides:
$\frac{A_1 + A_2}{A_1 - A_2} = \frac{3}{2}$
Cross-multiplying gives:
$2(A_1 + A_2) = 3(A_1 - A_2)$
$2A_1 + 2A_2 = 3A_1 - 3A_2$
$5A_2 = A_1$
Therefore,the ratio $\frac{A_2}{A_1} = \frac{1}{5} = 0.2$.

(A) Initially,for an open pipe of length '$L$',the fundamental frequency is given by:
$f = \frac{V}{2L}$
When the tube is dipped vertically in water such that half of its length is submerged,the remaining length of the air column above the water surface is '$L/2$'.
The water surface acts as a closed end. Thus,the tube now behaves as a pipe closed at one end with an effective length '$L' = L/2$'.
The fundamental frequency of a pipe closed at one end is given by:
$f' = \frac{V}{4L'}$
Substituting '$L' = L/2$':
$f' = \frac{V}{4(L/2)} = \frac{V}{2L} = f$
Therefore,the new fundamental frequency remains '$f$'.

(B) The general equation of a progressive wave is $y(x, t) = A \sin(kx - \omega t)$.
The wave speed $v$ is given by the ratio of the angular frequency $\omega$ to the wave number $k$:
$v = \frac{\omega}{k} = \frac{\text{coefficient of } t}{\text{coefficient of } x}$.
Calculating the speed for each option:
$(a)$ $v = \frac{2}{2} = 1.0 \text{ m/s}$
$(b)$ $v = \frac{3}{2} = 1.5 \text{ m/s}$
$(c)$ $v = \frac{2}{3} \approx 0.67 \text{ m/s}$
$(d)$ $v = \frac{2}{5} = 0.4 \text{ m/s}$
Comparing these values,the wave speed is largest for the equation in option $(b)$.

(C) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Initially, $L_1 = 0.4 \,m$, $f_1 = 250 \,Hz$, and tension is $T_1 = T$.
So, $250 = \frac{1}{2 \times 0.4} \sqrt{\frac{T}{\mu}} \quad ...(i)$
Finally, $L_2 = 0.4 + 0.1 = 0.5 \,m$, and the new tension is $T_2 = T/4$.
So, $f_2 = \frac{1}{2 \times 0.5} \sqrt{\frac{T/4}{\mu}} \quad ...(ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{f_2}{250} = \frac{\frac{1}{2 \times 0.5} \sqrt{\frac{T}{4\mu}}}{\frac{1}{2 \times 0.4} \sqrt{\frac{T}{\mu}}} = \frac{0.4}{0.5} \times \sqrt{\frac{1}{4}} = \frac{4}{5} \times \frac{1}{2} = \frac{2}{5} = 0.4$
$f_2 = 250 \times 0.4 = 100 \,Hz$.

(D) Initial speed $V = 36 \,km/h = 36 \times \frac{5}{18} = 10 \,m/s$.
Radial acceleration $a_r = \frac{V^2}{R} = \frac{10^2}{50} = \frac{100}{50} = 2 \,m/s^2$.
Tangential acceleration $a_t = 0.5 \,m/s^2$.
Net acceleration $a_{net} = \sqrt{a_r^2 + a_t^2} = \sqrt{2^2 + 0.5^2} = \sqrt{4 + 0.25} = \sqrt{4.25} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \,m/s^2$.
Direction $\theta$ with respect to the radial vector is given by $\tan \theta = \frac{a_t}{a_r} = \frac{0.5}{2} = \frac{1}{4}$.
However, the angle with respect to the tangential vector is $\tan \phi = \frac{a_r}{a_t} = \frac{2}{0.5} = 4$, so $\phi = \tan^{-1}(4)$.

(B) Given:
Volume of water $V = 36 \,m^3$
Time $t = 30 \,min = 30 \times 60 \,s = 1800 \,s$
Height $h = 50 \,m$
Input Power $P_{\text{in}} = 40 \,kW = 40,000 \,W$
Density of water $\rho = 1000 \,kg/m^3$
Acceleration due to gravity $g = 10 \,m/s^2$
Mass of water $m = V \times \rho = 36 \times 1000 = 36,000 \,kg$
Work done to lift the water $W = mgh = 36,000 \times 10 \times 50 = 18,000,000 \,J$
Output Power $P_{\text{out}} = \frac{W}{t} = \frac{18,000,000}{1800} = 10,000 \,W = 10 \,kW$
Efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{10,000}{40,000} \times 100\% = 25\%$

(B) For a conservative force,the relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
Given $U(x) = 5x^2 - 4x^3$.
To find the position where the force is zero,we set $F = 0$:
$0 = -\frac{d}{dx}(5x^2 - 4x^3)$
$0 = -(10x - 12x^2)$
$12x^2 - 10x = 0$
$2x(6x - 5) = 0$
This gives two solutions: $x = 0 \ m$ or $x = 5/6 \ m$.
Comparing this with the given options,the correct position is $5/6 \ m$.

(C) According to the Work-Energy Theorem,the work done by the net force is equal to the change in kinetic energy:
$W = \Delta K = K_f - K_i$
Given $m = 1 \ kg$,$c = 1$,$v = x^\alpha$,$x_i = 0 \ m$,$x_f = 4 \ m$,and $W = 128 \ J$.
At $x = 0$,$v_i = 1 \times (0)^\alpha = 0$.
At $x = 4$,$v_f = 1 \times (4)^\alpha = 4^\alpha$.
Substituting these into the Work-Energy equation:
$128 = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$
$128 = \frac{1}{2} (1) (4^\alpha)^2 - 0$
$128 = \frac{1}{2} (4^{2\alpha})$
$256 = 4^{2\alpha}$
Since $256 = 2^8 = (2^2)^4 = 4^4$,we have:
$4^4 = 4^{2\alpha}$
$2\alpha = 4$
$\alpha = 2$.

(C) To complete a vertical circle,the minimum velocity at the bottom must be $v = \sqrt{5Rg}$,where $R$ is the radius of the circle.
Applying the law of conservation of mechanical energy between the top of the incline and the bottom:
$P_1 + K_1 = P_2 + K_2$
$mgh + 0 = 0 + \frac{1}{2} m v^2$
$mgh = \frac{1}{2} m (\sqrt{5Rg})^2$
$mgh = \frac{1}{2} m (5Rg)$
$h = \frac{5R}{2}$
Given $R = 2 \ m$,we have:
$h = \frac{5 \times 2}{2} = 5 \ m$

(C) By the law of conservation of energy,the total initial mechanical energy equals the total final mechanical energy at the highest point.
Initial energy: $E_i = K.E_{trans} + K.E_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,$I = \frac{2}{5}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.
$E_i = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
At the highest point,the final kinetic energy is zero,and the potential energy is $mgh$.
So,$\frac{7}{10}mv^2 = mgh \Rightarrow h = \frac{7v^2}{10g}$
Given $v = 4 \ m/s$ and $g = 10 \ m/s^2$,$h = \frac{7 \times 4^2}{10 \times 10} = \frac{7 \times 16}{100} = 1.12 \ m$.
The distance $\ell$ along the plane is given by $\ell = \frac{h}{\sin 30^{\circ}} = \frac{1.12}{0.5} = 2.24 \ m = 224 \ cm$.

(C) Analysis of Statement $(I)$: The kinetic energy $K = \frac{1}{2}mv^2$. The slope of the $K-x$ curve is $\frac{dK}{dx} = \frac{d}{dx}(\frac{1}{2}mv^2) = mv \frac{dv}{dx} = m \cdot a$. Since $m$ is constant,the slope is directly proportional to acceleration $a$. Thus,Statement $(I)$ is true.
Analysis of Statement $(II)$: Initial velocity $u = 30 \ m/s$,height $h = 15 \ m$,$g = 10 \ m/s^2$. Velocity just before hitting the ground $v^2 = u^2 + 2gh = 30^2 + 2(10)(15) = 900 + 300 = 1200 \ (m/s)^2$. Kinetic energy before impact $E_i = \frac{1}{2}m(1200)$. After impact,it rises to $15 \ m$,so velocity after impact $v' = \sqrt{2gh} = \sqrt{2(10)(15)} = \sqrt{300} \ m/s$. Kinetic energy after impact $E_f = \frac{1}{2}m(300)$. Loss in energy $\Delta E = E_i - E_f = \frac{1}{2}m(1200 - 300) = \frac{1}{2}m(900)$. Percentage loss $= (\frac{\Delta E}{E_i}) \times 100 = (\frac{900}{1200}) \times 100 = 75 \%$. Thus,Statement $(II)$ is false.
Analysis of Statement $(III)$: From $v^2 = u^2 + 2as$,where $u=0$ and $a = F/m$,we get $v^2 = 2(F/m)s$. Thus $v = \sqrt{2Fs/m}$. Velocity is inversely proportional to $\sqrt{m}$,not directly proportional to $m$. Thus,Statement $(III)$ is false.

(A) First, calculate the acceleration of the boat: $a = \frac{v-u}{t} = \frac{20-0}{5} = 4 \,m/s^2$.
Next, calculate the distance traveled by the boat: $S = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 4 \times 5^2 = 50 \,m$.
The average power is given by $P_{av} = \frac{W}{t} = \frac{F_{boat} \times S}{t}$.
Substituting the values: $45000 = \frac{F_{boat} \times 50}{5} = F_{boat} \times 10$.
Thus, the force exerted by the boat engine is $F_{boat} = 4500 \,N$.
According to Newton's second law, the net force is $F_{net} = F_{boat} - F_{drag} = ma$.
Substituting the values: $4500 - F_{drag} = 1000 \times 4 = 4000 \,N$.
Therefore, the drag force is $F_{drag} = 4500 - 4000 = 500 \,N$.

(B) Power $P$ is defined as the rate of doing work,given by $P = F \cdot v$.
Since $F = m \cdot a = m \frac{dv}{dt}$,we have $P = m \frac{dv}{dt} \cdot v$.
Rearranging the terms,we get $P \cdot dt = m \cdot v \cdot dv$.
Integrating both sides,$\int_{0}^{t} P \cdot dt = \int_{0}^{v} m \cdot v \cdot dv$.
Since $P$ and $m$ are constant,$P \cdot t = m \cdot \frac{v^2}{2}$.
Solving for $v$,we get $v^2 = \frac{2Pt}{m}$,which implies $v = \sqrt{\frac{2P}{m}} \cdot t^{\frac{1}{2}}$.
Therefore,$v \propto t^{\frac{1}{2}}$.

(B) Given mass $m = 250 \ g = 0.25 \ kg$,length of string $l = 50 \ cm = 0.5 \ m$,and maximum tension $T_{\max} = 72 \ N$.
For a horizontal circular motion,the tension in the string provides the necessary centripetal force.
$T = m \omega^2 R$.
Here,the radius $R$ of the circular path is equal to the length of the string $l$ (assuming the string remains horizontal).
$T_{\max} = m \omega_{\max}^2 l$.
Substituting the values:
$72 = 0.25 \times \omega_{\max}^2 \times 0.5$.
$72 = 0.125 \times \omega_{\max}^2$.
$\omega_{\max}^2 = \frac{72}{0.125} = 576$.
$\omega_{\max} = \sqrt{576} = 24 \ rad/s$.

(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
For a $p-n$ junction to detect a signal,the energy of the incident photon must be greater than or equal to the band gap energy $(E_g = 2.8 \ eV)$.
If $E < E_g$,the photon cannot excite an electron from the valence band to the conduction band,and thus cannot be detected.
Using $h = 6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$,we have $hc = 18 \times 10^{-26} \ J \cdot m$.
Converting $E_g$ to Joules: $2.8 \ eV = 2.8 \times 1.6 \times 10^{-19} \ J = 4.48 \times 10^{-19} \ J$.
The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{E_g} = \frac{18 \times 10^{-26}}{4.48 \times 10^{-19}} \approx 4.017 \times 10^{-7} \ m = 401.7 \ \text{nm}$.
Any wavelength $\lambda > \lambda_0$ will have energy $E < E_g$ and will not be detected.
Among the given options,$600 \ nm$ is the only wavelength greater than $401.7 \ nm$.

(C) The energy of a photon required to create an electron-hole pair in a semiconductor must be at least equal to the band gap energy $(E_g)$.
$E_g = 0.6 eV$
We know that the energy of a photon is given by $E = \frac{hc}{\lambda}$.
To find the maximum wavelength $(\lambda_{max})$,we set the photon energy equal to the band gap energy:
$E_g = \frac{hc}{\lambda_{max}}$
$\lambda_{max} = \frac{hc}{E_g}$
Given $hc = 1242 eV-nm$ and $E_g = 0.6 eV$:
$\lambda_{max} = \frac{1242}{0.6} nm = 2070 nm$
If the wavelength is greater than $2070 nm$,the photon energy will be less than $0.6 eV$,which is insufficient to excite an electron from the valence band to the conduction band. Therefore,$2070 nm$ is the maximum wavelength.

(C) In an $n-p-n$ transistor,the emitter is heavily doped to provide a large number of charge carriers. The base is very thin and lightly doped to allow most of the charge carriers from the emitter to pass through to the collector. The collector is moderately doped and is the largest in size to dissipate the heat generated during operation. Therefore,the statement that the collector is lightly doped is incorrect.

(D) The output of an $\text{AND}$ gate is $Y = A \cdot B$.
$A$ $\text{NAND}$ gate is an $\text{AND}$ gate followed by a $\text{NOT}$ gate. Therefore,the output of a $\text{NAND}$ gate is $Y = \overline{A \cdot B}$.
According to De Morgan's theorem,$Y = \overline{A} + \overline{B}$.
The truth table for the $\text{NAND}$ gate is as follows:

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

Comparing this with the given options,when $A=1$ and $B=1$,the output $Y=0$.

(A) Let the inputs be $A$ and $B$. The circuit consists of an $AND$ gate and an $OR$ gate whose outputs are fed into a final $OR$ gate.
Let the output of the $AND$ gate be $Y_1 = A \cdot B$.
Let the output of the first $OR$ gate be $Y_2 = A + B$.
The final output $X$ is the $OR$ operation of $Y_1$ and $Y_2$:
$X = Y_1 + Y_2 = (A \cdot B) + (A + B)$.
Using the Boolean identity $(A \cdot B) + A + B = A + B$,we get $X = A + B$.
Truth Table:

$A, B$	$Y_1 = A \cdot B$	$Y_2 = A + B$	$X = Y_1 + Y_2$
$0, 0$	$0$	$0$	$0$
$0, 1$	$0$	$1$	$1$
$1, 0$	$0$	$1$	$1$
$1, 1$	$1$	$1$	$1$

Since the output $X$ follows the truth table of an $OR$ gate,the circuit behaves like an $OR$ gate.

(D) The given circuit consists of two buffers (or $NOT$ gates connected in series to act as buffers) followed by a $NAND$ gate.
$1$. The input $A$ passes through a buffer,so the output is $A$.
$2$. The input $B$ passes through a buffer,so the output is $B$.
$3$. These two outputs $A$ and $B$ are fed into a $NAND$ gate.
$4$. The output of a $NAND$ gate with inputs $A$ and $B$ is $Y = \overline{A \cdot B}$.
$5$. This is the definition of a $NAND$ gate.
Therefore,the circuit is equivalent to a $NAND$ gate.

(D) The given circuit consists of two $NAND$ gates acting as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.
Let the inputs be $A$ and $B$.
The first two $NAND$ gates invert the inputs to produce $\bar{A}$ and $\bar{B}$.
The final $NAND$ gate takes these as inputs,so the output $Y$ is given by $Y = \overline{\bar{A} \cdot \bar{B}}$.
Using De Morgan's theorem,$\overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.
This is the Boolean expression for an $OR$ gate.
Therefore,the circuit behaves as an $OR$ gate.

(A) In a $p-$type semiconductor,trivalent impurities (such as Boron,Aluminum,etc.) are added to an intrinsic semiconductor.
These trivalent atoms create an excess of holes in the valence band.
Therefore,holes act as the majority charge carriers,while electrons act as the minority charge carriers.

(A) The resistivity of materials is categorized as follows:
$1$. Conductors (Metals): $10^{-8} \Omega \cdot m$ to $10^{-6} \Omega \cdot m$.
$2$. Semiconductors: $10^{-5} \Omega \cdot m$ to $10^6 \Omega \cdot m$.
$3$. Insulators: $10^6 \Omega \cdot m$ to $10^{18} \Omega \cdot m$.
Given the resistivity of the material is $10^8 \Omega \cdot m$,which falls within the range of insulators.
Therefore,the material is an insulator.

(C) Given that the concentration of donor atoms (Arsenic) is $N_D = 4.5 \times 10^{21} \ m^{-3}$.
Since Arsenic is a pentavalent impurity,the number of electrons $n_e \approx N_D = 4.5 \times 10^{21} \ m^{-3}$.
The intrinsic carrier concentration is $n_i = 1.5 \times 10^{16} \ m^{-3}$.
Using the law of mass action,$n_e \cdot n_h = n_i^2$,the number of holes $n_h$ is given by:
$n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{21}} = \frac{2.25 \times 10^{32}}{4.5 \times 10^{21}} = 0.5 \times 10^{11} = 5 \times 10^{10} \ m^{-3}$.
The ratio of the number of electrons to the number of holes is:
$\frac{n_e}{n_h} = \frac{4.5 \times 10^{21}}{5 \times 10^{10}} = 0.9 \times 10^{11} = 9 \times 10^{10}$.

(D) In an $RC$ circuit,the charge $q$ on the capacitor at time $t$ is given by $q(t) = q_0(1 - e^{-t/RC})$.
Since the exponent of the exponential function must be dimensionless,the term $t/RC$ must be dimensionless.
Therefore,the dimensions of $RC$ must be equal to the dimensions of time $t$.
Thus,the time constant of an $RC$ circuit is $\tau = RC$,which has the dimension of time.

(B) Mobility $\mu$ is defined as the ratio of drift velocity $v_d$ to the applied electric field $E$, given by the formula $\mu = \frac{e \tau}{m}$.
Here, $e$ is the charge (unit: $A \cdot s$), $\tau$ is the relaxation time (unit: $s$), and $m$ is the mass (unit: $kg$).
Substituting these units into the formula: $\text{Unit of } \mu = \frac{(A \cdot s) \cdot s}{kg} = \frac{A \cdot s^2}{kg}$.
This can be written as $kg^{-1} \,s^2 \,A$.

(B) In nuclear physics,$\beta$-decay is a type of radioactive decay in which a beta particle (fast energetic electron or positron) is emitted from an atomic nucleus,transforming the original nuclide into an isobar of that nuclide.
Beta decay is a fundamental process governed by the weak nuclear force,which allows for the transmutation of quarks (e.g.,a down quark changing into an up quark) within nucleons.

(C) When a light wave enters a medium of refractive index $\mu$,its wavelength becomes $\lambda^{\prime} = \frac{\lambda}{\mu}$.
Fringe width in $\text{YDSE}$ is given by $\beta = \frac{\lambda^{\prime} D}{d}$.
Given: $\lambda = 550 \times 10^{-9} \ m$,$D = 0.6 \ m$,$d = 0.2 \times 10^{-3} \ m$,and $\mu = \frac{11}{9}$.
Substituting the values:
$\beta = \frac{(\lambda / \mu) D}{d} = \frac{\lambda D}{\mu d} = \frac{550 \times 10^{-9} \times 0.6}{(11/9) \times 0.2 \times 10^{-3}}$.
$\beta = \frac{550 \times 10^{-9} \times 0.6 \times 9}{11 \times 0.2 \times 10^{-3}} = \frac{330 \times 10^{-9} \times 9}{2.2 \times 10^{-3}} = \frac{2970 \times 10^{-9}}{2.2 \times 10^{-3}} = 1350 \times 10^{-6} \ m = 1.35 \ mm$.

(A) The angular fringe width $\beta_{\theta}$ is given by the formula $\beta_{\theta} = \frac{\lambda}{d}$,where $\lambda$ is the wavelength of light and $d$ is the slit separation.
Since $\beta_{\theta} \propto \lambda$,when the system is immersed in water,the wavelength changes to $\lambda' = \frac{\lambda}{\mu}$.
Therefore,the new angular width $\beta_{\theta}'$ is $\beta_{\theta}' = \frac{\beta_{\theta}}{\mu}$.
Given $\beta_{\theta} = 0.2^{\circ}$ and $\mu = \frac{4}{3}$,the new angular width is $\beta_{\theta}' = 0.2^{\circ} \times \frac{3}{4} = 0.15^{\circ}$.
The change in angular width is $\Delta \beta_{\theta} = \beta_{\theta} - \beta_{\theta}' = 0.2^{\circ} - 0.15^{\circ} = 0.05^{\circ}$.

(B) The condition for the position of the $m^{\text{th}}$ order maximum in Young's double slit experiment is given by $y = \frac{m \lambda D}{d}$.
For the $m^{\text{th}}$ order of red light and $(m+1)^{\text{th}}$ order of blue light to coincide,their positions must be equal:
$y_{\text{red}} = y_{\text{blue}}$
$\frac{m \lambda_1 D}{d} = \frac{(m+1) \lambda_2 D}{d}$
$m \lambda_1 = (m+1) \lambda_2$
Substituting the given values:
$m(780 \ nm) = (m+1)(520 \ nm)$
$780m = 520m + 520$
$780m - 520m = 520$
$260m = 520$
$m = \frac{520}{260} = 2$
Thus,the $2^{\text{nd}}$ order of red light coincides with the $3^{\text{rd}}$ order of blue light.

(A) The angular width of a fringe in a double-slit experiment is given by $\theta = \frac{\lambda}{d}$,where $\lambda$ is the wavelength of light and $d$ is the slit separation.
Since $d$ remains constant,the angular width is directly proportional to the wavelength: $\theta \propto \lambda$.
When the apparatus is immersed in a medium of refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Therefore,the new angular width $\theta'$ is given by $\theta' = \frac{\theta}{\mu}$.
Given $\theta = 0.15^{\circ}$ and $\mu = \frac{5}{3}$,we have:
$\theta' = \frac{0.15^{\circ}}{5/3} = 0.15^{\circ} \times \frac{3}{5} = 0.03^{\circ} \times 3 = 0.09^{\circ}$.

(A) The position of the $n^{\text{th}}$ order maximum in Young's double slit experiment is given by the formula: $Y_n = \frac{n \lambda D}{d}$.
Since the position $Y_n$ remains the same for both light sources,we can equate the expressions:
$Y_n = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$.
This simplifies to: $n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 12$,$\lambda_1 = 6000 \ Å$,and $\lambda_2 = 4800 \ Å$,we substitute these values:
$12 \times 6000 = n_2 \times 4800$.
$n_2 = \frac{12 \times 6000}{4800} = \frac{72000}{4800} = 15$.
Therefore,the $15^{\text{th}}$ order maximum will be visible at the same point.

(C) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
From the formula,we can see that $\beta \propto \frac{\lambda}{d}$.
Let the initial fringe width be $\beta_1 = \frac{\lambda_1 D}{d_1}$.
According to the problem,the new wavelength $\lambda_2 = 4\lambda_1$ and the new slit distance $d_2 = \frac{d_1}{2}$.
The new fringe width $\beta_2$ is given by: $\beta_2 = \frac{\lambda_2 D}{d_2} = \frac{(4\lambda_1) D}{(d_1/2)} = 8 \times \frac{\lambda_1 D}{d_1} = 8\beta_1$.
Therefore,the distance between two maxima will become $8$ times the original value.