TS EAMCET 2023 Physics Question Paper with Answer and Solution

(C) The relative strengths of the four fundamental forces in nature are as follows:
$1$. Strong nuclear force: Relative strength = $1$
$2$. Electromagnetic force: Relative strength = $10^{-2}$
$3$. Weak nuclear force: Relative strength = $10^{-13}$
$4$. Gravitational force: Relative strength = $10^{-39}$
Comparing these with the given options:
$(A)$ Strong nuclear force matches with $(f)$ $1$.
$(B)$ Weak nuclear force matches with $(h)$ $10^{-13}$.
$(C)$ Electromagnetic force matches with $(e)$ $10^{-2}$.
$(D)$ Gravitational force matches with $(i)$ $10^{-39}$.
Therefore,the correct matching is $A-f, B-h, C-e, D-i$.

(C) Transverse waves require a medium with shear modulus to propagate,which is why they can propagate through solids and on the surface of liquids. They cannot propagate through gases or the interior of liquids. Longitudinal waves can propagate through all states of matter (solids,liquids,and gases) but cannot propagate through a vacuum. Therefore,the statement that transverse waves can propagate through solids is correct.

(A) Given: Frequency of tuning fork $A$ $(f_A)$ = $250 \,Hz$. Frequency of tuning fork $B$ $(f_B)$ = $x \,Hz$.
Initially, the beat frequency is $5 \,Hz$, so $|f_A - f_B| = 5$.
This implies $250 - x = 5$ or $x - 250 = 5$, giving $x = 245 \,Hz$ or $x = 255 \,Hz$.
When tuning fork $B$ is waxed, its frequency decreases $(f_B' < f_B)$.
After waxing, the new beat frequency is $3 \,Hz$.
If $x = 255 \,Hz$, then $f_B$ decreases towards $250 \,Hz$, so the beat frequency would increase from $5$ to a higher value or pass through $0$ and then increase. This does not match the observation of $3 \,Hz$.
If $x = 245 \,Hz$, then $f_B$ decreases further away from $250 \,Hz$ (e.g., to $244 \,Hz$), which would increase the beat frequency to $6 \,Hz$. However, if we consider the frequency of $B$ was $245 \,Hz$ and it was slightly loaded, the beat frequency decreases from $5$ to $3$ only if the frequency of $B$ was originally $255 \,Hz$ and it decreased to $253 \,Hz$ ($255 - 253 = 2$ - incorrect) or if $x = 245 \,Hz$ and the beat frequency changed. Let's re-evaluate: If $x = 255 \,Hz$, loading $B$ makes $f_B$ approach $250 \,Hz$, decreasing the beat frequency. Thus, $x = 255 \,Hz$ is the correct value.

(A) Let the speed of sound be $v$. The observer moves towards the stationary source with a speed $v_0 = \frac{v}{5}$.
According to the Doppler effect,when an observer moves towards a stationary source,the apparent frequency $f'$ is given by:
$f' = f \left( \frac{v + v_0}{v} \right)$
Substituting the given values:
$f' = f \left( \frac{v + \frac{v}{5}}{v} \right)$
$f' = f \left( \frac{\frac{6v}{5}}{v} \right)$
$f' = f \left( \frac{6}{5} \right) = 1.2 f$
Therefore,the apparent frequency recorded by the observer is $1.2 f$.

(D) According to the Doppler effect,the observed frequency $f^{\prime}$ is given by $f^{\prime} = f_0 \left( \frac{v + v_0}{v - v_s} \right)$.
Here,$v$ is the speed of sound,$v_0$ is the speed of the observer,and $v_s$ is the speed of the source.
Given that the source is stationary,$v_s = 0$.
The observer moves towards the source with a speed $v_0 = \frac{v}{5}$.
Substituting these values into the formula: $f^{\prime} = f_0 \left( \frac{v + v/5}{v} \right) = f_0 \left( \frac{6v/5}{v} \right) = 1.2 f_0$.
The fractional increase in frequency is $\frac{f^{\prime} - f_0}{f_0} = \frac{1.2 f_0 - f_0}{f_0} = 0.2$.
To express this as a percentage: $0.2 \times 100 \% = 20 \%$.

(B) According to the Doppler effect formula for a source and observer moving away from each other:
$f' = f_0 \left( \frac{v - v_o}{v + v_s} \right)$
Here, the observed frequency $f' = 1980 \,Hz$.
The speed of sound in air $v = 340 \,ms^{-1}$.
The velocity of the observer $v_o = 10 \,ms^{-1}$ (moving away, so negative sign in numerator).
The velocity of the source $v_s = 10 \,ms^{-1}$ (moving away, so positive sign in denominator).
Substituting the values:
$1980 = f_0 \left( \frac{340 - 10}{340 + 10} \right)$
$1980 = f_0 \left( \frac{330}{350} \right)$
$f_0 = 1980 \times \frac{350}{330}$
$f_0 = 1980 \times \frac{35}{33}$
$f_0 = 60 \times 35 = 2100 \,Hz$.

(D) For a closed organ pipe,the frequency of the $n^{th}$ harmonic is given by $f_c = \frac{nv}{4L_c}$. For the $7^{th}$ harmonic,$n = 7$,so $f_c = \frac{7v}{4L_c}$.
For an open organ pipe,the frequency of the $n^{th}$ harmonic is given by $f_o = \frac{nv}{2L_o}$. For the $4^{th}$ harmonic,$n = 4$,so $f_o = \frac{4v}{2L_o}$.
Given that the two frequencies are in unison,$f_c = f_o$.
Therefore,$\frac{7v}{4L_c} = \frac{4v}{2L_o}$.
Simplifying the equation: $\frac{7}{4L_c} = \frac{2}{L_o}$.
Rearranging for the ratio of lengths: $\frac{L_c}{L_o} = \frac{7}{4 \times 2} = \frac{7}{8}$.

(C) The given equation of the wave is $y = 0.02 \sin(\pi x - 8 \pi t)$.
Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$:
Here,the angular wave number $k = \pi \ m^{-1}$ and the angular frequency $\omega = 8 \pi \ rad/s$.
The velocity of the wave $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,we get $v = \frac{8 \pi}{\pi} = 8 \ m/s$.

(A) The given wave equation is $y = 5 \times 10^{-3} \sin \left(12.5 \pi x - \frac{\pi}{2} t\right)$.
Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$,we get:
Wave number $k = 12.5 \pi$ and angular frequency $\omega = \frac{\pi}{2}$.
We know that $k = \frac{2 \pi}{\lambda}$,so $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{12.5 \pi} = \frac{2}{12.5} = 0.16 \ m$.
We also know that $\omega = \frac{2 \pi}{T}$,so $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{\pi / 2} = 4 \ s$.
Therefore,the wavelength is $0.16 \ m$ and the time period is $4 \ s$.

(D) The speed of a transverse wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the mass per unit length.
At the lower end of the string, the tension $T$ is provided by the load of mass $M$. Let the mass of the string be $m$ and its length be $L$. The tension at the lower end is $T = Mg$.
When the lift accelerates upwards with $a_1 = 2.1 \,ms^{-2}$, the effective acceleration is $g_{eff1} = g + a_1 = 10 + 2.1 = 12.1 \,ms^{-2}$. The tension is $T_1 = M(g + a_1)$.
The wave speed is $v_1 = \sqrt{\frac{M(g+a_1)}{\mu}} = 88 \,ms^{-1}$.
When the lift accelerates downwards with $a_2 = 1.9 \,ms^{-2}$, the effective acceleration is $g_{eff2} = g - a_2 = 10 - 1.9 = 8.1 \,ms^{-2}$. The tension is $T_2 = M(g - a_2)$.
The wave speed is $v_2 = \sqrt{\frac{M(g-a_2)}{\mu}}$.
Taking the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{g-a_2}{g+a_1}} = \sqrt{\frac{8.1}{12.1}} = \sqrt{\frac{81}{121}} = \frac{9}{11}$.
Therefore, $v_2 = v_1 \times \frac{9}{11} = 88 \times \frac{9}{11} = 8 \times 9 = 72 \,ms^{-1}$.

(D) Given: Angular speed $\omega = 40 \ rad \ s^{-1}$,radius $r = 50 \ cm = 0.5 \ m$.
The linear speed of the source is $v_s = r \omega = 0.5 \times 40 = 20 \ m \ s^{-1}$.
The Doppler effect formula for maximum frequency (when the source moves towards the observer) is $n_{\max} = n \left( \frac{v}{v - v_s} \right)$.
The Doppler effect formula for minimum frequency (when the source moves away from the observer) is $n_{\min} = n \left( \frac{v}{v + v_s} \right)$.
The ratio of maximum to minimum frequency is $\frac{n_{\max}}{n_{\min}} = \frac{v + v_s}{v - v_s}$.
Substituting the values $v = 340 \ m \ s^{-1}$ and $v_s = 20 \ m \ s^{-1}$:
$\frac{n_{\max}}{n_{\min}} = \frac{340 + 20}{340 - 20} = \frac{360}{320} = \frac{9}{8}$.
Thus,the ratio is $9: 8$.

(C) Given: Masses $m_1 = 1 \,g$ and $m_2 = 4 \,g$. Kinetic energies are equal, $K_1 = K_2$.
We know the relation between kinetic energy $K$ and linear momentum $p$ is $K = \frac{p^2}{2m}$, which implies $p = \sqrt{2mK}$.
Since $K_1 = K_2$, the ratio of momenta is $\frac{p_1}{p_2} = \frac{\sqrt{2m_1K_1}}{\sqrt{2m_2K_2}} = \sqrt{\frac{m_1}{m_2}}$.
Substituting the values: $\frac{p_1}{p_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus, the ratio of their linear momenta is $1:2$.

(D) Let the mass of the first piece be $m_1 = 4 \,kg$ and its velocity be $v_1$.
Let the mass of the second piece be $m_2 = 12 \,kg$ and its velocity be $v_2 = 4 \,ms^{-1}$.
According to the law of conservation of linear momentum, the initial momentum of the bomb is zero (assuming it was at rest).
Therefore, $m_1 v_1 + m_2 v_2 = 0$.
Taking magnitudes, $m_1 v_1 = m_2 v_2$.
$4 \times v_1 = 12 \times 4$.
$v_1 = 12 \,ms^{-1}$.
The kinetic energy of the $4 \,kg$ piece is $KE_1 = \frac{1}{2} m_1 v_1^2$.
$KE_1 = \frac{1}{2} \times 4 \times (12)^2 = 2 \times 144 = 288 \,J$.

(A) The engine is moving a mass up an inclined plane at a constant velocity. The force required to move the mass up the plane must balance the component of the gravitational force acting down the plane.
Force $F = mg \sin \theta$.
Given that the inclination is $1$ in $50$,we have $\sin \theta = 1/50$.
Taking $g = 10 \ ms^{-2}$,the force is $F = 5000 \times 10 \times (1/50) = 1000 \ N$.
The power $P$ is given by $P = F \times v$,where $v = 5 \ ms^{-1}$.
$P = 1000 \ N \times 5 \ ms^{-1} = 5000 \ W = 5 \ kW$.

(D) Given: Mass $m = 3 \,kg$,displacement $s = \frac{t^3}{3}$.
Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(\frac{t^3}{3}) = t^2$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(t^2) = 2t$.
Force $F = ma = 3 \times 2t = 6t$.
Work done $W = \int F \cdot ds = \int_0^2 F \cdot v dt = \int_0^2 (6t)(t^2) dt$.
$W = \int_0^2 6t^3 dt = 6 \left[ \frac{t^4}{4} \right]_0^2$.
$W = 6 \times \frac{16}{4} = 6 \times 4 = 24 \,J$.

(C) The work done in blowing a soap bubble is equal to the surface energy stored in the bubble.
Since a soap bubble has two surfaces (inner and outer), the work done $W$ is given by:
$W = T \times \Delta A = T \times 2 \times (4 \pi R^2) = 8 \pi R^2 T$
From this expression, we can see that $W \propto R^2$.
Let $W_1$ be the work done for radius $R$, so $W_1 = W$.
Let $W_2$ be the work done for radius $2R$.
Then, $\frac{W_2}{W_1} = \frac{(2R)^2}{R^2} = \frac{4R^2}{R^2} = 4$.
Therefore, $W_2 = 4W$.

(D) Given: Mass $m = 3 \text{ kg}$, Displacement $x = \frac{t^3}{3} \text{ m}$.
First, find the velocity $v$ by differentiating displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2 \text{ m/s}$.
At $t = 0$, $v_i = 0^2 = 0 \text{ m/s}$.
At $t = 2 \text{ s}$, $v_f = 2^2 = 4 \text{ m/s}$.
According to the Work-Energy Theorem, the work done $W$ is equal to the change in kinetic energy:
$W = \Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
Substituting the values:
$W = \frac{1}{2} \times 3 \times (4)^2 - \frac{1}{2} \times 3 \times (0)^2 = \frac{1}{2} \times 3 \times 16 = 24 \text{ J}$.

(C) Power $P$ is given by $P = Fv$, where $F$ is force and $v$ is velocity.
Since $P$ is constant, $Fv = \text{constant}$.
Using Newton's second law, $F = ma$, so $(ma)v = \text{constant}$.
Since $a = \frac{dv}{dt}$ and $v = \frac{ds}{dt}$, we have $m \left(\frac{dv}{dt}\right)v = P$.
$mv \, dv = P \, dt$.
Integrating both sides: $\int mv \, dv = \int P \, dt \implies \frac{1}{2}mv^2 = Pt$.
Thus, $v^2 \propto t$, which means $v \propto t^{1/2}$.
Since $v = \frac{ds}{dt}$, we have $\frac{ds}{dt} \propto t^{1/2}$.
Integrating with respect to time: $s \propto \int t^{1/2} \, dt$.
$s \propto t^{3/2}$.

(B) $1$. $A$ point source of light emits light in all directions,forming spherical wavefronts as the light propagates outward.
$2$. When these spherical wavefronts strike a concave mirror,the rays are reflected parallel to the principal axis (for paraxial rays).
$3$. $A$ set of parallel rays corresponds to a plane wavefront.
$4$. Therefore,the incident wavefront is spherical,and the reflected wavefront is planar.

(A) Given: Focal length of objective $f_o = 2 \,cm$, focal length of eyepiece $f_e = 3 \,cm$, and distance between lenses $L = 15 \,cm$.
Since the final image is formed at infinity, the image formed by the objective lens must lie at the principal focus of the eyepiece.
Therefore, the image distance from the eyepiece is $v_e = f_e = 3 \,cm$.
The image distance from the objective lens is $v_o = L - f_e = 15 \,cm - 3 \,cm = 12 \,cm$.
Using the lens formula for the objective lens: $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$
Substituting the values: $\frac{1}{2} = \frac{1}{12} - \frac{1}{u_o}$
$\frac{1}{u_o} = \frac{1}{12} - \frac{1}{2} = \frac{1 - 6}{12} = -\frac{5}{12}$
$u_o = -\frac{12}{5} = -2.4 \,cm$.
The magnitude of the object distance is $2.4 \,cm$ and the image distance is $12 \,cm$.

(B) For a light ray passing through a prism,the angle of deviation $\delta$ at the first surface is $\theta_1 = i - r_1$ and at the second surface is $\theta_2 = e - r_2$.
The total deviation $\delta$ is the sum of deviations at both surfaces:
$\delta = \theta_1 + \theta_2$
$\delta = (i - r_1) + (e - r_2)$
$\delta = (i + e) - (r_1 + r_2)$
Since the prism angle $A = r_1 + r_2$,we substitute this into the equation:
$\delta = i + e - A$

(A) Given,angle of incidence $i = 2r$,where $r$ is the angle of refraction.
According to Snell's law,$\mu_1 \sin i = \mu_2 \sin r$.
Assuming the light travels from air $(\mu_1 = 1)$ to a medium with refractive index $\mu_2 = \mu$,we have:
$1 \cdot \sin(2r) = \mu \sin r$
Using the trigonometric identity $\sin(2r) = 2 \sin r \cos r$,we get:
$2 \sin r \cos r = \mu \sin r$
Since $\sin r \neq 0$,we can divide both sides by $\sin r$:
$2 \cos r = \mu \implies \cos r = \frac{\mu}{2}$
Thus,$r = \cos^{-1}\left(\frac{\mu}{2}\right)$.
Since $i = 2r$,the angle of incidence is $i = 2 \cos^{-1}\left(\frac{\mu}{2}\right)$.

(C) The built-in potential (barrier potential) of a $p-n$ junction diode is denoted by $V_B = 0.7 \,V$.
When a $p-n$ junction diode is forward biased with an external voltage $V_f$, the effective barrier height (or effective potential barrier) decreases.
The formula for the effective barrier height $V_{eff}$ is given by $V_{eff} = V_B - V_f$.
Given $V_B = 0.7 \,V$ and $V_f = 0.3 \,V$.
Substituting the values, we get $V_{eff} = 0.7 \,V - 0.3 \,V = 0.4 \,V$.
Therefore, the effective barrier height is $0.4 \,V$.

(A) Given: Base current,$i_B = 10 \mu A$.
Emitter current,$i_E = 1 \text{ mA} = 1000 \mu A$.
According to the fundamental relation for a transistor,the emitter current is the sum of the base current and the collector current: $i_E = i_B + i_C$.
Rearranging the formula to find the collector current: $i_C = i_E - i_B$.
Substituting the values: $i_C = 1000 \mu A - 10 \mu A = 990 \mu A$.

(B) In the circuit symbol of a transistor,the arrow on the emitter terminal indicates the direction of conventional current flow.
For an $n-p-n$ transistor,the current flows from the base to the emitter,so the arrow on the emitter points outwards.
For a $p-n-p$ transistor,the current flows from the emitter to the base,so the arrow on the emitter points inwards.
In the given symbol,the arrow on the emitter is pointing outwards,which confirms that it is an $n-p-n$ transistor.

(A) $NAND$ gate performs the operation $Y = \overline{A \cdot B}$.
When this output is fed into a $NOT$ gate, the final output becomes $Y' = \overline{Y} = \overline{(\overline{A \cdot B})}$.
By the law of double negation, $\overline{(\overline{X})} = X$. Therefore, $Y' = A \cdot B$.
This is the Boolean expression for an $AND$ gate.

Input $(A, B)$	$NAND$ Output $(\overline{A \cdot B})$	$NOT$ Output $(A \cdot B)$
$0, 0$	$1$	$0$
$0, 1$	$1$	$0$
$1, 0$	$1$	$0$
$1, 1$	$0$	$1$

(A) Given inputs are $X=1$ and $Y=1$.
$1$. The input $X$ passes through a $NOT$ gate,so the input to the $OR$ gate is $\bar{X} = 0$.
$2$. The input $Y$ is directly connected to the $OR$ gate,so the input is $1$.
$3$. The output $P$ of the $OR$ gate is $P = \bar{X} + Y = 0 + 1 = 1$.
$4$. The input $X$ is directly connected to the $NAND$ gate,so the input is $1$.
$5$. The input $Y$ passes through a $NOT$ gate,so the input to the $NAND$ gate is $\bar{Y} = 0$.
$6$. The output $Q$ of the $NAND$ gate is $Q = \overline{X \cdot \bar{Y}} = \overline{1 \cdot 0} = \overline{0} = 1$.
$7$. Finally,$P$ and $Q$ are inputs to a $NOR$ gate to give output $R$.
$8$. $R = \overline{P + Q} = \overline{1 + 1} = \overline{1} = 0$.
Thus,$P=1, Q=1, R=0$.

(A) The circuit consists of two $NOT$ gates, two $AND$ gates, and one final $AND$ gate.
The inputs to the first $AND$ gate are $A$ and $\overline{B}$, so its output is $A \cdot \overline{B}$.
The inputs to the second $AND$ gate are $\overline{A}$ and $B$, so its output is $\overline{A} \cdot B$.
The final output $Y$ is the $AND$ operation of these two outputs: $Y = (A \cdot \overline{B}) \cdot (\overline{A} \cdot B)$.
Using the associative and commutative properties of Boolean algebra: $Y = A \cdot \overline{A} \cdot B \cdot \overline{B}$.
Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$, the output $Y = 0 \cdot 0 = 0$ for all input combinations.
Therefore, the output $Y$ is always $0$.

(A) donor impurity atom provides an extra electron to the semiconductor lattice.
When a donor impurity is added,the number of free electrons in the conduction band increases significantly.
According to the law of mass action,$n_e \cdot n_h = n_i^2$,where $n_e$ is the electron concentration,$n_h$ is the hole concentration,and $n_i$ is the intrinsic carrier concentration.
Since $n_e$ increases,the hole concentration $n_h$ must decrease to maintain the product $n_i^2$ as a constant at a given temperature.
Therefore,the electron concentration increases and the hole concentration decreases.
Thus,option $A$ is the correct answer.

(B) The current in a semiconductor is given by the formula $I = n e A v_d$,where $n$ is the charge carrier concentration,$e$ is the elementary charge,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.
For electrons and holes,the currents are $I_e = n_e e A v_e$ and $I_h = n_h e A v_h$.
The ratio of currents is given by $\frac{I_e}{I_h} = \frac{n_e e A v_e}{n_h e A v_h} = \frac{n_e v_e}{n_h v_h} = \frac{7}{4}$.
We are given the ratio of drift velocities as $\frac{v_e}{v_h} = \frac{5}{4}$.
Substituting this into the ratio equation: $\frac{n_e}{n_h} \times \frac{5}{4} = \frac{7}{4}$.
Solving for the ratio of concentrations: $\frac{n_e}{n_h} = \frac{7}{4} \times \frac{4}{5} = \frac{7}{5}$.
Therefore,the ratio of concentrations of electrons and holes is $7: 5$.

(D) When an $n$-type semiconductor is heated,the thermal energy provides enough energy to break covalent bonds in the crystal lattice.
This process generates electron-hole pairs.
Since each electron-hole pair consists of one free electron and one hole,the number of electrons and the number of holes increase by the same amount.
Therefore,the number of holes and electrons increases equally.

(D) An $LED$ (Light Emitting Diode) is a $p-n$ junction diode that emits light when forward biased.
Key characteristics of LEDs include:
$1$. They operate at low voltages compared to incandescent bulbs.
$2$. They do not require any warm-up time.
$3$. They have very fast on-off switching capabilities,making them ideal for high-speed communication and display technologies.
$4$. The bandwidth of light emitted is typically narrow,not $4000 \ Å - 7000 \ Å$ (which covers the entire visible spectrum).
Therefore,the correct statement is that they have fast on-off switching.

(C) In a photodiode,the reverse saturation current is primarily due to the flow of minority charge carriers. When light is incident on the photodiode,it generates electron-hole pairs,which increases the number of minority charge carriers. Because the initial number of minority carriers is very small,even a small amount of incident light causes a significant fractional change in their concentration. This leads to a noticeable change in the reverse current,making the photodiode highly sensitive to light. In contrast,in forward bias,the current is dominated by majority carriers,which are already present in large numbers,making the fractional change due to incident light negligible.

(A) Intensity of radiation is defined as the energy incident per unit area per unit time: $I = \frac{E}{A \cdot t}$.
For a photon of frequency $f$,the energy is $E = hf$. Thus,the total energy incident is $E_{total} = N \cdot hf$.
For surface $A$: $I_A = \frac{n \cdot hf}{A \cdot t}$.
For surface $B$: $I_B = \frac{(2n) \cdot h(3f)}{A \cdot (4t)} = \frac{6nhf}{4At} = \frac{3nhf}{2At}$.
Taking the ratio: $\frac{I_A}{I_B} = \frac{nhf}{At} \cdot \frac{2At}{3nhf} = \frac{2}{3}$.
Therefore,the ratio is $2: 3$.

(C) The relative strength of the gravitational force between two protons is approximately $10^{-39}$ times the strength of the electromagnetic force between them.
Therefore,the ratio of the gravitational force to the electromagnetic force is $10^{-39} / 10^{-2} = 10^{-37}$.
Thus,the correct option is $C$.

(A) The relative strength of the strong nuclear force is approximately $1$.
The relative strength of the weak nuclear force is approximately $10^{-13}$.
Therefore,the ratio of the relative strengths of strong and weak nuclear forces is given by:
Ratio $= \frac{\text{Strong nuclear force}}{\text{Weak nuclear force}} = \frac{1}{10^{-13}} = 10^{13}$.

(B) The formula for Fresnel distance $(Z_F)$ is given by $Z_F = \frac{a^2}{\lambda}$, where $a$ is the slit width and $\lambda$ is the wavelength of light.
Given:
Slit width $a = 2 \, mm = 2 \times 10^{-3} \, m$
Wavelength $\lambda = 4000 \, Å = 4 \times 10^{-7} \, m$
Substituting these values into the formula:
$Z_F = \frac{(2 \times 10^{-3})^2}{4 \times 10^{-7}}$
$Z_F = \frac{4 \times 10^{-6}}{4 \times 10^{-7}}$
$Z_F = 10 \, m$
Therefore, the Fresnel distance is $10 \, m$.

(B) The angular width of the central maximum in single slit diffraction is given by $\theta = \frac{2\lambda}{a}$,where $a$ is the slit width.
Thus,$\theta \propto \lambda$.
Let the initial wavelength be $\lambda_1 = 6000 \text{ Å}$ and the initial angular width be $\theta_1 = \theta$.
Then $\theta_1 = k \lambda_1$ $(i)$
When the wavelength is changed to $\lambda_2 = \lambda$,the angular width decreases by $30 \%$.
So,$\theta_2 = \theta_1 - 0.30 \theta_1 = 0.70 \theta_1$.
Since $\theta_2 = k \lambda_2$,we have $0.70 \theta_1 = k \lambda_2$ (ii)
Dividing equation (ii) by equation $(i)$:
$\frac{k \lambda_2}{k \lambda_1} = \frac{0.70 \theta_1}{\theta_1}$
$\frac{\lambda_2}{6000 \text{ Å}} = 0.70$
$\lambda_2 = 0.70 \times 6000 \text{ Å} = 4200 \text{ Å}$.

(C) In $YDSE$,the position of the $n^{\text{th}}$ maxima is given by $y_n = \frac{n \lambda D}{d} + y_0$,where $y_0$ is the position of the central maxima.
Given in air: $y_0 = 2 \text{ cm}$ and $y_{10} = 5 \text{ cm}$.
The fringe width $\beta = y_{10} - y_0 = 5 \text{ cm} - 2 \text{ cm} = 3 \text{ cm}$.
We know $\beta = \frac{\lambda D}{d}$. Thus,$\frac{\lambda D}{d} = 3 \text{ cm}$.
When the apparatus is immersed in a liquid of refractive index $\mu = 1.5$,the wavelength becomes $\lambda' = \frac{\lambda}{\mu}$.
The new fringe width $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu} = \frac{3 \text{ cm}}{1.5} = 2 \text{ cm}$.
The central maxima position $y_0'$ remains unchanged because it is independent of the wavelength,so $y_0' = 2 \text{ cm}$.
The new position of the $10^{\text{th}}$ maxima is $y_{10}' = y_0' + 10 \beta' = 2 \text{ cm} + 10 \times \frac{3 \text{ cm}}{10 \times 1.5} = 2 \text{ cm} + 2 \text{ cm} = 4 \text{ cm}$.
Alternatively,$y_{10}' = y_0 + 10 \beta' = 2 \text{ cm} + 10 \times (0.2 \text{ cm}) = 4 \text{ cm}$.

(B) The separation between the two slits is $d = 1 \,mm = 10^{-3} \,m$.
The distance between the screen and the slits is $D = 1 \,m$.
The wavelength of the light used is $\lambda = 500 \,nm = 500 \times 10^{-9} \,m$.
The formula for fringe width $\beta$ is given by $\beta = \frac{D \lambda}{d}$.
Substituting the given values:
$\beta = \frac{1 \times 500 \times 10^{-9}}{10^{-3}} = 500 \times 10^{-6} \,m = 0.5 \times 10^{-3} \,m = 0.5 \,mm$.

(B) Given: Distance between slits $d = 1 \,mm = 10^{-3} \,m$, wavelength $\lambda = 6.5 \times 10^{-7} \,m$, distance of screen $D = 1 \,m$.
For the $n^{th}$ dark fringe, the position is given by $y_n = (n - 0.5) \frac{\lambda D}{d}$.
For the $3^{rd}$ dark fringe $(n=3)$: $y_3 = (3 - 0.5) \frac{6.5 \times 10^{-7} \times 1}{10^{-3}} = 2.5 \times 6.5 \times 10^{-4} = 16.25 \times 10^{-4} \,m$.
For the $n^{th}$ bright fringe, the position is given by $y_n = \frac{n \lambda D}{d}$.
For the $5^{th}$ bright fringe $(n=5)$: $y_5 = \frac{5 \times 6.5 \times 10^{-7} \times 1}{10^{-3}} = 32.5 \times 10^{-4} \,m$.
The distance between the $3^{rd}$ dark fringe and the $5^{th}$ bright fringe is $\Delta y = y_5 - y_3 = 32.5 \times 10^{-4} - 16.25 \times 10^{-4} = 16.25 \times 10^{-4} \,m = 1.625 \,mm$.