AIPMT 1990 Physics Question Paper with Answer and Solution

(D) The dimensional formula for frequency $f$ is $[M^0 L^0 T^{-1}]$.
The dimensional formula for mass $m$ is $[M^1 L^0 T^0]$.
The dimensional formula for spring constant $K$ is $[M^1 L^0 T^{-2}]$.
Given the relation $f = C m^x K^y$,substituting the dimensions on both sides:
$[M^0 L^0 T^{-1}] = [M^1]^x [M^1 T^{-2}]^y$
$[M^0 L^0 T^{-1}] = [M^{x+y} T^{-2y}]$
Comparing the powers of $M$ and $T$ on both sides:
For $M$: $x + y = 0$
For $T$: $-2y = -1 \implies y = \frac{1}{2}$
Substituting $y = \frac{1}{2}$ into $x + y = 0$,we get $x + \frac{1}{2} = 0 \implies x = -\frac{1}{2}$.
Thus,$x = -\frac{1}{2}$ and $y = \frac{1}{2}$.

(C) Pressure is defined as the force applied per unit area.
$P = \frac{F}{A}$
The dimensional formula for force $(F)$ is $[MLT^{-2}]$.
The dimensional formula for area $(A)$ is $[L^2]$.
Therefore,the dimensions of pressure are:
$[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.

(B) The formula for viscous force is $F = - \eta A \frac{\Delta v}{\Delta z}$.
Rearranging for $\eta$,we get $\eta = \frac{F}{A (\Delta v / \Delta z)}$.
Substituting the dimensions of each quantity:
Force $F = [M L T^{-2}]$
Area $A = [L^2]$
Velocity gradient $\frac{\Delta v}{\Delta z} = \frac{[L T^{-1}]}{[L]} = [T^{-1}]$
Now,calculating the dimensions of $\eta$:
$[\eta] = \frac{[M L T^{-2}]}{[L^2] [T^{-1}]} = [M L^{1-2} T^{-2+1}] = [M L^{-1} T^{-1}]$.
Thus,the correct option is $B$.

(C) Let the total distance be $2d$. The car travels the first half distance $d$ with velocity $v_1 = 40 \, km/h$ and the second half distance $d$ with velocity $v_2 = 60 \, km/h$.
Time taken for the first half is $t_1 = \frac{d}{v_1} = \frac{d}{40}$.
Time taken for the second half is $t_2 = \frac{d}{v_2} = \frac{d}{60}$.
Average velocity $v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{t_1 + t_2} = \frac{2d}{\frac{d}{40} + \frac{d}{60}}$.
$v_{av} = \frac{2}{\frac{1}{40} + \frac{1}{60}} = \frac{2}{\frac{3+2}{120}} = \frac{2 \times 120}{5} = \frac{240}{5} = 48 \, km/h$.

(B) The acceleration of a point on the tip of the blade is the centripetal acceleration,given by $a = \omega^2 R$.
First,convert the angular velocity from $r.p.m.$ to $rad/s$:
$\omega = 2 \pi f = 2 \pi \times \frac{1200}{60} = 40 \pi \, rad/s$.
Given the radius $R = 30 \, cm = 0.3 \, m$.
Now,calculate the acceleration:
$a = (40 \pi)^2 \times 0.3$
$a = 1600 \times \pi^2 \times 0.3$
Using $\pi^2 \approx 9.87$:
$a = 1600 \times 9.87 \times 0.3 = 4737.6 \, m/s^2$.
Rounding to the nearest value,we get $a \approx 4740 \, m/s^2$.

(B) The maximum horizontal range $(R_{\max})$ of a projectile is given by the formula: $R_{\max} = \frac{u^2}{g}$,where $u$ is the muzzle velocity and $g$ is the acceleration due to gravity.
Given: $R_{\max} = 16 \, km = 16,000 \, m$ and $g = 10 \, m/s^2$.
Substituting the values into the formula:
$16,000 = \frac{u^2}{10}$
$u^2 = 16,000 \times 10 = 160,000$
$u = \sqrt{160,000} = 400 \, m/s$.
Therefore,the muzzle velocity of the shell is $400 \, m/s$.

(D) The power $P$ is given by the formula $P = \frac{W}{t} = \frac{mgh}{t}$.
Given: $P = 2 \, kW = 2000 \, W$,$g = 10 \, m/s^2$,$h = 10 \, m$,and $t = 1 \, \text{minute} = 60 \, s$.
Rearranging the formula to solve for mass $m$:
$m = \frac{P \times t}{g \times h} = \frac{2000 \times 60}{10 \times 10} = \frac{120000}{100} = 1200 \, kg$.
Since the density of water is $1000 \, kg/m^3$,the volume $V$ in cubic meters is $V = \frac{m}{\rho} = \frac{1200 \, kg}{1000 \, kg/m^3} = 1.2 \, m^3$.
Since $1 \, m^3 = 1000 \, liters$,the volume in liters is $1.2 \times 1000 = 1200 \, liters$.

(C) bimetallic strip consists of two different metals with different coefficients of linear expansion ($\alpha_A$ and $\alpha_B$).
When the strip is heated, both metals expand, but the metal with the higher coefficient of linear expansion expands more than the other.
As shown in the figure, if $\alpha_A > \alpha_B$, metal $A$ will expand more than metal $B$.
To accommodate this difference in length while remaining bonded together, the strip must bend.
The metal with the higher expansion coefficient $(A)$ forms the outer side of the arc, while the metal with the lower expansion coefficient $(B)$ forms the inner side.
Therefore, the correct option is $C$.

(D) Thermal capacity is defined as the product of the mass of the substance $(m)$ and its specific heat capacity $(c)$.
Given:
Mass $(m)$ $= 40\, g$
Specific heat $(c)$ $= 0.2\, cal/g/^{\circ}C$
Thermal capacity $= m \times c$
Thermal capacity $= 40\, g \times 0.2\, cal/g/^{\circ}C = 8\, cal/^{\circ}C$.
Therefore, the correct option is $D$.

(D) The root mean square velocity $(v_{rms})$ of a gas molecule is given by the formula:
$v_{rms} = \sqrt{\frac{3kT}{m}}$
where $k$ is the Boltzmann constant,$T$ is the absolute temperature,and $m$ is the mass of the molecule.
From this expression,it is clear that $v_{rms}$ is inversely proportional to the square root of the mass of the molecule.
Therefore,$v_{rms} \propto \frac{1}{\sqrt{m}}$.

(C) According to the kinetic theory of gases,the root mean square velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
At absolute temperature,$T = 0 \ K$.
Substituting this value into the formula,we get $v_{rms} = \sqrt{\frac{3R(0)}{M}} = 0$.
Since the root mean square velocity represents the average motion of the molecules,a velocity of $0$ implies that all molecular motion ceases at absolute zero.

(B) We know that the molar specific heat at constant volume is given by ${C_V} = \frac{R}{J(\gamma - 1)}$.
Substituting $\gamma = 1.5$,we get ${C_V} = \frac{R}{J(1.5 - 1)} = \frac{R}{J(0.5)} = \frac{2R}{J}$.
Using Mayer's relation,${C_P} - {C_V} = \frac{R}{J}$,we have ${C_P} = {C_V} + \frac{R}{J}$.
Substituting the value of ${C_V}$,we get ${C_P} = \frac{2R}{J} + \frac{R}{J} = \frac{3R}{J}$.
Thus,the correct option is ${C_P} = \frac{3R}{J}$.

(D) The heat required at constant pressure is given by $(\Delta Q)_P = n C_P \Delta T$.
Given $n = 1 \, mol$,$(\Delta Q)_P = 207 \, J$,and $\Delta T = 10 \, K$.
$207 = 1 \times C_P \times 10 \implies C_P = 20.7 \, J/mol \cdot K$.
Using Mayer's relation,$C_P - C_V = R$,we find $C_V = C_P - R$.
$C_V = 20.7 - 8.3 = 12.4 \, J/mol \cdot K$.
The heat required at constant volume is $(\Delta Q)_V = n C_V \Delta T$.
$(\Delta Q)_V = 1 \times 12.4 \times 10 = 124 \, J$.

(B) According to the first law of thermodynamics,
$Q = \Delta U + W$
where $Q$ is the heat supplied to the system,$\Delta U$ is the change in internal energy,and $W$ is the work done by the system.
This law is a statement of the law of conservation of energy,which states that energy can neither be created nor destroyed,only transformed from one form to another.

(C) When two springs with force constants $K_1$ and $K_2$ are connected in series,the equivalent force constant $K_{eq}$ is given by the formula:
$\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$
$\frac{1}{K_{eq}} = \frac{K_1 + K_2}{K_1 K_2}$
$K_{eq} = \frac{K_1 K_2}{K_1 + K_2}$
The time period $T$ of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{K_{eq}}}$.
Substituting the value of $K_{eq}$:
$T = 2\pi \sqrt{\frac{m}{\frac{K_1 K_2}{K_1 + K_2}}}$
$T = 2\pi \sqrt{\frac{m(K_1 + K_2)}{K_1 K_2}}$

(A) Let the two simple harmonic motions (SHMs) be represented as:
$x = a_1 \sin(\omega t)$
$y = a_2 \sin(\omega t + \pi)$
Since $\sin(\omega t + \pi) = -\sin(\omega t)$,we have:
$y = -a_2 \sin(\omega t)$
From the first equation,$\sin(\omega t) = \frac{x}{a_1}$.
Substituting this into the second equation:
$y = -a_2 \left(\frac{x}{a_1}\right)$
$y = -\left(\frac{a_2}{a_1}\right)x$
This is the equation of a straight line passing through the origin with a negative slope. Thus,the particle moves along a straight line.

(C) The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by: $\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$.
Given $\Delta \phi = 1.6 \pi$ and $\Delta x = 40 \; cm = 0.4 \; m$.
Substituting these values: $1.6 \pi = \frac{2\pi}{\lambda} \times 0.4$.
Solving for wavelength $(\lambda)$: $\lambda = \frac{2 \times 0.4}{1.6} = \frac{0.8}{1.6} = 0.5 \; m$.
Using the wave equation $v = f \lambda$,where $v = 330 \; m/s$ and $\lambda = 0.5 \; m$:
$330 = f \times 0.5$.
Therefore,$f = \frac{330}{0.5} = 660 \; Hz$.

(B) The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4}MR^2$.
Given that this value is $I$,we have $MR^2 = 4I$.
To find the moment of inertia about an axis perpendicular to the plane and passing through a point on the rim,we use the parallel axis theorem.
First,the moment of inertia about an axis perpendicular to the plane and passing through the center of mass is $I_{cm} = \frac{1}{2}MR^2$.
By the parallel axis theorem,$I_{rim} = I_{cm} + Md^2$,where $d = R$ is the distance between the axes.
$I_{rim} = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Substituting $MR^2 = 4I$ into the expression:
$I_{rim} = \frac{3}{2}(4I) = 6I$.

(B) Given: Moment of inertia $I = 1.2 \; kg \cdot m^2$,initial angular velocity $\omega_0 = 0$,rotational kinetic energy $K_r = 1500 \; J$,and angular acceleration $\alpha = 25 \; rad/s^2$.
The formula for rotational kinetic energy is $K_r = \frac{1}{2} I \omega^2$.
Substituting the values: $1500 = \frac{1}{2} \times 1.2 \times \omega^2$.
$1500 = 0.6 \times \omega^2 \Rightarrow \omega^2 = \frac{1500}{0.6} = 2500$.
Thus,the final angular velocity $\omega = \sqrt{2500} = 50 \; rad/s$.
Using the kinematic equation for rotation: $\omega = \omega_0 + \alpha t$.
$50 = 0 + 25 \times t$.
$t = \frac{50}{25} = 2 \; s$.

(C) The rotational kinetic energy $(K)$ of a body rotating about a fixed axis is given by the formula: $K = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular speed.
Given:
Kinetic energy $(K)$ = $360 \ J$
Angular speed $(\omega)$ = $30 \ rad/s$
Substituting the values into the formula:
$360 = \frac{1}{2} \times I \times (30)^2$
$360 = \frac{1}{2} \times I \times 900$
$360 = 450 \times I$
$I = \frac{360}{450}$
$I = 0.8 \ kg \ m^2$
Therefore,the moment of inertia of the wheel is $0.8 \ kg \ m^2$.

(B) The thrust force $F$ required to overcome the weight of the rocket is given by $F = mg$.
The thrust force produced by the rocket engine is given by $F = v \frac{dm}{dt}$,where $v$ is the exhaust speed and $\frac{dm}{dt}$ is the rate of mass ejection.
Equating the two forces: $mg = v \frac{dm}{dt}$.
Rearranging to solve for the rate of mass ejection: $\frac{dm}{dt} = \frac{mg}{v}$.
Given $m = 600\; kg$,$g = 10\; m/s^2$,and $v = 1000\; m/s$:
$\frac{dm}{dt} = \frac{600 \times 10}{1000} = \frac{6000}{1000} = 6\; kg/s$.

(B) The initial temperature of boiling water is $100\,^{\circ}C$ or $212\,^{\circ}F$.
The final temperature in Fahrenheit is $F = 140\,^{\circ}F$.
Using the conversion formula $\frac{C}{100} = \frac{F-32}{180}$,we find the final temperature in Celsius:
$\frac{C}{100} = \frac{140-32}{180} = \frac{108}{180} = 0.6$
$C = 60\,^{\circ}C$.
The fall in temperature on the Centigrade scale is $\Delta C = 100\,^{\circ}C - 60\,^{\circ}C = 40\,^{\circ}C$.

(A) Impulse is defined as the change in momentum of a body.
Mathematically,Impulse $J = \Delta p = p_f - p_i$.
Given the initial velocity is $v_1$ and the final velocity is $v_2$,the initial momentum is $p_i = m v_1$ and the final momentum is $p_f = m v_2$.
Therefore,Impulse $J = m v_2 - m v_1 = m(v_2 - v_1)$.

(D) $1$. In an elliptical orbit,the planet's distance from the sun changes,causing the speed to vary. Therefore,kinetic energy $T$ is not conserved.
$2$. Gravitational potential energy $V$ is defined as $V = -GMm/r$,which is always negative for bound states.
$3$. Angular momentum $L$ is conserved in magnitude and direction because the gravitational force is a central force,meaning the torque $\tau = r \times F = 0$. Since the motion is confined to a plane,the direction of the angular momentum vector remains constant.
$4$. For a bound orbit (elliptical),the total mechanical energy $E = T + V$ is always negative,representing the energy required to escape the gravitational field.

(D) For a body sliding down an inclined plane of length $L$ starting from rest,the time taken is given by $L = \frac{1}{2} a t^2$,where $a$ is the acceleration.
For the smooth surface,acceleration $a_S = g \sin \theta$. Thus,$L = \frac{1}{2} (g \sin \theta) t_S^2$.
For the rough surface,acceleration $a_R = g \sin \theta - \mu_k g \cos \theta$. Thus,$L = \frac{1}{2} (g \sin \theta - \mu_k g \cos \theta) t_R^2$.
Since $L$ is the same for both,$\frac{1}{2} a_S t_S^2 = \frac{1}{2} a_R t_R^2$,which implies $\frac{a_R}{a_S} = \left(\frac{t_S}{t_R}\right)^2$.
Given $t_R = 2 t_S$,we have $\frac{a_R}{a_S} = \left(\frac{t_S}{2 t_S}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Substituting the expressions for acceleration: $\frac{g \sin \theta - \mu_k g \cos \theta}{g \sin \theta} = \frac{1}{4}$.
Since $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
Therefore,$\frac{\frac{1}{\sqrt{2}} - \mu_k \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \frac{1}{4}$.
$1 - \mu_k = \frac{1}{4} \Rightarrow \mu_k = 1 - 0.25 = 0.75$.

(C) Let $n$ be the total number of resistors used. Let the equivalent resistance be $R_{eq} = 5\,\Omega$ and the maximum current capacity be $I_{total} = 4\,A$.
Each resistor has $R = 10\,\Omega$ and $I_{max} = 1\,A$.
The power dissipated by the combination is $P = I_{total}^2 \times R_{eq} = 4^2 \times 5 = 16 \times 5 = 80\,W$.
The power dissipated by each individual resistor is $P_{res} = I_{max}^2 \times R = 1^2 \times 10 = 10\,W$.
Since the total power is the sum of power dissipated by each resistor,we have $n = P / P_{res} = 80 / 10 = 8$.
Thus,a minimum of $8$ resistors is required.

(D) The circuit can be simplified by identifying series and parallel combinations.
$1$. The two resistors connected to point $C$ form a delta-star or can be simplified by recognizing symmetry. Alternatively,observe the top triangle: the two resistors of resistance $r$ are in series,giving $2r$. This $2r$ is in parallel with the third resistor $r$ of the top triangle.
$2$. The equivalent resistance of the top part is $R_{top} = (2r \times r) / (2r + r) = (2/3)r$.
$3$. Now,this $R_{top}$ is in series with the two resistors connected to $A$ and $B$ respectively,but looking at the structure,we can simplify the network into a parallel combination of branches.
$4$. The total resistance between $A$ and $B$ is found by reducing the network step-by-step to a single equivalent resistor,which results in $R_{eq} = 8r / 7$.

(B) For two identical batteries of $e.m.f.$ $E = 2\,V$ and internal resistance $r = 1.0\,\Omega$, we can connect them in series or parallel.
Case $1$: Series connection.
Total $e.m.f.$ $E_{eq} = 2E = 4\,V$.
Total internal resistance $r_{eq} = 2r = 2.0\,\Omega$.
Current $I = E_{eq} / (R + r_{eq}) = 4 / (0.5 + 2.0) = 4 / 2.5 = 1.6\,A$.
Power $P = I^2 R = (1.6)^2 \times 0.5 = 2.56 \times 0.5 = 1.28\,W$.
Case $2$: Parallel connection.
Total $e.m.f.$ $E_{eq} = E = 2\,V$.
Total internal resistance $r_{eq} = r/2 = 0.5\,\Omega$.
Current $I = E_{eq} / (R + r_{eq}) = 2 / (0.5 + 0.5) = 2 / 1.0 = 2.0\,A$.
Power $P = I^2 R = (2.0)^2 \times 0.5 = 4.0 \times 0.5 = 2.0\,W$.
Comparing both cases, the maximum power is $2.0\,W$.

(A) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula: $B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{r}$.
Since the current $i$ is constant,the magnetic field is inversely proportional to the distance $r$,i.e.,$B \propto \frac{1}{r}$.
Therefore,we can write the ratio: $\frac{B_1}{B_2} = \frac{r_2}{r_1}$.
Given: $B_1 = 10^{-8} \, T$,$r_1 = 4 \, cm$,and $r_2 = 12 \, cm$.
Substituting the values: $\frac{10^{-8}}{B_2} = \frac{12}{4}$.
$\frac{10^{-8}}{B_2} = 3$.
$B_2 = \frac{10^{-8}}{3} = 3.33 \times 10^{-9} \, T$.

(A) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$,where $L$ is the self-inductance and $i$ is the current flowing through the inductor.
This energy is associated with the magnetic field produced by the current in the inductor.
Therefore,the energy is stored in the form of a magnetic field.

(C) The magnetic flux through the coil at any time $t$ is given by $\phi = NBA \cos(\omega t)$,where $\omega$ is the angular frequency.
The induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt} = NBA\omega \sin(\omega t)$.
The amplitude of the induced $EMF$ is $e_0 = NBA\omega$.
The amplitude of the induced current is $i_0 = \frac{e_0}{R} = \frac{NBA\omega}{R}$.
Given: $B = 10^{-2} \, T$,$r = 0.3 \, m$,$A = \pi r^2 = \pi(0.3)^2 = 0.09\pi \, m^2$,$R = \pi^2 \, \Omega$,and frequency $f = \frac{200}{60} = \frac{10}{3} \, Hz$.
Angular frequency $\omega = 2\pi f = 2\pi \times \frac{10}{3} = \frac{20\pi}{3} \, rad/s$.
Assuming $N=1$ (single turn coil):
$i_0 = \frac{1 \times 10^{-2} \times (0.09\pi) \times (20\pi/3)}{\pi^2} = \frac{10^{-2} \times 0.03 \times 20 \times \pi^2}{\pi^2} = 0.006 \, A = 6 \, mA$.

(D) The momentum $p$ of a photon is related to its frequency $\nu$ by the formula: $p = \frac{h\nu}{c}$.
Rearranging for frequency $\nu$,we get: $\nu = \frac{pc}{h}$.
Given: $p = 3.3 \times 10^{-29} \ kg \ m/s$,$c = 3 \times 10^8 \ m/s$,and Planck's constant $h = 6.6 \times 10^{-34} \ J \ s$.
Substituting the values: $\nu = \frac{3.3 \times 10^{-29} \times 3 \times 10^8}{6.6 \times 10^{-34}}$.
$\nu = \frac{9.9 \times 10^{-21}}{6.6 \times 10^{-34}} = 1.5 \times 10^{13} \ Hz$.

(A) The power $P$ of the transmitter is the energy emitted per second. The energy of a single photon is given by $E = h\nu$,where $h = 6.6 \times 10^{-34} \, J \cdot s$ is Planck's constant and $\nu$ is the frequency.
If $n$ is the number of photons emitted per second,then the total power is $P = n \times E = n \times h\nu$.
Therefore,$n = \frac{P}{h\nu}$.
Given: $P = 10 \, kW = 10^4 \, W$ and $\nu = 880 \, kHz = 880 \times 10^3 \, Hz$.
Substituting the values:
$n = \frac{10^4}{(6.6 \times 10^{-34}) \times (880 \times 10^3)}$
$n = \frac{10^4}{5808 \times 10^{-31}}$
$n = \frac{10^4}{5.808 \times 10^{-28}}$
$n \approx 1.72 \times 10^{31}$ photons per second.

(B) Let the energies of the states $A, B,$ and $C$ be $E_A, E_B,$ and $E_C$ respectively.
From the principle of conservation of energy,the energy of the transition from $C$ to $A$ is equal to the sum of the energies of the transitions from $C$ to $B$ and $B$ to $A$.
Thus,$(E_C - E_A) = (E_C - E_B) + (E_B - E_A)$.
Using the relation $E = \frac{hc}{\lambda}$,we can write:
$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$
Dividing both sides by $hc$,we get:
$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$
$\frac{1}{\lambda_3} = \frac{\lambda_1 + \lambda_2}{\lambda_1 \lambda_2}$
Therefore,$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

(D) According to Bohr's postulate for the quantization of angular momentum,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $mvr = n\frac{h}{2\pi}$.
Rearranging this equation,we get $2\pi r = n\left(\frac{h}{mv}\right)$.
Since the de Broglie wavelength $\lambda$ is defined as $\lambda = \frac{h}{p} = \frac{h}{mv}$,we can substitute this into the equation.
Therefore,the circumference of the orbit is $2\pi r = n\lambda$.

(A) Nuclear forces are the strong forces that hold nucleons (protons and neutrons) together inside the nucleus.
$1$. They are short-ranged,acting only over distances of the order of $10^{-15} \ m$ (femtometers).
$2$. They are primarily attractive in nature,which overcomes the electrostatic repulsion between protons.
$3$. They are charge independent,meaning the force between a proton-proton,neutron-neutron,or proton-neutron pair is approximately the same,provided the separation distance is the same.
Therefore,the correct option is $A$.

(B) The nuclear reaction is given by:
$_6^{12}C + _0^1n \rightarrow _6^{13}C + \gamma$ (capture of neutron)
Then,the unstable $_6^{13}C$ nucleus undergoes beta decay:
$_6^{13}C \rightarrow _7^{13}N + _{-1}^0\beta + \bar{\nu}$
Thus,the resulting nucleus is $_7^{13}N$.

(A) In a common base $(CB)$ amplifier configuration,the input signal is applied between the emitter and the base,and the output is taken between the collector and the base.
Since the base is common to both the input and output circuits,the input voltage and output voltage signals are in the same phase.
Therefore,the phase difference between the input signal voltage and the output voltage is $0$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
In air,$\frac{1}{f_a} = (_{a}\mu_{g} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Given $f_a = 20 \ cm$,so $0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{20}$,which means $\left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{10}$.
In water,$\frac{1}{f_w} = (_{w}\mu_{g} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where $_{w}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{w}} = \frac{1.5}{4/3} = \frac{1.5 \times 3}{4} = \frac{4.5}{4} = 1.125$.
So,$\frac{1}{f_w} = (1.125 - 1) \times \frac{1}{10} = 0.125 \times \frac{1}{10} = \frac{0.125}{10} = \frac{1}{80}$.
Therefore,$f_w = 80 \ cm$.

(C) The distance of the $n^{th}$ bright fringe from the central maximum is given by the formula: $y_n = \frac{n \lambda D}{d}$.
Since $n$,$D$,and $d$ are constant for both cases,we have $y_n \propto \lambda$.
Therefore,the ratio of the distances for the $4^{th}$ fringe is: $\frac{x(\text{Blue})}{x(\text{Green})} = \frac{\lambda(\text{Blue})}{\lambda(\text{Green})} = \frac{4360}{5460}$.
Since $4360 < 5460$,it follows that $x(\text{Blue}) < x(\text{Green})$.

(A) The fringe width in a Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.
Here,$\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
When the apparatus is immersed in a medium of refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.
Since $D$ and $d$ remain unchanged,the new fringe width $\beta'$ is given by $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.
Given $\beta = 0.4 \, mm$ and $\mu = 4/3$,we have $\beta' = \frac{0.4}{4/3} = 0.4 \times \frac{3}{4} = 0.3 \, mm$.

(D) The electromagnetic spectrum is ordered by wavelength. The order of wavelengths for the given radiations is $\lambda_{\gamma} < \lambda_{X-ray} < \lambda_{Blue} < \lambda_{Red}$.
Among the given options,Red light has the longest wavelength,while $\gamma$-rays have the shortest wavelength.
Therefore,the correct option is $(d)$.

(B) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
For the $^{27}_{13}Al$ nucleus,the mass number $A_{Al} = 27$. Thus,$R_{Al} = R_0 (27)^{1/3} = 3 R_0$.
For the $^{125}_{53}Te$ nucleus,the mass number $A_{Te} = 125$. Thus,$R_{Te} = R_0 (125)^{1/3} = 5 R_0$.
Taking the ratio of the two radii:
$\frac{R_{Te}}{R_{Al}} = \frac{5 R_0}{3 R_0} = \frac{5}{3}$.
Therefore,$R_{Te} = \frac{5}{3} R_{Al}$.

(D) To determine the relationship between the nuclei ${ }_{6} C^{13}$ and ${ }_{7} N^{14}$,we calculate the number of neutrons $(N)$ in each.
For ${ }_{6} C^{13}$,the atomic number $Z = 6$ and the mass number $A = 13$. The number of neutrons is $N = A - Z = 13 - 6 = 7$.
For ${ }_{7} N^{14}$,the atomic number $Z = 7$ and the mass number $A = 14$. The number of neutrons is $N = A - Z = 14 - 7 = 7$.
Since both nuclei have the same number of neutrons $(N = 7)$,they are called isotones.

(B) $X$-rays emission: These are produced due to transitions in the inner energy levels of an atom,not by the emission of electrons from the surface.
Photoelectric emission: This is the emission of electrons from a metal surface upon irradiation with radiation of a suitable frequency.
Secondary emission: When an incident electron strikes the surface of a metallic plate,it causes the emission of other electrons from the surface.
Thermionic emission: When a metal is heated to a high temperature,the free electrons gain sufficient kinetic energy to escape from the surface of the metal.

(D) The induced $e.m.f.$ in an inductor is given by the formula: $e.m.f. = L \frac{dI}{dt}$.
Here,$L = 40\; mH = 40 \times 10^{-3}\; H$.
The change in current is $dI = 11\; A - 1\; A = 10\; A$.
The time interval is $dt = 4\; ms = 4 \times 10^{-3}\; s$.
Substituting these values into the formula:
$e.m.f. = (40 \times 10^{-3}) \times \frac{10}{4 \times 10^{-3}}$.
$e.m.f. = 40 \times \frac{10}{4} = 10 \times 10 = 100\; V$.