AIPMT 1992 Physics Question Paper with Answer and Solution

(B) According to Newton's law of universal gravitation,the force $F$ between two masses $m_1$ and $m_2$ separated by a distance $d$ is given by $F = \frac{G m_1 m_2}{d^2}$.
Rearranging the formula to solve for the gravitational constant $G$,we get $G = \frac{F d^2}{m_1 m_2}$.
The dimensional formula for force $F$ is $[MLT^{-2}]$,for distance $d$ is $[L]$,and for mass $m$ is $[M]$.
Substituting these into the expression for $G$: $[G] = \frac{[MLT^{-2}][L^2]}{[M][M]} = \frac{[ML^3T^{-2}]}{[M^2]} = [M^{-1}L^3T^{-2}]$.

(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{v^2 \sin(2\theta)}{g}$.
For body $A$,the angle of projection is $\theta_A = 30^{\circ}$. Thus,$R_A = \frac{v^2 \sin(2 \times 30^{\circ})}{g} = \frac{v^2 \sin(60^{\circ})}{g}$.
For body $B$,the angle of projection is $\theta_B = 60^{\circ}$. Thus,$R_B = \frac{v^2 \sin(2 \times 60^{\circ})}{g} = \frac{v^2 \sin(120^{\circ})}{g}$.
Since $\sin(120^{\circ}) = \sin(180^{\circ} - 60^{\circ}) = \sin(60^{\circ})$,it follows that $R_A = R_B$.
Therefore,the ratio of the horizontal range of $A$ to $B$ is $R_A : R_B = 1:1$.

(B) First,convert the speed from $km/h$ to $m/s$:
$u = 72 \times \frac{5}{18} = 20 \, m/s$.
When the car stops,the final velocity $v = 0$.
The retarding force is the kinetic friction $f_k = \mu_k N = \mu_k mg$.
Using Newton's second law,$ma = -f_k = -\mu_k mg$,so the deceleration $a = -\mu_k g$.
Using the kinematic equation $v^2 - u^2 = 2as$:
$0^2 - (20)^2 = 2(-\mu_k g)s$.
$s = \frac{u^2}{2 \mu_k g} = \frac{20^2}{2 \times 0.5 \times 10} = \frac{400}{10} = 40 \, m$.

(D) The work done by a variable force is given by the integral of the force with respect to displacement: $W = \int_{x_1}^{x_2} F(x) \, dx$.
Given $F(x) = 7 - 2x + 3x^2$,$x_1 = 0$,and $x_2 = 5$.
$W = \int_{0}^{5} (7 - 2x + 3x^2) \, dx$.
Integrating term by term:
$W = [7x - x^2 + x^3]_{0}^{5}$.
Substituting the limits:
$W = (7(5) - (5)^2 + (5)^3) - (7(0) - (0)^2 + (0)^3)$.
$W = (35 - 25 + 125) - 0$.
$W = 135 \, J$.

(D) The escape velocity of a body from the surface of the Earth is given by the formula $v_e = \sqrt{2gR_e}$,where $g$ is the acceleration due to gravity and $R_e$ is the radius of the Earth.
This expression shows that the escape velocity depends only on the mass and radius of the planet (or the gravitational potential at the point of projection).
It is independent of the direction or the angle at which the body is projected.
Therefore,if the body is projected at an angle of $45^o$ with the vertical,the escape velocity remains the same,which is $11.2 \ km/s$.

(D) For a geostationary satellite,the orbital period $T$ is equal to the rotational period of the earth,which is $T = 2\pi/\omega$.
According to Kepler's third law,the square of the orbital period is proportional to the cube of the orbital radius $r$,given by $T^2 = (4\pi^2/GM)r^3$.
We know that the acceleration due to gravity at the surface is $g = GM/R^2$,which implies $GM = gR^2$.
Substituting $GM$ into the Kepler's law equation: $(2\pi/\omega)^2 = (4\pi^2 / gR^2)r^3$.
Simplifying this: $4\pi^2/\omega^2 = (4\pi^2 / gR^2)r^3$.
Canceling $4\pi^2$ from both sides: $1/\omega^2 = r^3 / (gR^2)$.
Therefore,$r^3 = gR^2/\omega^2$.

(C) The boiling point of mercury is approximately $357^oC$.
$A$ mercury thermometer operates on the principle of thermal expansion of liquid mercury with a rise in temperature.
It is typically used to measure temperatures in the range of $-30^oC$ to $357^oC$.
Therefore,among the given options,the mercury thermometer can be used to measure temperatures up to $360^oC$ (as it is the closest practical upper limit provided).

(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.

(A) From the given $P-V$ graph:
Process $AB$ is an isochoric process (volume is constant),so work done $W_{AB} = 0$.
Process $BC$ is an isobaric process (pressure is constant at $P_B = 8 \times 10^4 \, Pa$),so work done $W_{BC} = P_B(V_C - V_B)$.
Since $V_C = V_D = 5 \times 10^{-3} \, m^3$ and $V_B = V_A = 2 \times 10^{-3} \, m^3$,we have:
$W_{BC} = 8 \times 10^4 \times (5 - 2) \times 10^{-3} = 8 \times 10^4 \times 3 \times 10^{-3} = 240 \, J$.
Total work done in process $AC$ (path $A \rightarrow B \rightarrow C$) is $W_{AC} = W_{AB} + W_{BC} = 0 + 240 = 240 \, J$.
Total heat added in process $AC$ is $\Delta Q_{AC} = \Delta Q_{AB} + \Delta Q_{BC} = 600 + 200 = 800 \, J$.
Using the First Law of Thermodynamics,$\Delta Q_{AC} = \Delta U_{AC} + W_{AC}$.
Substituting the values: $800 = \Delta U_{AC} + 240$.
Therefore,$\Delta U_{AC} = 800 - 240 = 560 \, J$.

(C) In a $PV$ diagram,the work done by a thermodynamic system during a process is equal to the area under the curve.
For a complete cyclic process,the net work done is equal to the area enclosed by the closed loop of the $PV$ diagram.
In the given diagram,the system follows the path $A \rightarrow C \rightarrow B$ and then returns to $A$ via $B \rightarrow D \rightarrow A$.
The closed loop formed by these paths is $ACBDA$.
Therefore,the net work done during the complete cycle is given by the area enclosed by the loop $ACBDA$.

(A) The displacement of a simple harmonic oscillator starting from the extreme position ($x = a$ at $t = 0$) is given by $x = a \cos(\omega t)$.
We need to find the time $t$ when $x = a/2$.
Substituting the value: $a/2 = a \cos(\omega t)$.
This simplifies to $\cos(\omega t) = 1/2$.
Therefore,$\omega t = \pi/3$.
Since the angular frequency $\omega = 2\pi/T$,we substitute this into the equation:
$(2\pi/T) \cdot t = \pi/3$.
Solving for $t$,we get $t = T/6$.

(A) For a simple harmonic oscillator,the magnitude of acceleration $a$ is given by the relation $a = \omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.
Given: $x = 0.02\, m$ and $a = 2.0\, m\, s^{-2}$.
Rearranging the formula for angular frequency: $\omega = \sqrt{\frac{a}{x}}$.
Substituting the values: $\omega = \sqrt{\frac{2.0}{0.02}} = \sqrt{\frac{200}{2}} = \sqrt{100}$.
Therefore,$\omega = 10\, rad\, s^{-1}$.

(D) wave is defined as a disturbance that travels through a medium,transporting energy and momentum from one point to another without the net transport of matter.
In a longitudinal wave,particles of the medium oscillate back and forth about their mean positions in the direction of wave propagation.
While the particles themselves do not travel with the wave (there is no net mass transfer),the wave carries energy and linear momentum through the medium.
Therefore,the correct quantities transmitted are energy and linear momentum.

(D) The standard form of a sinusoidal wave equation is $y = a \cos(\omega t + kx + \phi)$.
Comparing the given equation $y = 0.40 \cos(2000t + 0.80x)$ with the standard form,we identify the angular frequency $\omega = 2000 \text{ rad/s}$.
The relationship between angular frequency $\omega$ and frequency $f$ is given by $\omega = 2\pi f$.
Substituting the value of $\omega$,we get $2000 = 2\pi f$.
Solving for $f$,we find $f = \frac{2000}{2\pi} = \frac{1000}{\pi} \text{ Hz}$.

(D) Beats are the periodic variation in the intensity of sound heard when two sound waves of slightly different frequencies,but similar amplitudes,are superimposed on each other.
If the frequencies of the two waves are $f_1$ and $f_2$,the beat frequency is given by $f_{beat} = |f_1 - f_2|$.
Therefore,the condition for producing beats is that the waves must have different frequencies.

(C) The strength of bonding in solids is determined by the energy required to break the bonds.
$1$. Ionic bonds are electrostatic attractions between ions,which are very strong.
$2$. Metallic bonds involve the sharing of free electrons among a lattice of positive ions,which are strong.
$3$. Covalent bonds involve the sharing of electron pairs between atoms,which are very strong.
$4$. Van der Waals forces are weak intermolecular forces arising from temporary or permanent dipole-dipole interactions.
Therefore,Van der Waals bonding is the weakest among the given options.

(B) The solar system consists of $8$ planets. Mercury is the closest planet to the Sun,followed by Venus,Earth,Mars,Jupiter,Saturn,Uranus,and Neptune. Therefore,the correct option is $B$.

(A) For a solid sphere,the moment of inertia about its center is $I = \frac{2}{5} m R^2$.
According to the work-energy theorem,the potential energy lost equals the sum of translational and rotational kinetic energy: $mgh = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$.
Since the sphere rolls without sliding,the condition for pure rolling is $v = R\omega$,which implies $\omega = \frac{v}{R}$.
Substituting the values into the energy equation: $mgh = \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$.
Simplifying the expression: $mgh = \frac{1}{2} m v^2 + \frac{1}{5} m v^2$.
$mgh = (\frac{1}{2} + \frac{1}{5}) m v^2 = \frac{7}{10} m v^2$.
Solving for $v$: $v^2 = \frac{10}{7} gh$,therefore $v = \sqrt{\frac{10}{7} gh}$.

(B) Let $h$ be the height of the tower and $t$ be the total time taken by the body to reach the ground.
Given initial velocity $u = 0$ and acceleration $a = g = 10 \; m/s^2$.
The total height is given by $h = \frac{1}{2} g t^2$.
The distance covered in the last two seconds is the difference between the total height and the height covered in $(t-2)$ seconds.
$40 = \frac{1}{2} g t^2 - \frac{1}{2} g (t-2)^2$.
Substituting $g = 10 \; m/s^2$:
$40 = 5 t^2 - 5 (t^2 - 4t + 4)$.
$40 = 5 t^2 - 5 t^2 + 20 t - 20$.
$40 = 20 t - 20$.
$60 = 20 t \implies t = 3 \; s$.
Now,calculate the height $h$:
$h = \frac{1}{2} \times 10 \times (3)^2 = 5 \times 9 = 45 \; m$.

(B) In a velocity-time $(v-t)$ graph,for a given instant of time $t$,there should be only one unique value of velocity $v$.
If we draw a vertical line at any time $t$ on the graph,it should intersect the curve at only one point.
In option $(B)$,the curve is a circle. $A$ vertical line drawn at any time $t$ (within the range of the circle) will intersect the circle at two points,meaning the particle has two different velocities at the same instant of time,which is physically impossible for motion in one dimension.
Therefore,the circular graph does not represent motion in one dimension.

(B) The cross product of any vector $\vec{A}$ with a zero vector $\vec{0}$ is defined as $\vec{A} \times \vec{0} = \vec{0}$.
Since the cross product of two vectors results in another vector,the result of $\vec{A} \times 0$ is a zero vector,which is denoted as $\vec{0}$.
Therefore,the correct option is the zero vector.

(C) Given: $\frac{R}{C_{v}} = 0.67$.
We know that the gas constant $R = C_{p} - C_{v}$.
Substituting this into the given equation: $\frac{C_{p} - C_{v}}{C_{v}} = 0.67$.
$\frac{C_{p}}{C_{v}} - 1 = 0.67$.
Since the adiabatic index $\gamma = \frac{C_{p}}{C_{v}}$,we have $\gamma - 1 = 0.67$,which implies $\gamma = 1.67$.
For a monoatomic gas,the degrees of freedom $f = 3$,so $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = 1 + 0.666... \approx 1.67$.
Therefore,the gas is monoatomic.

(B) The particle moves in a circle with uniform speed $v$. Let the initial velocity be $\vec{v}_i = v \hat{i}$.
At the diametrically opposite point,the velocity is $\vec{v}_f = -v \hat{i}$.
The change in momentum is $\Delta \vec{p} = m \vec{v}_f - m \vec{v}_i = M(-v \hat{i}) - M(v \hat{i}) = -2 M v \hat{i}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2 M v$.
Since the speed $v$ is uniform,the kinetic energy $K = \frac{1}{2} M v^2$ remains constant throughout the motion.
Therefore,the change in kinetic energy is $0$.

(B) The dimensions of the given quantities are:
Radiation pressure $P = [M L^{-1} T^{-2}]$
Radiation energy per unit area per second $Q = [M T^{-3}]$
Speed of light $c = [L T^{-1}]$
For the expression $P^x Q^y c^z$ to be dimensionless,we have:
$[M L^{-1} T^{-2}]^x [M T^{-3}]^y [L T^{-1}]^z = [M^0 L^0 T^0]$
Equating the powers of $M, L,$ and $T$ on both sides:
For $M: x + y = 0 \implies y = -x$
For $L: -x + z = 0 \implies z = x$
For $T: -2x - 3y - z = 0$
Substituting $y = -x$ and $z = x$ into the $T$ equation:
$-2x - 3(-x) - x = -2x + 3x - x = 0$
This identity holds for any non-zero $x$. If we choose $x = 1$,then $y = -1$ and $z = 1$. Thus,the set of integers is $x = 1, y = -1, z = 1$.

(A) Kirchhoff's first law,also known as the Kirchhoff's Current Law $(KCL)$,states that the algebraic sum of currents meeting at a junction in an electric circuit is zero,i.e.,$\Sigma i = 0$.
This law implies that the total charge entering a junction must equal the total charge leaving the junction in the same time interval.
Since electric charge cannot be created or destroyed at a junction,this law is a direct consequence of the law of conservation of charge.

(D) The power $P$ delivered to a variable resistance $R$ connected to a battery of $e.m.f.$ $E$ and internal resistance $r$ is given by the formula $P = I^2 R$,where $I = \frac{E}{R+r}$.
Substituting $I$,we get $P = \left( \frac{E}{R+r} \right)^2 R$.
To find the value of $R$ for maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = 0$.
This condition leads to the Maximum Power Transfer Theorem,which states that the power delivered to the load is maximum when the load resistance $R$ is equal to the internal resistance $r$ of the source.
Given $r = 0.5\, \Omega$,the power delivered is maximum when $R = r = 0.5\, \Omega$.

(A) The magnetic field $B$ at a distance $r$ from a long straight current-carrying wire is given by the formula $B = \frac{\mu_0 i}{2\pi r}$.
From this expression,we can see that the magnetic field is inversely proportional to the distance,i.e.,$B \propto \frac{1}{r}$.
Let $B_1 = 0.4 \ T$ at distance $r_1 = r$.
We need to find the magnetic field $B_2$ at distance $r_2 = 2r$.
Using the proportionality $B_1 r_1 = B_2 r_2$,we get:
$0.4 \times r = B_2 \times (2r)$.
$B_2 = \frac{0.4 \times r}{2r} = 0.2 \ T$.

(B) The force $F$ on a current-carrying wire in a magnetic field is given by the formula $F = BIl \sin(\theta)$.
Here,$B = 2\,T$ (magnetic field induction),
$I = 1.2\,A$ (current),
$l = 0.5\,m$ (length of the wire),
and $\theta = 90^\circ$ (since the magnetic field is perpendicular to the wire),so $\sin(90^\circ) = 1$.
Substituting the values into the formula:
$F = 2 \times 1.2 \times 0.5 \times 1$
$F = 1.2\,N$.
Therefore,the force on the wire is $1.2\,N$.

(C) Given:
Number of turns $N = 20$
Area $A = 25 \, cm^2 = 25 \times 10^{-4} \, m^2$
Resistance $R = 100 \, \Omega$
Rate of change of magnetic field $\frac{dB}{dt} = 1000 \, T/s$
Since the magnetic field is perpendicular to the plane of the coil,the angle $\theta$ between the area vector and the magnetic field is $0^\circ$.
According to Faraday's law,the induced electromotive force $(EMF)$ is given by $e = -N \frac{d\phi}{dt} = -N A \cos \theta \frac{dB}{dt}$.
The magnitude of the induced $EMF$ is $|e| = N A \frac{dB}{dt} \cos 0^\circ = N A \frac{dB}{dt}$.
Substituting the values:
$|e| = 20 \times (25 \times 10^{-4}) \times 1000 = 20 \times 25 \times 10^{-1} = 50 \times 10^{-1} = 5 \, V$.
The current $i$ in the coil is given by $i = \frac{|e|}{R}$.
$i = \frac{5}{100} = 0.05 \, A$. Wait,recalculating: $20 \times 25 \times 10^{-4} \times 1000 = 20 \times 25 \times 10^{-1} = 50 \times 0.1 = 5 \, V$. Then $i = 5 / 100 = 0.05 \, A$.
Re-evaluating the provided solution: The provided solution states $0.5 \, A$. Let's check: $20/100 * 1000 * 25 * 10^{-4} = 0.2 * 1000 * 0.0025 = 200 * 0.0025 = 0.5 \, A$. The calculation $20 \times 25 \times 10^{-4} \times 1000 = 500000 \times 10^{-4} = 50$. $50 / 100 = 0.5 \, A$. Correct.

(A) The photoelectric current in a cell depends on the number of photoelectrons emitted per unit time.
This number is directly proportional to the intensity of the incident light,provided the frequency of the incident light is greater than the threshold frequency $(h\nu > W)$.
The work function $(W)$ determines the threshold frequency $(W = h\nu_0)$,but it does not affect the number of photoelectrons emitted for a given intensity of light.
Since the intensity of the incident light remains unchanged and the condition $h\nu > W_2$ is satisfied,the number of photoelectrons emitted per unit time remains the same.
Therefore,the photoelectric current remains unchanged,i.e.,$I_1 = I_2$.

(C) According to Bohr's model,the radius of the $n^{th}$ orbit of a hydrogen atom is given by the formula $r_n = n^2 a_0$,where $a_0$ is the Bohr radius.
For the second Bohr orbit,the principal quantum number $n = 2$.
Substituting $n = 2$ into the formula,we get $r_2 = (2)^2 a_0 = 4 a_0$.
Therefore,the radius of the second Bohr orbit is $4 a_0$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} \, eV$.
For the $3^{rd}$ orbit $(n=3)$: $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \, eV$.
For the $4^{th}$ orbit $(n=4)$: $E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, eV$.
The energy difference for a transition between the $4^{th}$ and $3^{rd}$ orbit is: $\Delta E = E_4 - E_3 = -0.85 - (-1.51) = 0.66 \, eV$.

(B) The correct option is $B$.
The nucleus of an atom is composed of protons and neutrons,which are collectively known as nucleons.
Protons are positively charged particles,and neutrons are electrically neutral particles.
Electrons are subatomic particles that revolve around the nucleus in extranuclear orbits and are not constituents of the nucleus.

(C) $1 \, a.m.u. = 1.66 \times 10^{-27} \, kg$.
According to Einstein's mass-energy equivalence,$E = mc^2$,where $c$ is the speed of light $(c \approx 3 \times 10^8 \, m/s)$.
Substituting the values:
$E = 1.66 \times 10^{-27} \times (3 \times 10^8)^2 \, J$
$E = 1.66 \times 10^{-27} \times 9 \times 10^{16} \, J$
$E = 14.94 \times 10^{-11} \, J$.
To convert this energy into electron-volts $(eV)$,divide by $1.6 \times 10^{-19} \, J/eV$:
$E = \frac{14.94 \times 10^{-11}}{1.6 \times 10^{-19}} \, eV$
$E \approx 9.3375 \times 10^8 \, eV \approx 931 \times 10^6 \, eV$.
Since $10^6 \, eV = 1 \, MeV$,the equivalent energy is $931 \, MeV$.

(A) The mass of an $\alpha$-particle is less than the sum of the masses of its constituent particles (two protons and two neutrons).
This difference in mass is known as the mass defect $(\Delta m)$.
The mass defect is converted into binding energy,which is required to hold the nucleons together within the nucleus of the $\alpha$-particle.
Therefore,the correct option is $A$.

(B) The energy produced by the Sun is primarily due to nuclear fusion reactions occurring in its core.
In these reactions,hydrogen nuclei (protons) fuse together to form helium nuclei.
This process releases a tremendous amount of energy in the form of electromagnetic radiation,as the mass of the resulting helium nucleus is slightly less than the sum of the masses of the hydrogen nuclei that fused,with the mass defect being converted into energy according to Einstein's mass-energy equivalence principle,$E = \Delta mc^2$.
Therefore,the correct option is $B$.

(D) Copper $(Cu)$ is a conductor,and its resistance decreases as the temperature decreases because the scattering of electrons by lattice vibrations reduces.
Germanium $(Ge)$ is a semiconductor,and its resistance increases as the temperature decreases because the number of free charge carriers (electrons and holes) decreases exponentially with a decrease in temperature.
Therefore,the resistance of copper decreases while that of germanium increases.

(B) The formula for the position of the $n^{th}$ bright fringe (maxima) in Young's double slit experiment is given by $x_n = \frac{n \lambda D}{d}$.
Given values are:
$\lambda = 5000\;\mathring{A} = 5000 \times 10^{-10}\;m = 5 \times 10^{-7}\;m$
$d = 0.2\;mm = 0.2 \times 10^{-3}\;m = 2 \times 10^{-4}\;m$
$D = 200\;cm = 2\;m$
For the third maximum, $n = 3$.
Substituting these values into the formula:
$x_3 = \frac{3 \times (5 \times 10^{-7}\;m) \times (2\;m)}{2 \times 10^{-4}\;m}$
$x_3 = \frac{30 \times 10^{-7}}{2 \times 10^{-4}}\;m = 15 \times 10^{-3}\;m = 1.5 \times 10^{-2}\;m = 1.5\;cm$.
Therefore, the third maximum is at $x = 1.5\;cm$.

(C) The electromagnetic wave propagates in a direction perpendicular to both the electric field vector $\overrightarrow E$ and the magnetic field vector $\overrightarrow B$.
According to the properties of electromagnetic waves,the direction of wave propagation is given by the direction of the Poynting vector $\vec S$,which is proportional to $\overrightarrow E \times \overrightarrow B$.
Therefore,the direction of propagation is along $\overrightarrow E \times \overrightarrow B$.

(C) The Sun's energy is generated through nuclear fusion reactions in its core.
In these reactions,hydrogen nuclei (protons) fuse together to form helium nuclei.
This process releases a tremendous amount of energy in the form of electromagnetic radiation,as the mass of the resulting helium nucleus is slightly less than the sum of the masses of the hydrogen nuclei that formed it,with the mass difference being converted into energy according to Einstein's mass-energy equivalence principle,$E = \Delta mc^2$.

(C) The correct answer is $C$.
Although $Mercury$ is the closest planet to the $Sun$, $Venus$ is the hottest planet in the solar system.
This is because $Venus$ has a very thick atmosphere composed primarily of $CO_2$, which traps heat through a runaway greenhouse effect, leading to surface temperatures of approximately $464^{\circ}C$.

(B) For the light ray to retrace its path,it must strike the silvered surface normally (at an angle of $90^{\circ}$ to the surface).
Let the angle of incidence at the first surface be $i$ and the angle of refraction be $r_1$. The angle of the prism is $A = 30^{\circ}$.
Since the ray strikes the second surface normally,the angle of refraction at the second surface is $r_2 = 0^{\circ}$.
From the prism formula,$A = r_1 + r_2$,we have $30^{\circ} = r_1 + 0^{\circ}$,so $r_1 = 30^{\circ}$.
Applying Snell's law at the first surface:
$\mu = \frac{\sin i}{\sin r_1}$
$\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$
$\sin i = \sqrt{2} \times \sin 30^{\circ} = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$
Therefore,$i = 45^{\circ}$.

(D) To determine the electronic configuration,we count the total number of electrons in each species:
$1$. For $C$ (Carbon,atomic number $Z=6$),the number of electrons is $6$.
$2$. For $N^{+}$ (Nitrogen ion,atomic number $Z=7$),the number of electrons is $7 - 1 = 6$.
Since both $C$ and $N^{+}$ have $6$ electrons,they are isoelectronic and possess the same electronic configuration $(1s^{2} 2s^{2} 2p^{2})$.
Therefore,the correct pair is $C$ and $N^{+}$.

(D) The formula for fringe width in Young's double-slit experiment is $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the screen and the slits,and $d$ is the distance between the two slits.
Since the wavelength of blue light $(\lambda_{b})$ is less than the wavelength of yellow light $(\lambda_{y})$,i.e.,$\lambda_{b} < \lambda_{y}$.
As $\beta \propto \lambda$,the fringe width $\beta$ will decrease when yellow light is replaced by blue light.

(D) The relationship between the wavelength in vacuum $(\lambda_0)$ and the wavelength in a medium $(\lambda_m)$ is given by $\lambda_m = \frac{\lambda_0}{n}$,where $n$ is the refractive index of the medium.
Given that the refractive index $n = 1.5$,we have $\lambda_m = \frac{\lambda_0}{1.5}$.
Since $1.5 > 1$,it follows that $\lambda_m < \lambda_0$.
Therefore,the wavelength of the refracted light will be smaller than the wavelength in vacuum.

(A) To convert a galvanometer into an ammeter,a low resistance,known as a shunt $(S)$,must be connected in parallel with the galvanometer coil.
This arrangement ensures that the majority of the current in the circuit passes through the shunt,protecting the sensitive galvanometer coil from damage due to high current.
Additionally,it reduces the overall resistance of the device,allowing it to measure current accurately without significantly affecting the circuit's current flow.

(D) $X$-rays have a wavelength on the order of the inter-atomic spacing in solid crystals $(10^{-10} \ m)$.
Because the wavelength of $X$-rays is comparable to the inter-atomic distances,they undergo diffraction when interacting with the crystal lattice.
Therefore,$X$-rays are the most suitable electromagnetic waves for investigating the structure of solids.

(C) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0$ is a constant.
The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
The mass of the nucleus is approximately $M = A \times m_p$,where $m_p$ is the mass of a nucleon.
The mass density $\rho$ is defined as $\rho = \frac{M}{V} = \frac{A \times m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{m_p}{\frac{4}{3} \pi R_0^3}$.
Since $m_p$,$R_0$,and $\pi$ are constants,the mass density $\rho$ is independent of the mass number $A$ and remains constant for all nuclei.

(C) According to Faraday's law of induction,the induced electromotive force $(e)$ is given by $e = -\frac{d\Phi}{dt}$.
From Ohm's law,the induced current $(I)$ is $I = \frac{e}{R} = -\frac{1}{R} \frac{d\Phi}{dt}$,where $R$ is the resistance of the loop.
The charge $(q)$ that flows through the loop is given by $q = \int I dt$.
Substituting the expression for $I$,we get $q = \int -\frac{1}{R} \frac{d\Phi}{dt} dt = -\frac{1}{R} \int d\Phi$.
Therefore,$q = -\frac{\Delta\Phi}{R}$,where $\Delta\Phi$ is the total change in magnetic flux.
Thus,the total charge induced depends on the total change in magnetic flux.