AIIMS 2003 Physics Question Paper with Answer and Solution

(A) The force per unit length between two parallel current-carrying wires is given by the formula: $F/l = \frac{\mu_0 I_1 I_2}{2\pi r}$.
Rearranging for $\mu_0$,we get: $\mu_0 = \frac{2\pi r F}{l I_1 I_2}$.
The dimensional formula for force $F$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for length $l$ is $[L^1]$.
The dimensional formula for current $I$ is $[A^1]$.
The dimensional formula for distance $r$ is $[L^1]$.
Substituting these into the formula: $[\mu_0] = \frac{[L^1] [M^1 L^1 T^{-2}]}{[L^1] [A^1] [A^1]}$.
Simplifying the expression: $[\mu_0] = [M^1 L^1 T^{-2} A^{-2}]$.

(B) When a ball is thrown vertically upwards,air resistance acts in the direction opposite to the motion.
During the upward motion,both gravity $(g)$ and air resistance $(a)$ act downwards. Thus,the net acceleration is $a_{up} = -(g + a)$,which is constant and directed downwards. Since acceleration is constant,the speed decreases linearly with time.
During the downward motion,gravity $(g)$ acts downwards and air resistance $(a)$ acts upwards. Thus,the net acceleration is $a_{down} = (g - a)$,which is also constant and directed downwards. Since acceleration is constant,the speed increases linearly with time.
However,because $|a_{up}| > |a_{down}|$,the slope of the speed-time graph (which represents the magnitude of acceleration) will be steeper during the upward motion than during the downward motion. This corresponds to the graph shown in option $B$.

(B) For a head-on elastic collision,the fractional energy loss of the projectile (mass $m_1$) colliding with a stationary target (mass $m_2$) is given by the formula:
$\frac{\Delta K}{K} = 1 - \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2$
Here,the mass of a neutron $m_1 = 1 \text{ u}$ and the mass of a deuteron $m_2 = 2 \text{ u}$.
Substituting these values into the formula:
$\frac{\Delta K}{K} = 1 - \left( \frac{1 - 2}{1 + 2} \right)^2$
$\frac{\Delta K}{K} = 1 - \left( \frac{-1}{3} \right)^2$
$\frac{\Delta K}{K} = 1 - \frac{1}{9}$
$\frac{\Delta K}{K} = \frac{8}{9}$
Thus,the fractional energy loss of the neutron is $8/9$.

(B) The escape velocity is the minimum initial velocity required for an object to escape the gravitational field of a planet and never return.
For an object of mass $m$ to escape from the surface of the Earth (mass $M_e$,radius $R_e$),its initial kinetic energy must be equal to the magnitude of its gravitational potential energy.
$\frac{1}{2} m v_e^2 = \frac{G M_e m}{R_e}$
Solving for $v_e$,we get:
$v_e = \sqrt{\frac{2 G M_e}{R_e}}$
From this formula,it is clear that the escape velocity $v_e$ depends only on the gravitational constant $G$,the mass of the Earth $M_e$,and the radius of the Earth $R_e$.
It is independent of the mass of the projectile $m$. Therefore,the correct option is $B$.

(B) Let the radius of each small drop be $R$. The volume of two small drops is $V_{initial} = 2 \times (\frac{4}{3}\pi R^3) = \frac{8}{3}\pi R^3$.
Let the radius of the large drop be $R'$. Since the volume remains constant,$\frac{4}{3}\pi (R')^3 = \frac{8}{3}\pi R^3$,which gives $R' = 2^{1/3}R$.
The surface energy of a drop is given by $E = T \times A$,where $T$ is surface tension and $A$ is surface area.
Initial surface energy $E_i = 2 \times (T \times 4\pi R^2) = 8\pi R^2 T$.
Final surface energy $E_f = T \times 4\pi (R')^2 = 4\pi (2^{1/3}R)^2 T = 4\pi 2^{2/3} R^2 T$.
The ratio of initial to final surface energy is $\frac{E_i}{E_f} = \frac{8\pi R^2 T}{4\pi 2^{2/3} R^2 T} = \frac{2}{2^{2/3}} = 2^{1 - 2/3} = 2^{1/3}$.
Thus,the ratio is $2^{1/3}:1$.

(A) When a lead shot falls through a viscous liquid like glycerine, it is acted upon by three forces: gravity $(mg)$ acting downwards, and both buoyancy $(F_B)$ and viscous drag $(F_v = 6\pi \eta rv)$ acting upwards. The net force is $F_{net} = mg - F_B - 6\pi \eta rv$. Initially, the velocity $v$ is zero, so the viscous drag is zero, and the acceleration is maximum. As the velocity increases, the viscous drag increases, causing the acceleration to decrease. Eventually, the net force becomes zero when the viscous drag balances the effective weight, and the lead shot attains a constant terminal velocity $(v_t)$. The velocity-distance graph starts from the origin, increases at a decreasing rate, and asymptotically approaches the terminal velocity value. This behavior is correctly represented by plot $(a)$.

(C) According to Stefan-Boltzmann Law,the power radiated by a black body is given by $P = A\sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature in Kelvin.
Given:
$P_1 = 20\,W$
$T_1 = 227^{\circ}C = 227 + 273 = 500\,K$
$T_2 = 727^{\circ}C = 727 + 273 = 1000\,K$
Since $P \propto T^4$,we have:
$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{P_2}{20} = \left( \frac{1000}{500} \right)^4$
$\frac{P_2}{20} = (2)^4$
$\frac{P_2}{20} = 16$
$P_2 = 16 \times 20 = 320\,W$
Therefore,the radiating power will be $320\,W$.

(A) According to Wien's displacement law,the wavelength corresponding to maximum intensity $\lambda_m$ is inversely proportional to the absolute temperature $T$,i.e.,$\lambda_m \propto \frac{1}{T}$.
Given $T_2 > T_1$,it follows that $\lambda_{m_2} < \lambda_{m_1}$.
This means the peak of the intensity-wavelength curve for the higher temperature $T_2$ shifts towards a shorter wavelength (left side) compared to the curve for $T_1$.
Additionally,according to the Stefan-Boltzmann law,the total energy emitted per unit area per unit time is proportional to $T^4$,so the intensity $I$ for $T_2$ will be higher than for $T_1$ at all wavelengths.
Therefore,the correct plot is the one where the curve for $T_2$ is higher and its peak is shifted to the left relative to the curve for $T_1$.

(C) When a mass $m$ is connected between two springs in parallel (as shown in the figure),the effective spring constant $K_{eff}$ is the sum of the individual spring constants.
$K_{eff} = K_1 + K_2 = K + 2K = 3K$
The frequency of oscillation $f$ for a spring-mass system is given by the formula:
$f = \frac{1}{2\pi} \sqrt{\frac{K_{eff}}{m}}$
Substituting the value of $K_{eff}$:
$f = \frac{1}{2\pi} \sqrt{\frac{3K}{m}}$
Thus,the correct option is $C$.

(C) Let $d$ be the distance of the epicenter from the seismograph.
Let $v_P = 8.0 \, km/s$ and $v_S = 4.5 \, km/s$ be the speeds of $P$ and $S$ waves respectively.
The time taken by $P$ waves is $t_P = d / v_P$ and by $S$ waves is $t_S = d / v_S$.
Given the time difference $\Delta t = t_S - t_P = 4.0 \, min = 240 \, s$.
Substituting the expressions for time: $d / v_S - d / v_P = 240$.
$d (1 / 4.5 - 1 / 8.0) = 240$.
$d ((8.0 - 4.5) / (4.5 \times 8.0)) = 240$.
$d (3.5 / 36) = 240$.
$d = (240 \times 36) / 3.5 \approx 2468.6 \, km$.
Rounding to the nearest given option,the distance is approximately $2500 \, km$.

(B) mean solar day is the time taken by the Earth to rotate once on its axis relative to the Sun,which is approximately $24$ hours.
$A$ sidereal day is the time taken by the Earth to rotate once on its axis relative to the distant stars,which is approximately $23$ hours and $56$ minutes.
The difference between a mean solar day ($24$ hours) and a sidereal day ($23$ hours $56$ minutes) is $4$ minutes.
Therefore,the correct option is $B$.

(C) The gravitational force exerted by the Sun on a planet acts along the line joining the Sun and the planet.
This force is a central force,meaning its torque about the Sun is zero.
According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the angular momentum of the system remains constant.
Therefore,the motion of planets in the solar system is an example of the conservation of angular momentum.

(D) The angular momentum of the Earth-Moon system is conserved because no external torque acts on the system.
From the relation $\tau = \frac{dL}{dt}$,if $\tau = 0$,then $\frac{dL}{dt} = 0$,which implies $L$ is constant.
Therefore,the $Reason$ is incorrect.
Due to tidal friction,the Earth's rotation slows down,meaning its angular velocity $\omega_1$ decreases. Since the total angular momentum $L = I_1\omega_1 + I_2\omega_2$ must remain constant,the angular momentum of the Moon's orbit must increase to compensate for the decrease in the Earth's rotational angular momentum.
As the Moon gains angular momentum,its orbital radius $r_2$ increases,causing the Moon to move away from the Earth. Thus,the $Assertion$ is also incorrect.

(C) The length of the day is slowly increasing due to tidal friction caused by the gravitational interaction between the Earth and the Moon,as well as viscous forces between the Earth and its atmosphere.
This process causes a gradual dissipation of the Earth's rotational energy,leading to a slowdown in its rotation.
The gravitational pull of other planets in the solar system is negligible in this context compared to tidal effects.
Therefore,the $Assertion$ is correct,but the $Reason$ is incorrect.

(C) The $Assertion$ is correct because the air directly above the fire becomes heated,expands,and rises due to convection currents.
This process carries heat energy upwards,making the region above the fire significantly hotter than the sides.
The $Reason$ is incorrect because air is a poor conductor of heat. The heat transfer in this scenario is primarily due to convection,not conduction.

(A) When a bottle of cold carbonated drink is opened,the high-pressure gas inside the bottle expands rapidly into the atmosphere.
This rapid expansion is an adiabatic process because it happens so quickly that there is no time for heat exchange with the surroundings.
According to the first law of thermodynamics,for an adiabatic expansion $(dQ = 0)$,the work done by the gas $(dW > 0)$ results in a decrease in internal energy $(dU < 0)$,which leads to a significant drop in temperature.
Due to this sudden cooling,the water vapour present in the air near the opening condenses into tiny liquid droplets,forming a visible fog.

(C) The amplitude of an oscillating pendulum decreases with time due to air resistance (damping). This is known as damped oscillation.
However,the frequency of an oscillating pendulum depends on its length $L$ and acceleration due to gravity $g$,given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$.
Since $g$ and $L$ remain constant during the motion,the frequency of the pendulum remains constant over time.
Therefore,the Assertion is correct,but the Reason is incorrect.

(A) When a beetle moves along the sand,it creates disturbances that travel as waves on the sand's surface.
These disturbances consist of two types of pulses: longitudinal waves (which travel faster) and transverse waves (which travel slower).
The sand scorpion is highly sensitive to these vibrations. By sensing the time interval between the arrival of the longitudinal waves and the transverse waves,the scorpion can calculate the distance to the beetle.
Furthermore,by comparing the signals received by its different legs,the scorpion can determine the direction of the beetle.
Thus,the Reason correctly explains how the scorpion detects and locates the beetle.

(C) From the third equation of kinematics,we have $v^2 - u^2 = 2as$.
Assuming the object starts from rest,the initial velocity $u = 0$.
Substituting this into the equation,we get $v^2 = 2as$,which can be rearranged as $s = \frac{v^2}{2a}$.
Since $a$ is a uniform (constant) acceleration,the relationship between displacement $s$ and velocity $v$ is $s \propto v^2$.
This represents a parabola opening along the $s$-axis.
Graph $C$ shows a parabolic curve where $s$ increases with the square of $v$,starting from the origin $(0, 0)$,which matches the derived relationship $s = \frac{1}{2a} v^2$.

(C) Let the charges at $A, B,$ and $C$ be $+Q, -Q,$ and $+Q$ respectively. The distance between any two vertices is $a$.
The force on charge at $A$ due to charge at $B$ is $F_B = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$ (attractive,directed towards $B$).
The force on charge at $A$ due to charge at $C$ is $F_C = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$ (repulsive,directed away from $C$).
Resolving these forces into components parallel and normal to $BC$:
$1$. The components of $F_B$ and $F_C$ parallel to $BC$ are $F_B \cos 60^\circ$ and $F_C \cos 60^\circ$ respectively,both directed towards the left.
$2$. The components of $F_B$ and $F_C$ normal to $BC$ are $F_B \sin 60^\circ$ (downwards) and $F_C \sin 60^\circ$ (upwards) respectively.
Since $|F_B| = |F_C|$,the net force in the direction normal to $BC$ is $F_C \sin 60^\circ - F_B \sin 60^\circ = 0$.

(C) The kinetic energy $(KE)$ gained by a charged particle when accelerated through a potential difference $(V)$ is given by the formula: $KE = qV$.
Here,the charge of a proton $(q)$ is equal to the elementary charge $(e)$.
The potential difference $(V)$ is given as $1\, kV$.
Substituting these values into the formula:
$KE = e \times 1\, kV = 1\, keV$.
Therefore,the kinetic energy of the proton is $1\, keV$.

(A) In a non-uniform electric field,the electric field intensity at the position of the two charges of the dipole ($-q$ and $+q$) is different.
Since $F = qE$,the forces acting on the two charges are $F_1 = -qE_1$ and $F_2 = qE_2$.
Because $E_1
eq E_2$,the net force $F_{net} = F_1 + F_2
eq 0$.
Additionally,since these forces act at different points and are not collinear,they exert a net torque on the dipole.
Therefore,the dipole experiences both a net force and a net torque.

(B) According to Gauss's Law,the electric flux $\phi$ through any closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.
Mathematically,$\phi = \frac{Q_{enc}}{\varepsilon_0}$.
In the given figure,the surface $S$ encloses two charges,each of magnitude $+q$.
Therefore,the total enclosed charge $Q_{enc} = +q + q = 2q$.
Substituting this into Gauss's Law,we get $\phi = \frac{2q}{\varepsilon_0}$.

(B) The magnetic field $B$ at the center of a circular loop carrying current $i$ is given by the formula: $B = \frac{\mu_0 i}{2r}$.
Given:
$B = 0.5 \times 10^{-5} \, T$ (since $1 \, Wb/m^2 = 1 \, T$)
$r = 5.0 \, cm = 5.0 \times 10^{-2} \, m$
$\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$
Substituting the values into the formula:
$0.5 \times 10^{-5} = \frac{4\pi \times 10^{-7} \times i}{2 \times 5.0 \times 10^{-2}}$
$0.5 \times 10^{-5} = \frac{2\pi \times 10^{-7} \times i}{5.0 \times 10^{-2}}$
$i = \frac{0.5 \times 10^{-5} \times 5.0 \times 10^{-2}}{2 \times 3.14 \times 10^{-7}}$
$i = \frac{2.5 \times 10^{-7}}{6.28 \times 10^{-7}}$
$i \approx 0.398 \, A \approx 0.4 \, A$.

(C) The magnetic field produced by the long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.
For the side of the loop closer to the wire at distance $r_1$,the current flows in the same direction as the wire. According to Fleming's left-hand rule,the force $F_1$ on this side is attractive (towards the wire).
For the side of the loop farther from the wire at distance $r_2$,the current flows in the opposite direction to the wire. The force $F_2$ on this side is repulsive (away from the wire).
Since $r_1 < r_2$,the magnetic field $B_1$ at the closer side is stronger than the magnetic field $B_2$ at the farther side $(B_1 > B_2)$.
Consequently,the attractive force $F_1$ is greater than the repulsive force $F_2$ $(F_1 > F_2)$.
Therefore,the net force $F_{net} = F_1 - F_2$ acts towards the wire,and the loop will move towards the wire.

(B) The phenomenon of levitation of a frog in a strong magnetic field is a classic demonstration of diamagnetism.
Biological tissues,including the body of a frog,consist largely of water and organic molecules,which exhibit diamagnetic properties.
Diamagnetic substances are weakly repelled by magnetic fields.
When a strong magnetic field is applied,the diamagnetic repulsion force acting on the frog's body can balance the gravitational force,allowing the frog to levitate.
Therefore,the correct option is $B$.

(B) According to the de-Broglie hypothesis,the momentum $p$ of a particle is related to its wavelength $\lambda$ by the equation $p = \frac{h}{\lambda}$,where $h$ is Planck's constant.
Since both the electron and the photon have the same wavelength $\lambda$,and $h$ is a universal constant,their momenta must be equal.
Therefore,they have the same momentum.

(B) Characteristic $X$-rays are produced when an element is bombarded with high-energy particles such as photons,electrons,or ions.
When an incident particle strikes a bound electron in an atom,the bound electron is ejected from an inner shell of the atom.
After the electron is ejected,the atom is left with a vacant energy level,known as a core hole.
Electrons from outer shells then fall into the inner shell,emitting quantized photons with energy equivalent to the difference between the higher and lower energy states.
Since each element has a unique set of energy levels,the transitions from higher to lower energy levels produce $X$-rays with frequencies that are characteristic to each specific element.

(B) When a positron $(e^+)$ encounters an electron $(e^-)$,they annihilate each other because they are particle-antiparticle pairs.
This annihilation process results in the conversion of their mass-energy into two gamma-ray photons,typically emitted in opposite directions to conserve momentum.
This physical phenomenon is the fundamental principle behind Positron Emission Tomography,commonly abbreviated as $PET$ scan,which is used in medical diagnostics to image metabolic processes in the body.

(A) The radioactive decay law is given by $\frac{N}{N_0} = \left( \frac{1}{2} \right)^n$,where $n$ is the number of half-lives.
Given $\frac{N}{N_0} = \frac{1}{16}$,we can write $\frac{1}{16} = \left( \frac{1}{2} \right)^4$.
Comparing the two expressions,we get $n = 4$.
The number of half-lives $n$ is related to the total time $t$ and half-life $T_{1/2}$ by $n = \frac{t}{T_{1/2}}$.
Given $t = 40$ days and $n = 4$,we have $4 = \frac{40}{T_{1/2}}$.
Therefore,$T_{1/2} = \frac{40}{4} = 10$ days.

(C) Radioactive nuclei injected into a patient accumulate at specific sites within the body.
These nuclei undergo radioactive decay and emit electromagnetic radiation (typically gamma rays).
These radiations are detected by external detectors to map the distribution of the tracer.
This diagnostic procedure is known as the Radiotracer technique.

(B) Germanium $(Ge)$ is a group $14$ element with $4$ valence electrons.
Gallium $(Ga)$ is a group $13$ element,which means it has $3$ valence electrons.
When a trivalent impurity like Gallium is added to a tetravalent semiconductor like Germanium,it creates a vacancy or a 'hole' in the crystal lattice.
Since the majority charge carriers in this doped semiconductor are holes (positive charge carriers),the material behaves as a $P-$ type semiconductor.

(C) Given: Input voltage $V_{in} = 1 \, mV = 10^{-3} \, V$, current gain $\beta = 100$, input resistance $R_{in} = 1 \, k\Omega = 10^3 \, \Omega$, and load resistance $R_L = 10 \, k\Omega = 10^4 \, \Omega$.
Voltage gain $A_v = \beta \times \frac{R_L}{R_{in}}$.
$A_v = 100 \times \frac{10 \, k\Omega}{1 \, k\Omega} = 100 \times 10 = 1000$.
Output voltage $V_{out} = A_v \times V_{in}$.
$V_{out} = 1000 \times 10^{-3} \, V = 1 \, V$.

(A) The critical frequency $(f_c)$ of the ionosphere is given by the formula: $f_c \approx 9 \times (N_{max})^{1/2}$, where $N_{max}$ is the maximum electron density in $\text{electrons/m}^3$.
Given $N_{max} = 10^{11} \text{ m}^{-3}$.
Substituting the value: $f_c \approx 9 \times (10^{11})^{1/2} = 9 \times \sqrt{10} \times 10^5 \approx 9 \times 3.16 \times 10^5 \approx 28.44 \times 10^5 \text{ Hz} \approx 2.84 \text{ MHz}$.
Since the wave is reflected back if its frequency is < or equal to the critical frequency, among the given options, $2 \text{ MHz}$ is the only frequency that will be reflected back by the ionosphere.

(A) The maximum line-of-sight distance $d$ for a $TV$ tower of height $h$ is given by the formula $d = \sqrt{2hR}$,where $R$ is the radius of the Earth.
Since $R$ is a constant,the relationship between the distance $d$ and the height $h$ is $d \propto \sqrt{h}$ or $d \propto h^{1/2}$.
Therefore,the maximum distance is proportional to $h^{1/2}$.

(D) The primary reason a laser beam is used in surgery is its ability to be focused into an extremely small,intense spot. This allows for precise cutting or ablation of biological tissue with minimal damage to the surrounding areas. While lasers are also monochromatic,coherent,and directional,these properties are secondary to the ability to be sharply focused for surgical applications.

(A) The critical frequency $(f_c)$ for sky wave propagation is related to the maximum electron density $(N_{\max})$ by the formula: $f_c = 9 \sqrt{N_{\max}}$.
Given $f_c = 10 \, MHz = 10 \times 10^6 \, Hz$.
Substituting the value: $10 \times 10^6 = 9 \sqrt{N_{\max}}$.
Squaring both sides: $(10^7)^2 = 81 \times N_{\max}$.
$N_{\max} = \frac{10^{14}}{81} \approx 1.23 \times 10^{12} \, m^{-3}$.
Thus, the minimum electron density required is approximately $1.2 \times 10^{12} \, m^{-3}$.

(C) The angular resolution limit of the human eye is given by the Rayleigh criterion: $\theta = \frac{1.22 \lambda}{d}$.
Here,$\lambda = 500 \times 10^{-9} \, m$ is the wavelength of light,$d = 5 \times 10^{-3} \, m$ is the diameter of the pupil,and $r = 400 \times 10^{3} \, m$ is the distance (altitude).
The linear resolution $x$ is given by $x = r \theta = \frac{1.22 \lambda r}{d}$.
Substituting the values:
$x = \frac{1.22 \times (500 \times 10^{-9} \, m) \times (400 \times 10^{3} \, m)}{5 \times 10^{-3} \, m}$.
$x = \frac{1.22 \times 5 \times 10^{-7} \times 4 \times 10^{5}}{5 \times 10^{-3}}$.
$x = \frac{1.22 \times 20 \times 10^{-2}}{5 \times 10^{-3}} = \frac{24.4 \times 10^{-2}}{5 \times 10^{-3}} = 4.88 \times 10^{1} \approx 50 \, m$.

(C) The mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
For a concave mirror,using sign convention,$u$ and $v$ are negative. Let their magnitudes be $u$ and $v$. Then $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,which can be written as $v = \frac{uf}{u-f}$.
As $u \to f$,$v \to \infty$.
As $u \to \infty$,$v \to f$.
This represents a rectangular hyperbola where $v$ decreases as $u$ increases,which corresponds to the curve shown in graph $C$.

(C) The accuracy of determining the position of an object using a beam of light is limited by the phenomenon of diffraction.
According to the Rayleigh criterion,the resolving power of an optical system is inversely proportional to the wavelength of the light used.
Mathematically,the limit of resolution or the minimum resolvable distance $d$ is given by $d \approx \frac{\lambda}{2 \cdot NA}$,where $\lambda$ is the wavelength and $NA$ is the numerical aperture.
To achieve maximum accuracy,the minimum resolvable distance $d$ must be as small as possible.
Since $d \propto \lambda$,a smaller wavelength $\lambda$ results in higher accuracy.
Therefore,the maximum accuracy is achieved if the light is of shorter wavelength.

(C) The path difference introduced by the thin film is $\Delta x = (\mu - 1)t$.
The shift in the central maximum in terms of fringe width $\beta$ is given by $n = \frac{(\mu - 1)t}{\lambda}$.
Substituting the given values: $n = \frac{(1.5 - 1) \times 2 \times 10^{-6} \, m}{500 \times 10^{-9} \, m} = \frac{0.5 \times 2 \times 10^{-6}}{500 \times 10^{-9}} = \frac{1 \times 10^{-6}}{5 \times 10^{-7}} = 2$.
Since the thin film is introduced in the path of the upper beam,the optical path length of the upper beam increases,causing the central maximum to shift upward towards the upper beam.

(A) The maximum line-of-sight distance $d$ for a $TV$ tower of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Since $R$ is a constant,the relationship between the distance $d$ and the height $h$ is $d \propto \sqrt{h}$.
Therefore,the maximum distance is proportional to $h^{1/2}$.

(C) In a tube light,the gas contains metal vapors. In metallic atoms,electronic transitions occur,resulting in the emission of light of specific wavelengths.
Emission of white light in a tube light is due to these electronic transitions and not due to thermal radiation (vibration of atoms) as seen in hot,incandescent substances.
Therefore,the Assertion is correct,but the Reason is incorrect.

(D) The gravitational force is the dominant force on a cosmic scale because it is always attractive and acts over long distances between massive objects,whereas the Coulomb force (electrostatic force) is often neutralized by the presence of both positive and negative charges in matter.
Therefore,the Assertion is incorrect.
Furthermore,the gravitational force is significantly weaker than the Coulomb force at the microscopic level (e.g.,between two electrons),but the Reason states that the Coulomb force is weaker than the gravitational force,which is incorrect.
Thus,both the Assertion and the Reason are incorrect.

(A) The phenomenon described is known as the Aurora.
In polar regions (high latitudes),high-energy charged particles (electrons and protons) emitted by the sun (solar wind) are guided by the Earth's magnetic field lines toward the poles.
As these particles enter the upper atmosphere,they collide with gas atoms and molecules,causing them to emit light,which appears as colourful curtains or streamers.
Therefore,the Assertion is correct,and the Reason correctly explains the mechanism behind this phenomenon.
In the northern hemisphere,this is called $Aurora$ $Borealis$,and in the southern hemisphere,it is called $Aurora$ $Australis$.

(B) The twinkling of stars is due to the atmospheric refraction caused by the changing refractive index of the atmosphere.
Stars appear as point-sized sources because they are very far away. Due to the small apparent size,the fluctuations in the path of light rays are significant,leading to the twinkling effect.
Planets are much closer to the Earth and appear as extended sources (collection of many point-sized sources). The total variation in the light coming from all these point sources averages out to zero,so they do not twinkle.
The Reason provided states that stars are bigger than planets,which is a fact,but it is not the reason why stars twinkle and planets do not. The actual reason is the difference in their apparent size (point-sized vs. extended source).

(B) According to classical physics,any charged particle undergoing acceleration must radiate electromagnetic energy. Since an electron revolving in an orbit around the nucleus is undergoing centripetal acceleration,classical electrodynamics predicts it should continuously lose energy and spiral into the nucleus.
To resolve this instability,Bohr postulated that electrons in specific 'stationary' orbits do not radiate energy.
Both the Assertion and the Reason are scientifically correct statements. However,the Reason describes the classical conflict,while the Assertion describes Bohr's specific quantum postulate to overcome that conflict. The Reason does not explain *why* Bohr's postulate is true; it only explains why the postulate was *necessary*. Therefore,the Reason is not the correct explanation of the Assertion.

(C) Radioactive nuclei do emit $\beta^-$ particles during the process of $\beta$-decay. Therefore,the Assertion is correct.
However,electrons do not exist inside the nucleus. The $\beta^-$ particle (electron) is created at the moment of decay when a neutron transforms into a proton,an electron,and an antineutrino $(n \rightarrow p + e^- + \bar{\nu}_e)$. Therefore,the Reason is incorrect.

(B) Neutrons penetrate matter more readily than protons because neutrons are electrically neutral,meaning they experience no electrostatic repulsion from the positively charged nucleus.
Protons,being positively charged,experience a strong electrostatic repulsive force from the nucleus,which hinders their penetration.
While it is true that the mass of a neutron $(m_n \approx 1.6749 \times 10^{-27} \ kg)$ is slightly greater than the mass of a proton $(m_p \approx 1.6726 \times 10^{-27} \ kg)$,this mass difference is not the reason for their higher penetrating power.
Therefore,both the Assertion and the Reason are correct,but the Reason is not the correct explanation for the Assertion.

(D) The Assertion is incorrect because the resistivity of a semiconductor decreases with an increase in temperature.
As temperature increases,more valence electrons gain sufficient thermal energy to jump across the forbidden energy gap into the conduction band.
This increase in the number of charge carriers (electrons and holes) leads to an increase in conductivity,which implies a decrease in resistivity.
The Reason is also incorrect because the mechanism described (vibration of atoms) is the primary cause for the increase in resistivity in metals,not semiconductors.

(B) The magnetic force on a charged particle moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by the Lorentz force formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
For an electron,the charge $q = -e$.
The velocity of the electron is $\overrightarrow{v} = v_x \hat{i}$ and the magnetic field is $\overrightarrow{B} = B_y \hat{j}$.
Substituting these into the force equation:
$\overrightarrow{F} = -e(v_x \hat{i} \times B_y \hat{j})$
$\overrightarrow{F} = -e v_x B_y (\hat{i} \times \hat{j})$
Since $\hat{i} \times \hat{j} = \hat{k}$,we get $\overrightarrow{F} = -e v_x B_y \hat{k}$.
The force is always perpendicular to both the velocity and the magnetic field. Since the velocity is in the $x$-direction and the magnetic field is in the $y$-direction,the force acts in the $z$-direction (specifically $-z$ for an electron). The particle will move in a circle in the plane perpendicular to the magnetic field,which is the $xz$-plane.