AIPMT 2005 Physics Question Paper with Answer and Solution

(C) Let the two vectors be $\overrightarrow{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ and $\overrightarrow{B} = -4\hat{i} + 4\hat{j} + \alpha\hat{k}$.
Since the vectors are perpendicular,their dot product must be zero,i.e.,$\overrightarrow{A} \cdot \overrightarrow{B} = 0$.
Calculating the dot product:
$(2\hat{i} + 3\hat{j} + 8\hat{k}) \cdot (-4\hat{i} + 4\hat{j} + \alpha\hat{k}) = 0$
$(2)(-4) + (3)(4) + (8)(\alpha) = 0$
$-8 + 12 + 8\alpha = 0$
$4 + 8\alpha = 0$
$8\alpha = -4$
$\alpha = -4/8 = -0.5$.

(B) The expression given is a scalar triple product of the form $\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{A})$.
By the properties of the cross product,the vector $\overrightarrow{C} = (\overrightarrow{B} \times \overrightarrow{A})$ is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$.
Since $\overrightarrow{C}$ is perpendicular to $\overrightarrow{A}$,the dot product of $\overrightarrow{A}$ and $\overrightarrow{C}$ must be zero.
Mathematically,$\overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{A}) = 0$ because the vector $(\overrightarrow{B} \times \overrightarrow{A})$ lies in a plane perpendicular to $\overrightarrow{A}$.

(A) The dimension of Planck's constant $(h)$ is $[ML^2T^{-1}]$.
The dimension of moment of inertia $(I)$ is $[ML^2]$.
Taking the ratio of the two:
$\frac{[h]}{[I]} = \frac{[ML^2T^{-1}]}{[ML^2]} = [T^{-1}]$.
Since the dimension $[T^{-1}]$ corresponds to frequency,the correct option is $A$.

(B) Let the two boys meet at point $C$ after time $t$ from the start.
In time $t$,the boy from $B$ covers distance $BC = v_1 t$.
The boy from $A$ covers distance $AC = v t$.
Since the boy at $B$ runs perpendicular to $AB$,$\triangle ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$.
Using the Pythagorean theorem: $(AC)^2 = (AB)^2 + (BC)^2$.
Substituting the values: $(vt)^2 = a^2 + (v_1 t)^2$.
$v^2 t^2 - v_1^2 t^2 = a^2$.
$t^2 (v^2 - v_1^2) = a^2$.
$t = \sqrt{\frac{a^2}{v^2 - v_1^2}}$.

(D) Given displacement $x = a e^{-\alpha t} + b e^{\beta t}$.
Velocity $v$ is the rate of change of displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}(a e^{-\alpha t} + b e^{\beta t})$
$v = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$.
To determine how velocity changes with time,we find the acceleration $a_{acc}$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(-a\alpha e^{-\alpha t} + b\beta e^{\beta t})$
$a_{acc} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$.
Since $a, b, \alpha, \beta$ are positive constants and the exponential functions $e^{-\alpha t}$ and $e^{\beta t}$ are always positive for any time $t$,the acceleration $a_{acc}$ is always positive.
Because the acceleration is positive,the velocity $v$ will go on increasing with time.

(B) Let the initial velocity be $u$ and the maximum height be $H$. The maximum height is given by $H = \frac{u^2}{2g}$.
At height $h = \frac{H}{2}$,the velocity $v = 10 \; m/s$.
Using the equation of motion $v^2 = u^2 - 2gh$:
$(10)^2 = u^2 - 2g \left( \frac{H}{2} \right)$
$100 = u^2 - gH$
Substitute $H = \frac{u^2}{2g}$ into the equation:
$100 = u^2 - g \left( \frac{u^2}{2g} \right)$
$100 = u^2 - \frac{u^2}{2} = \frac{u^2}{2}$
$u^2 = 200 \; m^2/s^2$.
Now,calculate the maximum height $H$:
$H = \frac{u^2}{2g} = \frac{200}{2 \times 10} = \frac{200}{20} = 10 \; m$.

(C) The length of the string is the radius of the circular path,$r = 1\,m$.
The frequency of revolution $n$ is given by $n = \frac{\text{number of revolutions}}{\text{time}} = \frac{22}{44} = 0.5\,Hz$.
The angular velocity $\omega$ is given by $\omega = 2\pi n = 2\pi(0.5) = \pi\,rad/s$.
The centripetal acceleration $a$ is given by $a = \omega^2 r$.
Substituting the values,$a = (\pi)^2 \times 1 = \pi^2\,m/s^2$.
In uniform circular motion,the acceleration is centripetal,meaning its direction is always along the radius and towards the centre.

(B) According to the law of conservation of linear momentum,the initial momentum of the bomb is zero.
Therefore,the final momentum of the two pieces must be equal in magnitude and opposite in direction.
Let $m_A = 18 \, kg$,$v_A = 6 \, m/s$ and $m_B = 12 \, kg$.
$m_A v_A = m_B v_B$
$18 \times 6 = 12 \times v_B$
$v_B = \frac{108}{12} = 9 \, m/s$
The kinetic energy of the $12 \, kg$ mass is given by:
$K.E. = \frac{1}{2} m_B v_B^2$
$K.E. = \frac{1}{2} \times 12 \times (9)^2$
$K.E. = 6 \times 81 = 486 \, J$

(B) The work done by a variable force is equal to the area under the $F-x$ graph.
From the graph,the area is a trapezoid with parallel sides of length $3\,m$ (from $x=0$ to $x=3$) and $6\,m$ (from $x=0$ to $x=6$),and height $3\,N$.
Alternatively,the area can be calculated as the sum of a rectangle (from $x=0$ to $x=3$) and a triangle (from $x=3$ to $x=6$).
Area of rectangle = $3\,m \times 3\,N = 9\,J$.
Area of triangle = $\frac{1}{2} \times (6-3)\,m \times 3\,N = \frac{1}{2} \times 3 \times 3 = 4.5\,J$.
Total work done = $9\,J + 4.5\,J = 13.5\,J$.

(B) The kinetic energy $(K)$ of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth (mass $M$) is given by $K = \frac{GMm}{2r}$.
The potential energy $(U)$ of the satellite is given by $U = -\frac{GMm}{r}$.
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{GMm/2r}{-GMm/r} = -\frac{1}{2}$.
Note: The magnitude ratio is $1/2$,but considering the sign,the ratio is $-1/2$.

(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = \frac{T_1 - T_2}{T_1} = \frac{W}{Q_1}$.
Here,$T_1 = 227^{\circ}C = 227 + 273 = 500 \text{ K}$ and $T_2 = 127^{\circ}C = 127 + 273 = 400 \text{ K}$.
The heat absorbed at the higher temperature is $Q_1 = 6 \times 10^4 \text{ cal}$.
Using the formula $W = Q_1 \left( \frac{T_1 - T_2}{T_1} \right)$:
$W = 6 \times 10^4 \times \left( \frac{500 - 400}{500} \right)$
$W = 6 \times 10^4 \times \left( \frac{100}{500} \right)$
$W = 6 \times 10^4 \times 0.2 = 1.2 \times 10^4 \text{ cal}$.
Thus,the amount of heat converted to work is $1.2 \times 10^4 \text{ cal}$.

(D) process is reversible if it occurs infinitely slowly such that the system remains in thermodynamic equilibrium with its surroundings at every stage.
$(a)$ Heat transfer by radiation involves a temperature gradient and is irreversible.
$(b)$ Electrical heating (Joule heating) is irreversible due to the dissipation of energy as heat.
$(c)$ Heat transfer by conduction requires a temperature difference and is irreversible.
$(d)$ An isothermal compression of an ideal gas,if performed infinitely slowly,allows the system to remain in equilibrium,making it a reversible process.

(B) The rate of heat conduction $H$ through a rod is given by the formula $H = \frac{kA \Delta T}{l}$,where $k$ is the thermal conductivity,$A$ is the cross-sectional area,$\Delta T$ is the temperature difference,and $l$ is the length.
Since the material is the same,$k$ is constant. Since the temperature difference $\Delta T$ is the same,$H \propto \frac{A}{l}$.
For a circular rod,the cross-sectional area $A = \pi r^2$,so $H \propto \frac{r^2}{l}$.
Evaluating the ratio $\frac{r^2}{l}$ for each option:
$(a) \frac{(2r_0)^2}{2l_0} = \frac{4r_0^2}{2l_0} = 2 \frac{r_0^2}{l_0}$
$(b) \frac{(2r_0)^2}{l_0} = \frac{4r_0^2}{l_0} = 4 \frac{r_0^2}{l_0}$
$(c) \frac{r_0^2}{l_0} = 1 \frac{r_0^2}{l_0}$
$(d) \frac{r_0^2}{2l_0} = 0.5 \frac{r_0^2}{l_0}$
Comparing these values,option $(b)$ has the highest ratio,meaning it conducts the most heat.

(B) When a particle moves in a circle with uniform speed,it covers equal distances in equal intervals of time along the circumference.
Since it returns to the same position after a fixed time interval (the time period),the motion is periodic.
However,simple harmonic motion $(SHM)$ requires a restoring force proportional to the displacement from the mean position $(F = -kx)$,which is not the case for uniform circular motion.
Therefore,the motion is periodic but not simple harmonic.

(D) The maximum speed of a particle in simple harmonic motion is given by the formula $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.
We know that $\omega = 2\pi n$,where $n$ is the frequency of oscillation.
Substituting the given values: $a = 5\, cm = 0.05\, m$ and $v_{\max} = 31.4\, cm/s = 0.314\, m/s$.
$v_{\max} = a \times 2\pi n$
$0.314 = 0.05 \times 2 \times 3.14 \times n$
$0.314 = 0.314 \times n$
$n = 1\, Hz$.

(A) The intensity $I$ of sound from a point source in a non-absorbing medium is inversely proportional to the square of the distance $r$ from the source,given by $I \propto \frac{1}{r^2}$.
Given distances are $r_P = 2 \ m$ and $r_Q = 3 \ m$.
The ratio of intensities at $P$ and $Q$ is $\frac{I_P}{I_Q} = \frac{r_Q^2}{r_P^2}$.
Substituting the values,we get $\frac{I_P}{I_Q} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
Thus,the ratio is $9:4$.

(B) The given equations for the progressive waves are $Y_1 = 4\sin(500\pi t)$ and $Y_2 = 2\sin(506\pi t)$.
Comparing these with the standard equation $Y = A\sin(\omega t)$,we get angular frequencies $\omega_1 = 500\pi \text{ rad/s}$ and $\omega_2 = 506\pi \text{ rad/s}$.
The frequencies $n_1$ and $n_2$ are calculated using $\omega = 2\pi n$:
$n_1 = \frac{500\pi}{2\pi} = 250 \text{ Hz}$
$n_2 = \frac{506\pi}{2\pi} = 253 \text{ Hz}$
The beat frequency is the difference between the two frequencies: $f_b = n_2 - n_1 = 253 - 250 = 3 \text{ beats per second}$.
To find the number of beats per minute,we multiply the beats per second by $60$:
$\text{Beats per minute} = 3 \times 60 = 180$.

(D) The rotational kinetic energy $(KE)$ is related to the angular momentum $(L)$ and moment of inertia $(I)$ by the formula: $KE = \frac{L^2}{2I}$.
Given that the rotational kinetic energies are equal $(KE_1 = KE_2)$,we have:
$\frac{L_1^2}{2I_1} = \frac{L_2^2}{2I_2}$
Substituting the given values $I_1 = I$ and $I_2 = 2I$:
$\frac{L_1^2}{2I} = \frac{L_2^2}{2(2I)}$
$\frac{L_1^2}{I} = \frac{L_2^2}{2I}$
$\frac{L_1^2}{L_2^2} = \frac{I}{2I} = \frac{1}{2}$
Taking the square root on both sides:
$\frac{L_1}{L_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio of their angular momenta is $1:\sqrt{2}$.

(D) When a body rolls down an inclined plane without slipping,the force of static friction acts up the incline.
This frictional force opposes the translational motion of the center of mass.
Simultaneously,this force provides a torque about the center of mass that causes the body to rotate.
Thus,the frictional force acts to convert some of the translational kinetic energy into rotational kinetic energy.

(A) The moment of inertia of a uniform circular disc about an axis passing through its centre and perpendicular to its plane is $I_{G} = \frac{1}{2}MR^2$.
By the theorem of parallel axes,the moment of inertia about an axis passing through the edge and normal to the disc is given by $I = I_{G} + Md^2$,where $d = R$ is the distance between the parallel axes.
Substituting the values,we get $I = \frac{1}{2}MR^2 + M(R)^2 = \frac{3}{2}MR^2$.

(D) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3} \pi R^3$,we can write $g = \frac{G}{R^2} \times \rho \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \rho G R$.
This shows that $g \propto R$ when the density $\rho$ is constant.
Given that the new planet has the same density as Earth but is $3$ times bigger in size (radius),let $R^{\prime} = 3R$.
Therefore,$\frac{g^{\prime}}{g} = \frac{R^{\prime}}{R} = \frac{3R}{R} = 3$.
Thus,$g^{\prime} = 3g$.

(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}$.
Point $A$ is at coordinates $(0, a)$,so its distance from the origin $O$ is $r_A = \sqrt{0^2 + a^2} = a$.
Point $B$ is at coordinates $(a, 0)$,so its distance from the origin $O$ is $r_B = \sqrt{a^2 + 0^2} = a$.
The potential at $A$ is $V_A = \frac{1}{4\pi \varepsilon_0} \frac{q}{a}$.
The potential at $B$ is $V_B = \frac{1}{4\pi \varepsilon_0} \frac{q}{a}$.
Since $V_A = V_B$,the potential difference $\Delta V = V_B - V_A = 0$.
The work done $W$ in moving a charge $-Q$ from $A$ to $B$ is given by $W = (-Q) \cdot \Delta V$.
Substituting $\Delta V = 0$,we get $W = (-Q) \cdot 0 = 0$.

(B) The circuit consists of two parallel branches connected to a battery of potential $V$.
Branch $1$ contains capacitors $C_1, C_2,$ and $C_3$ in series. The equivalent capacitance $C_{eq1}$ of this branch is given by:
$\frac{1}{C_{eq1}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C}$
Thus,$C_{eq1} = \frac{6C}{11}$.
The charge $Q_1$ on each capacitor in this series branch is $Q_1 = C_{eq1} V = \frac{6CV}{11}$.
Since $C_1, C_2,$ and $C_3$ are in series,the charge on $C_2$ is $Q_{C2} = Q_1 = \frac{6CV}{11}$.
Branch $2$ contains only capacitor $C_4$ connected directly across the battery. The charge on $C_4$ is $Q_{C4} = C_4 V = 4CV$.
The ratio of the charges on $C_2$ and $C_4$ is:
$\frac{Q_{C2}}{Q_{C4}} = \frac{\frac{6CV}{11}}{4CV} = \frac{6}{11 \times 4} = \frac{6}{44} = \frac{3}{22}$.

(B) The given circuit is a Wheatstone bridge. Let the nodes be $A, B, C, D$. The bridge is balanced if the ratio of resistances in opposite arms is equal. Here,the ratio of resistances is $\frac{4}{2} = 2$ and $\frac{6}{3} = 2$. Since the ratios are equal,the bridge is balanced.
In a balanced Wheatstone bridge,no current flows through the central $4\,\Omega$ resistor. Thus,it can be removed from the circuit.
Now,the circuit consists of two parallel branches: one with $(4\,\Omega + 2\,\Omega) = 6\,\Omega$ and the other with $(6\,\Omega + 3\,\Omega) = 9\,\Omega$.
The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{9} = \frac{3+2}{18} = \frac{5}{18}$.
Therefore,$R_{eq} = \frac{18}{5}\,\Omega$.
The total current $i$ from the battery is $i = \frac{V}{R_{eq}} = \frac{V}{18/5} = \frac{5V}{18}$.

(A) When a wire of resistance $R$ is bent into a circle,it is divided into two equal semicircular parts.
Each part has a resistance of $R' = \frac{R}{2}$.
These two semicircular parts are connected in parallel between the two diametrically opposite points.
The equivalent resistance $R_{eq}$ between these points is given by the parallel combination formula:
$\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{1}{R/2} + \frac{1}{R/2} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R}$.
Therefore,$R_{eq} = \frac{R}{4}$.

(C) When two cells with emf $E_1, E_2$ and internal resistances $r_1, r_2$ are connected in parallel,the equivalent emf $E_{eq}$ is given by the formula:
$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$
Given:
$E_1 = 18 \, V, r_1 = 2 \, \Omega$
$E_2 = 12 \, V, r_2 = 1 \, \Omega$
Substituting these values into the formula:
$E_{eq} = \frac{(18 \times 1) + (12 \times 2)}{2 + 1}$
$E_{eq} = \frac{18 + 24}{3}$
$E_{eq} = \frac{42}{3} = 14 \, V$
Since the voltmeter is connected across the parallel combination,it will measure the equivalent emf of the circuit.
Therefore,the reading of the voltmeter is $14 \, V$.

(D) $1$. According to the right-hand thumb rule,the magnetic field $\vec{B}$ produced by the current-carrying wire at point $P$ is directed perpendicular to the plane of the paper and inwards (into the page).
$2$. The force $\vec{F}$ on a moving charge $+Q$ in a magnetic field is given by the Lorentz force formula: $\vec{F} = Q(\vec{V} \times \vec{B})$.
$3$. Here,the velocity $\vec{V}$ is along the positive $X$-axis (to the right) and the magnetic field $\vec{B}$ is directed inwards (into the page,represented by $-\hat{k}$ direction).
$4$. Using the right-hand rule for the cross product $\vec{V} \times \vec{B}$,where $\vec{V}$ is along $\hat{i}$ and $\vec{B}$ is along $-\hat{k}$,the direction of the force is $\hat{i} \times (-\hat{k}) = -(\hat{i} \times \hat{k}) = -(-\hat{j}) = \hat{j}$.
$5$. This corresponds to the positive $Y$-axis direction,which is along $OY$.

(C) The electric current $i$ produced by an electron moving in a circular orbit of radius $r$ with speed $v$ is given by $i = \frac{ev}{2\pi r}$.
The magnetic field $B$ at the centre of the circular current loop is given by $B = \frac{\mu_0 i}{2r}$.
Substituting the value of $i$ into the expression for $B$,we get $B = \frac{\mu_0}{2r} \cdot \frac{ev}{2\pi r} = \frac{\mu_0 ev}{4\pi r^2}$.
Rearranging the formula to solve for $r^2$,we get $r^2 = \frac{\mu_0 ev}{4\pi B}$.
Taking the square root of both sides,we find $r = \sqrt{\frac{\mu_0 ev}{4\pi B}}$.
Since $\mu_0$,$e$,and $4\pi$ are constants,the radius $r$ is proportional to $\sqrt{v/B}$.

(C) In diamagnetic materials,the orbital and spin magnetic moments of electrons in an atom cancel each other out,resulting in a net magnetic dipole moment of $\mu_d = 0$ for each atom.
In paramagnetic materials,each atom possesses a permanent magnetic dipole moment due to unpaired electrons,so $\mu_p \neq 0$.
In ferromagnetic materials,each atom also possesses a permanent magnetic dipole moment,so $\mu_f \neq 0$.
Comparing the options,the statement $\mu_d = 0$ and $\mu_p \neq 0$ is correct.

(A) The phase angle $\phi$ in an $LCR$ series circuit is given by $\tan \phi = \frac{X_C - X_L}{R}$.
Since the current leads the voltage,the circuit is capacitive,and the phase angle is $\phi = -45^o$ (or we use the magnitude of the lead as $\tan(45^o) = \frac{X_C - X_L}{R}$).
Substituting the values: $\tan 45^o = \frac{\frac{1}{2\pi fC} - 2\pi fL}{R}$.
Since $\tan 45^o = 1$,we have $1 = \frac{\frac{1}{2\pi fC} - 2\pi fL}{R}$.
$R = \frac{1}{2\pi fC} - 2\pi fL$.
$\frac{1}{2\pi fC} = 2\pi fL + R$.
Therefore,$C = \frac{1}{2\pi f(2\pi fL + R)}$.

(C) The energy $E$ of the incident photon is given by the formula $E = \frac{hc}{\lambda}$.
Using the approximation $hc \approx 12400 \text{ eV} \cdot \mathring{A}$,we calculate the energy as:
$E = \frac{12400}{4100} \approx 3.02 \text{ eV}$.
Photoelectric emission occurs if the energy of the incident photon is greater than the work function $(\Phi)$ of the metal.
For metal $A$: $\Phi_A = 1.92 \text{ eV} < 3.02 \text{ eV}$ (Emission occurs).
For metal $B$: $\Phi_B = 2.0 \text{ eV} < 3.02 \text{ eV}$ (Emission occurs).
For metal $C$: $\Phi_C = 5.0 \text{ eV} > 3.02 \text{ eV}$ (No emission).
Therefore,only metals $A$ and $B$ will emit photoelectrons.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = E - W_0$,where $E$ is the energy of the incident photon and $W_0$ is the work function.
For the first case: $E_1 = 2h\nu_0$ and $W_0 = h\nu_0$.
$\frac{1}{2}mv_1^2 = 2h\nu_0 - h\nu_0 = h\nu_0$ ... $(i)$
For the second case: $E_2 = 5h\nu_0$ and $W_0 = h\nu_0$.
$\frac{1}{2}mv_2^2 = 5h\nu_0 - h\nu_0 = 4h\nu_0$ ... $(ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{4h\nu_0}{h\nu_0}$
$\left(\frac{v_2}{v_1}\right)^2 = 4$
$\frac{v_2}{v_1} = 2$
Given $v_1 = 4 \times 10^6 \, m/s$,we find:
$v_2 = 2 \times v_1 = 2 \times (4 \times 10^6 \, m/s) = 8 \times 10^6 \, m/s$.

(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,$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

(B) For an electron in a hydrogen atom,the total energy $(E)$,kinetic energy $(K)$,and potential energy $(U)$ are related as follows:
$E = -K$
$U = 2E = -2K$
Given that the total energy of the electron in the first excited state is $E = -3.4 \ eV$.
Therefore,the kinetic energy $K = -E = -(-3.4 \ eV) = +3.4 \ eV$.
Thus,the correct option is $B$.

(B) The energy released in a nuclear reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactants.
In the given reaction: $_1^2H + _1^3H \to _2^4He + _0^1n$.
The binding energy of the neutron $_0^1n$ is $0 \ MeV$.
The total binding energy of the reactants is $a + b$.
The total binding energy of the products is $c + 0 = c$.
Therefore, the energy released $Q = (\text{Total Binding Energy of Products}) - (\text{Total Binding Energy of Reactants})$.
$Q = c - (a + b) = c - a - b \ MeV$.

(A) Isotones are nuclei that have the same number of neutrons $(N)$ but different atomic numbers $(Z)$.
Neutron number is calculated as $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For option $A$: $_{34}Se^{74} \implies N = 74 - 34 = 40$ and $_{31}Ga^{71} \implies N = 71 - 31 = 40$.
Since both nuclei have $40$ neutrons,they are isotones.
For option $B$: $_{42}Mo^{92} \implies N = 92 - 42 = 50$ and $_{40}Zr^{92} \implies N = 92 - 40 = 52$. Not isotones.
For option $C$: These are isotopes (same $Z$,different $A$).
For option $D$: $_{20}Ca^{40} \implies N = 20$ and $_{16}S^{32} \implies N = 16$. Not isotones.
Therefore,the correct option is $A$.

(A) In a nuclear fission process,a heavy nucleus splits into two or more lighter nuclei.
According to Einstein's mass-energy equivalence principle,$E = mc^2$,the energy released during the process is derived from the mass defect.
Since energy is released,the total mass of the fission products must be less than the mass of the parent nucleus.
Therefore,the ratio $\frac{\text{mass of fission products}}{\text{mass of parent nucleus}} < 1$.

(A) The energy band gap $(E_g)$ is the energy difference between the valence band and the conduction band.
For carbon (diamond), the energy band gap is approximately $5.4 \text{ eV}$.
For silicon, the energy band gap is approximately $1.1 \text{ eV}$.
For germanium, the energy band gap is approximately $0.7 \text{ eV}$.
Comparing these values, we find that $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$.
Therefore, the correct relationship is $(E_g)_C > (E_g)_{Si}$.

(D) The angular resolution of a telescope is given by the formula: $d\theta = \frac{1.22 \lambda}{D}$.
Here,the wavelength $\lambda = 5000 \;\mathring A = 5000 \times 10^{-10} \;m = 5 \times 10^{-7} \;m$.
The diameter of the telescope aperture $D = 10 \;cm = 0.1 \;m = 10^{-1} \;m$.
Substituting these values into the formula:
$d\theta = \frac{1.22 \times 5 \times 10^{-7}}{10^{-1}}$
$d\theta = 1.22 \times 5 \times 10^{-6}$
$d\theta = 6.1 \times 10^{-6} \;rad$.
This value is of the order of $10^{-6} \;rad$.

(C) The electromagnetic spectrum in increasing order of frequency (or decreasing order of wavelength) is: Gamma rays > $X$-rays > Ultraviolet > Visible light > Infrared > Microwaves > Radio waves.
Therefore,the order of wavelengths is: $\lambda_{\text{microwaves}} > \lambda_{\text{visible}} > \lambda_{X\text{-rays}}$.
Given the symbols,this corresponds to: ${\lambda _m} > {\lambda _v} > {\lambda _x}$.

(A) When a $P-N$ junction is forward biased,the positive terminal of the external battery is connected to the $P$-side and the negative terminal to the $N$-side.
This setup opposes the internal electric field of the depletion region.
As a result,the potential barrier height decreases,and the width of the depletion zone decreases.
Since the electric field $E$ is related to the potential barrier $V_B$ and width $W$ by $E = V_B / W$,the reduction in the potential barrier and the narrowing of the depletion region lead to a decrease in the electric field within the depletion zone.

(B) Zener diode is specifically designed to operate in the reverse breakdown region. It is primarily used as a voltage regulator to maintain a constant output voltage across a load,which is known as voltage stabilisation. In contrast,a standard $p-n$ junction diode is typically used for rectification.

(A) The induced $e.m.f.$ $V$ in a closed loop is defined as the work done per unit charge by the induced electric field in moving a charge once around the loop.
By definition,$V = \frac{W}{Q}$,where $W$ is the work done and $Q$ is the charge.
Therefore,the work done in moving a charge $Q$ once along the loop is $W = QV$.
Since the induced electric field is non-conservative,the work done is non-zero and is equal to the product of the charge and the induced $e.m.f.$

(B) The torque $\tau$ on a current-carrying coil in a magnetic field is given by $\tau = \vec{m} \times \vec{B}$,where $\vec{m} = i \vec{A}$ is the magnetic moment.
Since $\vec{B}$ is in the plane of the coil,the angle between the area vector $\vec{A}$ and the magnetic field $\vec{B}$ is $90^{\circ}$.
Thus,the magnitude of the torque is $\tau = i A B \sin 90^{\circ} = i A B$.
The area $A$ of an equilateral triangle with side $l$ is $A = \frac{\sqrt{3}}{4} l^2$.
Substituting this into the torque equation: $\tau = i \left( \frac{\sqrt{3}}{4} l^2 \right) B$.
Rearranging for $l^2$: $l^2 = \frac{4 \tau}{\sqrt{3} i B}$.
Taking the square root: $l = \sqrt{\frac{4 \tau}{\sqrt{3} i B}} = 2 \left( \frac{\tau}{\sqrt{3} i B} \right)^{1/2}$.

(C) The power dissipated in a resistor is given by the formula $P = I^2 R$,where $P$ is the power,$I$ is the current,and $R$ is the resistance.
Given:
$P = 1\;W$
$I = 5\;A$
Substituting these values into the formula:
$1 = (5)^2 \times R$
$1 = 25 \times R$
$R = \frac{1}{25}\;\Omega$
$R = 0.04\;\Omega$
Therefore,the resistance of the fuse wire is $0.04\;\Omega$.

(B) Nuclear fission is a process in which a heavy nucleus splits into two lighter nuclei.
According to the binding energy per nucleon curve,for heavy nuclei (high mass number $A$),the binding energy per nucleon decreases as the mass number increases.
This means that heavy nuclei are less stable compared to intermediate-mass nuclei.
When a heavy nucleus splits into two lighter nuclei,the binding energy per nucleon of the resulting products is higher than that of the parent nucleus.
This increase in binding energy per nucleon results in the release of energy,making the fission process possible.
Therefore,the correct reason is that the binding energy per nucleon decreases with mass number at high mass numbers.

(C) In semiconductors, as the temperature increases, more charge carriers (electrons and holes) are generated due to thermal excitation. This leads to an increase in conductivity. Since conductivity $(\sigma)$ is the reciprocal of resistivity $(\rho)$, an increase in conductivity implies a decrease in resistivity. Therefore, the statement that resistivity increases with temperature is false.

(A) The potential at point $C$ due to charges $q_1$ and $q_2$ is:
$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AC} + \frac{q_2}{BC} \right)$
Given $AC = 40\ cm = 0.4\ m$. Since $AB = 30\ cm$ and $AC = 40\ cm$,by Pythagoras theorem,$BC = \sqrt{0.3^2 + 0.4^2} = 0.5\ m$.
So,$V_C = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right)$.
The potential at point $D$ due to charges $q_1$ and $q_2$ is:
$V_D = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{AD} + \frac{q_2}{BD} \right)$
Given $AD = 40\ cm = 0.4\ m$ (radius of the circular path) and $BD = AD - AB = 40\ cm - 30\ cm = 10\ cm = 0.1\ m$.
So,$V_D = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right)$.
The change in potential energy $\Delta U$ is given by:
$\Delta U = q_3 (V_D - V_C) = q_3 \left[ \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right) - \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right) \right]$
$\Delta U = \frac{q_3}{4 \pi \epsilon_0} \left( \frac{q_2}{0.1} - \frac{q_2}{0.5} \right) = \frac{q_3}{4 \pi \epsilon_0} (10 q_2 - 2 q_2) = \frac{8 q_2 q_3}{4 \pi \epsilon_0}$.
Comparing this with $\frac{q_3 K}{4 \pi \epsilon_0}$,we get $K = 8 q_2$.