WBJEE 2012 Physics Question Paper with Answer and Solution

(A) According to the law of conservation of linear momentum,the net external force on the system is zero,so the momentum along the $y$-axis (perpendicular to the initial direction of motion) must be conserved.
Initially,the system has no momentum along the $y$-axis.
After the collision,the components of momentum along the $y$-axis for balls $A$ and $B$ must be equal and opposite.
Let $m_1 = 4 \ kg$ and $m_2 = 1 \ kg$.
$m_1 v_1 \sin(30^{\circ}) = m_2 v_2 \sin(60^{\circ})$
$4 \cdot v_1 \cdot \frac{1}{2} = 1 \cdot v_2 \cdot \frac{\sqrt{3}}{2}$
$2 v_1 = \frac{\sqrt{3}}{2} v_2$
$\frac{v_1}{v_2} = \frac{\sqrt{3}}{4}$

(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ of a planet is directly proportional to the cube of its average distance $(R)$ from the Sun.
$T^{2} \propto R^{3}$
Let $T_{1} = D$ be the time period of the Earth at distance $L_{1}$,and $T_{2}$ be the time period of the other planet at distance $L_{2}$.
Then,$\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{L_{2}^{3}}{L_{1}^{3}}$
$\frac{T_{2}^{2}}{D^{2}} = \left(\frac{L_{2}}{L_{1}}\right)^{3}$
$T_{2}^{2} = D^{2} \left(\frac{L_{2}}{L_{1}}\right)^{3}$
Taking the square root on both sides:
$T_{2} = D \left(\frac{L_{2}}{L_{1}}\right)^{3/2} \text{ days}$.

(D) Given: Mass of the box $m = 2 kg$,coefficient of static friction $\mu = 0.2$,and acceleration due to gravity $g = 10 ms^{-2}$.
For the box to remain stationary on the roof of the car,the pseudo-force acting on the box must be balanced by the static frictional force.
The maximum static frictional force is given by $f_{max} = \mu N = \mu mg$.
The force required to accelerate the box with the car is $F = ma$.
Equating the two,we get $ma = \mu mg$.
Therefore,$a = \mu g$.
Substituting the values: $a = 0.2 \times 10 = 2 ms^{-2}$.
Thus,the maximum acceleration of the car is $2 ms^{-2}$.

(B) The total mass of the system is $M = 4 \ kg + 2 \ kg + 1 \ kg = 7 \ kg$.
Since the surface is frictionless,the acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{14 \ N}{7 \ kg} = 2 \ m/s^2$.
To find the contact force $N$ between the $4 \ kg$ and $2 \ kg$ blocks,we consider the motion of the combined $2 \ kg$ and $1 \ kg$ blocks (total mass $m' = 3 \ kg$).
The force $N$ is the only horizontal force acting on this combined system of $3 \ kg$ that causes it to accelerate at $2 \ m/s^2$.
Therefore,$N = m' \times a = 3 \ kg \times 2 \ m/s^2 = 6 \ N$.

(D) Let $m$ be the mass of the body in grams and $V$ be the volume of the body in $\text{cm}^3$. The density of water is $\rho_w = 1 \text{ g/cm}^3$. The apparent weight $W'$ is given by $W' = W_{actual} - F_B$,where $F_B$ is the buoyant force.
For the liquid with specific gravity $1.2$,the density is $\rho_l = 1.2 \text{ g/cm}^3$. The weight is $44 \text{ gwt}$,so:
$44 = m - 1.2V$ $(i)$
For water,the weight is $50 \text{ gwt}$,so:
$50 = m - V$ $(ii)$
Subtracting $(i)$ from $(ii)$:
$(50 - 44) = (m - V) - (m - 1.2V)$
$6 = 0.2V$
$V = 30 \text{ cm}^3$
Substituting $V = 30$ into $(ii)$:
$50 = m - 30$
$m = 80 \text{ g}$

(C) The velocity of a fluid,such as water,flowing through a tube,below which the flow remains streamline (or laminar) and above which it becomes turbulent,is defined as the critical velocity.

(C) The extension $\Delta L$ of a wire is given by the formula $\Delta L = \frac{F \cdot L}{A \cdot Y}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since $A = \pi \cdot (d/2)^2 = \frac{\pi d^2}{4}$,we can write $\Delta L = \frac{4 F L}{\pi d^2 Y}$.
Given that $F$ and $Y$ are constant for the same material and same tension,we have $\Delta L \propto \frac{L}{d^2}$.
Calculating the ratio $\frac{L}{d^2}$ for each case:
$A: \frac{200}{(0.5)^2} = \frac{200}{0.25} = 800$
$B: \frac{300}{(1.0)^2} = \frac{300}{1} = 300$
$C: \frac{50}{(0.05)^2} = \frac{50}{0.0025} = 20000$
$D: \frac{100}{(0.2)^2} = \frac{100}{0.04} = 2500$
Comparing these values,the ratio is maximum for option $C$.

(C) At the ground,the initial kinetic energy is given by $E = \frac{1}{2} m u^{2}$,where $u$ is the initial velocity.
At the highest point of the trajectory,the vertical component of the velocity becomes zero,and the velocity of the particle is only the horizontal component,which is $v_x = u \cos \theta$.
Given the angle of projection $\theta = 60^{\circ}$,the horizontal velocity at the highest point is $v_x = u \cos 60^{\circ} = u \times \frac{1}{2} = \frac{u}{2}$.
The kinetic energy at the highest point $E^{\prime}$ is given by $E^{\prime} = \frac{1}{2} m v_x^{2}$.
Substituting the value of $v_x$,we get $E^{\prime} = \frac{1}{2} m \left( \frac{u}{2} \right)^{2} = \frac{1}{2} m \left( \frac{u^{2}}{4} \right) = \frac{1}{4} \left( \frac{1}{2} m u^{2} \right)$.
Since $E = \frac{1}{2} m u^{2}$,we have $E^{\prime} = \frac{E}{4}$.

(C) Given: Height $h = 80 \ m$,initial horizontal velocity $v = 8 \ ms^{-1}$,and acceleration due to gravity $g = 10 \ ms^{-2}$.
For vertical motion,the time taken to reach the ground is given by the equation $h = \frac{1}{2}gt^2$.
Substituting the values: $80 = \frac{1}{2} \times 10 \times t^2$.
$80 = 5t^2 \implies t^2 = 16 \implies t = 4 \ s$.
For horizontal motion,the distance $d$ covered is given by $d = v \times t$.
Substituting the values: $d = 8 \times 4 = 32 \ m$.
Thus,the time $t$ is $4 \ s$ and the distance $d$ is $32 \ m$.

(B) Given equations for simple harmonic motion are:
$x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$
$x_{2}=A \cos (\omega t)$
To find the phase difference,we must express both equations in the same trigonometric function (sine).
Using the identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$,we can rewrite $x_{2}$ as:
$x_{2}=A \sin \left(\omega t+\frac{\pi}{2}\right)$
Now,the phase of the first motion is $\phi_{1} = \omega t + \frac{\pi}{6}$ and the phase of the second motion is $\phi_{2} = \omega t + \frac{\pi}{2}$.
The phase difference $\Delta \phi$ is given by:
$\Delta \phi = \phi_{2} - \phi_{1}$
$\Delta \phi = \left(\omega t + \frac{\pi}{2}\right) - \left(\omega t + \frac{\pi}{6}\right)$
$\Delta \phi = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$
Therefore,the phase difference is $\frac{\pi}{3}$.

(C) $1$. Heat required to melt $100 \ g$ of ice at $0^{\circ} C$ to water at $0^{\circ} C$: $Q_1 = m \times L_f = 100 \ g \times 80 \ cal/g = 8000 \ cal$.
$2$. Heat required to raise the temperature of $100 \ g$ of water from $0^{\circ} C$ to $100^{\circ} C$: $Q_2 = m \times s \times \Delta T = 100 \ g \times 1 \ cal/g^{\circ} C \times 100^{\circ} C = 10000 \ cal$.
$3$. Total heat used so far: $Q_{total} = Q_1 + Q_2 = 8000 + 10000 = 18000 \ cal$.
$4$. Remaining heat: $Q_{rem} = 22320 \ cal - 18000 \ cal = 4320 \ cal$.
$5$. This remaining heat will convert some water at $100^{\circ} C$ into steam: $m_{steam} = Q_{rem} / L_v = 4320 \ cal / 540 \ cal/g = 8 \ g$.
$6$. Final amount of water remaining: $100 \ g - 8 \ g = 92 \ g$ of water at $100^{\circ} C$.

(D) The relationship between any temperature scale and the Celsius scale is given by the formula: $\frac{X - \text{Lower Fixed Point}}{\text{Upper Fixed Point} - \text{Lower Fixed Point}} = \frac{C}{100}$.
Here, the lower fixed point is $10^{\circ}$ and the upper fixed point is $130^{\circ}$.
Substituting the given values: $\frac{X - 10}{130 - 10} = \frac{40}{100}$.
$\frac{X - 10}{120} = \frac{40}{100}$.
$X - 10 = \frac{40}{100} \times 120$.
$X - 10 = 0.4 \times 120 = 48$.
$X = 48 + 10 = 58^{\circ}$.

(C) According to the Doppler effect,when a source of sound moves towards a stationary observer,the apparent frequency $f'$ is given by:
$f' = f \left( \frac{v}{v - v_s} \right)$
Where:
$f = 640 \ Hz$ (source frequency)
$v = 340 \ ms^{-1}$ (speed of sound)
$v_s = 20 \ ms^{-1}$ (speed of the source/train)
Substituting the values:
$f' = 640 \left( \frac{340}{340 - 20} \right)$
$f' = 640 \left( \frac{340}{320} \right)$
$f' = 640 \times 1.0625 = 680 \ Hz$
Therefore,the frequency heard by the person on the platform is $680 \ Hz$.

(D) For a closed pipe,the frequency of the $n$-th overtone is given by $f_{c} = \frac{(2n+1)v}{4l_{1}}$,where $n$ is the overtone number. For the first overtone $(n=1)$,$f_{c} = \frac{3v}{4l_{1}}$.
For an open pipe,the frequency of the $n$-th overtone is given by $f_{o} = \frac{(n+1)v}{2l_{2}}$,where $n$ is the overtone number. For the first overtone $(n=1)$,$f_{o} = \frac{2v}{2l_{2}} = \frac{v}{l_{2}}$.
Given that the frequencies are equal: $\frac{3v}{4l_{1}} = \frac{v}{l_{2}}$.
Rearranging for the ratio: $\frac{l_{1}}{l_{2}} = \frac{3}{4}$.

(A) The given wave equation is $y = A \sin \left[ \frac{2 \pi}{\lambda} (vt - x) \right]$.
Comparing this with the standard form $y = A \sin (\omega t - kx)$,we have $\omega = \frac{2 \pi v}{\lambda}$ and $k = \frac{2 \pi}{\lambda}$.
The particle velocity $v_p$ is given by $\frac{\partial y}{\partial t} = A \omega \cos \left[ \frac{2 \pi}{\lambda} (vt - x) \right]$.
The maximum particle velocity is $(v_p)_{\max} = A \omega = A \left( \frac{2 \pi v}{\lambda} \right)$.
According to the problem,$(v_p)_{\max} = 3v$.
Therefore,$A \left( \frac{2 \pi v}{\lambda} \right) = 3v$.
Canceling $v$ from both sides,we get $\frac{2 \pi A}{\lambda} = 3$.
Solving for $\lambda$,we find $\lambda = \frac{2 \pi A}{3}$.

(B) The potential energy $U$ stored in a spring stretched by a distance $x$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the spring constant.
In the first case,the extension is $x_1 = 10 \ cm = 0.1 \ m$. Thus,$E = \frac{1}{2} k (0.1)^2 = 0.005 k$.
In the second case,the spring is stretched by $10 \ cm$ more,so the total extension is $x_2 = 10 \ cm + 10 \ cm = 20 \ cm = 0.2 \ m$.
The new potential energy $E'$ is $E' = \frac{1}{2} k (0.2)^2 = \frac{1}{2} k (4 \times 0.01) = 4 \times (\frac{1}{2} k (0.1)^2)$.
Therefore,$E' = 4 E$.

(D) According to Moseley's law for characteristic $X$-rays,the frequency $v$ of the emitted $X$-rays is related to the atomic number $Z$ as $v \propto (Z - b)^2$,where $b$ is a screening constant. For the $K_{\alpha}$ line,$b = 1$.
Assuming the same series for both elements,we have $v \propto Z^2$.
Given the ratio of atomic numbers $Z_{A} : Z_{B} = 1 : 2$.
Therefore,the ratio of frequencies is:
$\frac{v_{A}}{v_{B}} = \left( \frac{Z_{A}}{Z_{B}} \right)^2$
$\frac{v_{A}}{v_{B}} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$
Thus,$v_{A} : v_{B} = 1 : 4$.

(D) Let the potential difference across the capacitor $C$ (connected between $X$ and $Y$) be $V_{XY} = 60 \ V$.
The circuit consists of a capacitor $C$ in parallel with a series combination of capacitors $2C$,$C$,and $2C$.
Since the capacitors $2C$,$C$,and $2C$ are in series,the same charge $q$ flows through them.
The potential difference across the series branch is $V_{XY} = 60 \ V$.
The equivalent capacitance of the series branch is given by:
$\frac{1}{C_{eq}} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{2C} = \frac{1+2+1}{2C} = \frac{4}{2C} = \frac{2}{C}$
So,$C_{eq} = \frac{C}{2}$.
The charge $q$ on each capacitor in the series branch is:
$q = C_{eq} \times V_{XY} = \frac{C}{2} \times 60 \ V = 30C$.
The potential difference between points $M$ and $N$ is the potential difference across the capacitor $C$ located between them.
$V_{MN} = \frac{q}{C} = \frac{30C}{C} = 30 \ V$.

(A) The rate of heat production (power) in a resistor is given by the formula $H = \frac{V^2}{R}$.
In the first case,the power is $H = \frac{V^2}{R}$.
In the second case,the new voltage is $V' = \frac{V}{3}$ and the new resistance is $R' = 2R$.
The new rate of heat production $H'$ is given by:
$H' = \frac{(V')^2}{R'} = \frac{(\frac{V}{3})^2}{2R} = \frac{\frac{V^2}{9}}{2R} = \frac{V^2}{18R}$.
Since $H = \frac{V^2}{R}$,we can substitute this into the equation:
$H' = \frac{H}{18}$.

(B) For a metal wire,the resistance $R$ increases with an increase in temperature.
The slope of the $I-V$ graph is given by $\frac{I}{V} = \frac{1}{R}$.
From the figure,the slope of the line corresponding to $T_{1}$ is greater than the slope of the line corresponding to $T_{2}$.
Therefore,$\frac{1}{R_{1}} > \frac{1}{R_{2}}$,which implies $R_{1} < R_{2}$.
Since resistance increases with temperature,$R_{1} < R_{2}$ implies $T_{1} < T_{2}$.

(C) The resistance $R$ of the bulb is constant and is determined by its rated values:
$R = \frac{V_{rated}^2}{P_{rated}} = \frac{200^2}{50} = \frac{40000}{50} = 800 \, \Omega$
When connected to a new supply voltage $V' = 100 V$,the new power $P'$ consumed by the bulb is given by:
$P' = \frac{(V')^2}{R} = \frac{100^2}{800} = \frac{10000}{800} = 12.5 \, W$
Therefore,the present power of the bulb is $12.5 \, W$.

(A) First,find the equivalent resistance of the circuit. The $4 k\Omega$ and $2 k\Omega$ resistors are in series,so their equivalent resistance is $R_s = 4 k\Omega + 2 k\Omega = 6 k\Omega$.
This $6 k\Omega$ equivalent resistance is in parallel with the $3 k\Omega$ resistor. Their equivalent resistance $R_p$ is given by:
$\frac{1}{R_p} = \frac{1}{6 k\Omega} + \frac{1}{3 k\Omega} = \frac{1+2}{6 k\Omega} = \frac{3}{6 k\Omega} = \frac{1}{2 k\Omega}$
So,$R_p = 2 k\Omega$.
Now,the total resistance of the circuit is $R_{total} = 6 k\Omega + R_p = 6 k\Omega + 2 k\Omega = 8 k\Omega$.
The total current $I$ from the $72 V$ battery is:
$I = \frac{V}{R_{total}} = \frac{72 V}{8 k\Omega} = 9 mA$.
This current $I = 9 mA$ flows through the $6 k\Omega$ resistor and then splits at the junction into the $3 k\Omega$ branch and the branch containing the $4 k\Omega$ and $2 k\Omega$ resistors in series.
Let $i$ be the current through the $2 k\Omega$ resistor. The voltage across the parallel branches is the same:
$i \times (4 k\Omega + 2 k\Omega) = (I - i) \times 3 k\Omega$
$i \times 6 k\Omega = (9 mA - i) \times 3 k\Omega$
$2i = 9 mA - i$
$3i = 9 mA$
$i = 3 mA$.

(D) In a balanced Wheatstone bridge,the ratio of resistances in adjacent arms must be equal,i.e.,$\frac{P}{Q} = \frac{R}{S}$.
Let the four arms be $P = 10 \Omega$,$Q = 10 \Omega$,$R = 10 \Omega$,and $S = 30 \Omega$.
To balance the bridge,the effective resistance $S'$ in the fourth arm must satisfy $\frac{10}{10} = \frac{10}{S'}$,which gives $S' = 10 \Omega$.
Let a resistance $x$ be connected in parallel to the $30 \Omega$ resistor to make the equivalent resistance $10 \Omega$.
The formula for parallel resistance is $\frac{1}{S'} = \frac{1}{30} + \frac{1}{x}$.
Substituting the values: $\frac{1}{10} = \frac{1}{30} + \frac{1}{x}$.
$\frac{1}{x} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$.
Therefore,$x = 15 \Omega$.

(A) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{m_e v_e}$,where $v_e = c/2$. So,$\lambda_e = \frac{h}{m_e (c/2)} = \frac{2h}{m_e c}$.
For a photon,the wavelength is $\lambda_p = \frac{h}{p_p} = \frac{h}{E_p/c} = \frac{hc}{E_p}$,where $E_p$ is the energy of the photon.
Given $\lambda_e = \lambda_p$,we have $\frac{2h}{m_e c} = \frac{hc}{E_p}$.
This implies $E_p = \frac{m_e c^2}{2}$.
The kinetic energy of the electron is $K_e = \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_e (c/2)^2 = \frac{1}{8} m_e c^2$.
The ratio of kinetic energies is $\frac{K_e}{E_p} = \frac{\frac{1}{8} m_e c^2}{\frac{1}{2} m_e c^2} = \frac{1}{4}$.

(B) According to Einstein's photoelectric equation: $h\nu = h\nu_{0} + eV_{0}$,where $\nu_{0}$ is the threshold frequency.
For the first case:
$h\nu = h\nu_{0} + eV_{0}$ --- $(i)$
For the second case:
$h(\frac{\nu}{2}) = h\nu_{0} + e(\frac{V_{0}}{4})$ --- (ii)
From equation $(i)$,we have $eV_{0} = h\nu - h\nu_{0}$.
Substitute this into equation (ii):
$\frac{h\nu}{2} = h\nu_{0} + \frac{1}{4}(h\nu - h\nu_{0})$
Multiply the entire equation by $4$ to clear the denominator:
$2h\nu = 4h\nu_{0} + h\nu - h\nu_{0}$
$2h\nu = 3h\nu_{0} + h\nu$
$h\nu = 3h\nu_{0}$
$\nu_{0} = \frac{\nu}{3}$

(A) Given the magnetic flux $\phi = 4t^2 + 6t + 9 \text{ Wb}$.
According to Faraday's law of electromagnetic induction, the induced emf $\varepsilon$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Taking the magnitude, $\varepsilon = \left| \frac{d\phi}{dt} \right|$.
Differentiating $\phi$ with respect to $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(4t^2 + 6t + 9) = 8t + 6$.
At $t = 2 \text{ s}$, the induced emf is:
$\varepsilon = 8(2) + 6 = 16 + 6 = 22 \text{ V}$.

(B) The electric field due to an infinite plane sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_{0}}$.
For the positively charged plane $(+\sigma)$,the electric field $E_{+}$ points away from the plane.
For the negatively charged plane $(-\sigma)$,the electric field $E_{-}$ points towards the plane.
Between the two planes,both fields point in the same direction,i.e.,from the positive plane to the negative plane.
Therefore,the net electric field $E_{net} = E_{+} + E_{-} = \frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}} = \frac{\sigma}{\varepsilon_{0}}$.
The direction is from the positive plane towards the negative plane.

(D) Given,electric field intensity $E = (2i + 3j + k) \ NC^{-1}$.
Area vector $S = 10i \ m^2$.
The electric flux $\phi$ through a surface is defined by the dot product of the electric field vector and the area vector:
$\phi = E \cdot S$
Substituting the given values:
$\phi = (2i + 3j + k) \cdot (10i)$
Since $i \cdot i = 1$,$j \cdot i = 0$,and $k \cdot i = 0$:
$\phi = (2 \times 10) + (3 \times 0) + (1 \times 0)$
$\phi = 20 \ Nm^2 C^{-1}$.

(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{q}{4 \pi \varepsilon_{0} r}$.
Since the electric field is conservative,the work done in moving a charge $Q$ from point $A$ to point $B$ is independent of the path and is given by $W = Q(V_B - V_A)$.
The potential at point $A(a, 0)$ is $V_A = \frac{q}{4 \pi \varepsilon_{0} a}$.
The potential at point $B(0, b)$ is $V_B = \frac{q}{4 \pi \varepsilon_{0} b}$.
Therefore,the work done is $W = Q \left( \frac{q}{4 \pi \varepsilon_{0} b} - \frac{q}{4 \pi \varepsilon_{0} a} \right)$.
$W = \frac{q Q}{4 \pi \varepsilon_{0}} \left( \frac{1}{b} - \frac{1}{a} \right) = \frac{q Q}{4 \pi \varepsilon_{0}} \left( \frac{a - b}{a b} \right)$.

(D) The force $F$ on a current-carrying wire in a magnetic field is given by the formula $F = I L B \sin \theta$.
Given values are:
Current $I = 10 \ A$
Length $L = 2 \ m$
Magnetic field $B = 0.15 \ T$
Angle $\theta = 45^{\circ}$
Substituting these values into the formula:
$F = 10 \times 2 \times 0.15 \times \sin 45^{\circ}$
$F = 3 \times \frac{1}{\sqrt{2}}$
$F = \frac{3}{\sqrt{2}} \ N$.

(B) The work done in rotating a magnetic dipole in a uniform magnetic field is given by $W = MB(1 - \cos \theta)$.
Given $\theta = 60^{\circ}$,we have:
$W = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = \frac{MB}{2}$.
Therefore,$MB = 2W$.
The torque $\tau$ on the magnetic needle is given by $\tau = MB \sin \theta$.
Substituting the values,$\tau = (2W) \sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3} W$.

(B) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For substance $A$,$N_A = N_0 e^{-5 \lambda t}$.
For substance $B$,$N_B = N_0 e^{-\lambda t}$.
Given the ratio $\frac{N_A}{N_B} = (1/e)^2 = e^{-2}$.
Substituting the expressions:
$\frac{N_0 e^{-5 \lambda t}}{N_0 e^{-\lambda t}} = e^{-2}$
$e^{-5 \lambda t + \lambda t} = e^{-2}$
$e^{-4 \lambda t} = e^{-2}$
Equating the exponents:
$-4 \lambda t = -2$
$t = \frac{2}{4 \lambda} = \frac{1}{2 \lambda}$.

(D) For a concave mirror, the magnification $m$ for a real image is negative, so $m = -3$ and $m = -2$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and magnification $m = -\frac{v}{u}$, we have $v = -mu$.
Case $1$: $u_1 = -x$, $m_1 = -3$, so $v_1 = -(-3)(-x) = -3x$.
$\frac{1}{-3x} + \frac{1}{-x} = \frac{1}{f} \implies \frac{-1-3}{3x} = \frac{1}{f} \implies \frac{1}{f} = \frac{-4}{3x} \quad (i)$
Case $2$: $u_2 = -(x+5)$, $m_2 = -2$, so $v_2 = -(-2)(-(x+5)) = -2(x+5)$.
$\frac{1}{-2(x+5)} + \frac{1}{-(x+5)} = \frac{1}{f} \implies \frac{-1-2}{2(x+5)} = \frac{1}{f} \implies \frac{1}{f} = \frac{-3}{2(x+5)} \quad (ii)$
Equating $(i)$ and $(ii)$:
$\frac{-4}{3x} = \frac{-3}{2(x+5)} \implies 8(x+5) = 9x \implies 8x + 40 = 9x \implies x = 40 \ cm$.
Substituting $x = 40$ into $(i)$:
$\frac{1}{f} = \frac{-4}{3(40)} = \frac{-4}{120} = \frac{-1}{30} \implies f = -30 \ cm$.
The magnitude of the focal length is $30 \ cm$.

(A) Given: Velocity of light in vacuum $(c)$ = $3 \times 10^{8} \text{ m/s}$,thickness of glass plate $(d)$ = $10 \text{ cm} = 0.1 \text{ m}$,and refractive index $(\mu)$ = $1.5$.
The velocity of light in the glass plate $(v)$ is given by $v = \frac{c}{\mu} = \frac{3 \times 10^{8}}{1.5} = 2 \times 10^{8} \text{ m/s}$.
The time taken $(t)$ to travel through the glass plate is $t = \frac{d}{v}$.
Substituting the values: $t = \frac{0.1 \text{ m}}{2 \times 10^{8} \text{ m/s}} = 0.05 \times 10^{-8} \text{ s} = 0.5 \times 10^{-9} \text{ s}$.
Since $1 \text{ nanosecond} = 10^{-9} \text{ s}$,the time taken is $0.5 \text{ ns}$.