AIEEE 2007 Physics Question Paper with Answer and Solution

(C) We know that velocity $v = \frac{dx}{dt}$,which implies $dx = v \, dt$.
Integrating both sides with initial condition $x = 0$ at $t = 0$:
$x = \int_{0}^{t} v \, dt = \int_{0}^{t} (v_0 + gt + ft^2) \, dt$.
Performing the integration:
$x = \left[ v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3} \right]_{0}^{t} = v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3}$.
To find the displacement at unit time $(t = 1)$,substitute $t = 1$ into the expression:
$x = v_0(1) + \frac{g(1)^2}{2} + \frac{f(1)^3}{3} = v_0 + \frac{g}{2} + \frac{f}{3}$.

(D) Let the acceleration of the system be $a$. Since the blocks are connected by a spring and move together on a smooth horizontal surface,both blocks will have the same acceleration $a$.
For the system of two blocks,the net force is $F$. According to Newton's second law,$F = (m + M)a$,which gives $a = \frac{F}{m + M}$.
The force acting on the block of mass $m$ is the spring force $T$. Applying Newton's second law to the block of mass $m$,we get $T = ma$.
Substituting the value of $a$,we get $T = m \left( \frac{F}{m + M} \right) = \frac{mF}{m + M}$.

(A) Let the block compress the spring by $x$ meters before coming to rest.
According to the work-energy theorem,the initial kinetic energy of the block is equal to the sum of the potential energy stored in the spring and the work done against friction.
Initial Kinetic Energy $(K_i)$ = $\frac{1}{2} m v^2 = \frac{1}{2} \times 2 \times (4)^2 = 16 \ J$.
Work done against friction $(W_f)$ = $f_k \times x = 15x$.
Potential energy of the spring $(U_s)$ = $\frac{1}{2} k x^2 = \frac{1}{2} \times 10,000 \times x^2 = 5,000x^2$.
Applying the energy balance: $K_i = U_s + W_f$.
$16 = 5,000x^2 + 15x$.
$5,000x^2 + 15x - 16 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-15 + \sqrt{15^2 - 4(5,000)(-16)}}{2 \times 5,000} = \frac{-15 + \sqrt{225 + 320,000}}{10,000} = \frac{-15 + \sqrt{320,225}}{10,000} \approx \frac{-15 + 565.88}{10,000} \approx 0.055 \ m$.
$x = 5.5 \ cm$.

(A) In steady state,the rate of heat flow through both sections is the same.
Let $T$ be the temperature at the interface.
The rate of heat flow $H$ is given by $H = \frac{KA(T_{high} - T_{low})}{l}$.
For the first section: $H = \frac{K_1 A(T_1 - T)}{l_1}$.
For the second section: $H = \frac{K_2 A(T - T_2)}{l_2}$.
Equating the two rates: $\frac{K_1 A(T_1 - T)}{l_1} = \frac{K_2 A(T - T_2)}{l_2}$.
$\frac{K_1(T_1 - T)}{l_1} = \frac{K_2(T - T_2)}{l_2}$.
$K_1 l_2(T_1 - T) = K_2 l_1(T - T_2)$.
$K_1 l_2 T_1 - K_1 l_2 T = K_2 l_1 T - K_2 l_1 T_2$.
$K_1 l_2 T_1 + K_2 l_1 T_2 = T(K_1 l_2 + K_2 l_1)$.
$T = \frac{K_1 l_2 T_1 + K_2 l_1 T_2}{K_1 l_2 + K_2 l_1}$.

(B) For a body rolling down an inclined plane without slipping,the forces acting on it are the component of gravity $Mg \sin \theta$ down the plane and the static friction $f$ acting up the plane.
Applying Newton's second law for linear motion: $Mg \sin \theta - f = Ma$ (where $a$ is the linear acceleration).
Applying Newton's second law for rotational motion about the center of mass: $\tau = I \alpha = fR$,where $\alpha = a/R$ is the angular acceleration.
Substituting $f = I \alpha / R = I a / R^2$ into the linear equation:
$Mg \sin \theta - \frac{I a}{R^2} = Ma$
$Mg \sin \theta = Ma + \frac{I a}{R^2} = Ma(1 + \frac{I}{M R^2})$
Solving for $a$,we get: $a = \frac{g \sin \theta}{1 + \frac{I}{M R^2}}$.

(D) We know that the torque $\vec{\tau}$ is defined as the rate of change of angular momentum $\vec{L}$,given by $\vec{\tau} = \frac{d\vec{L}}{dt}$.
$A$ central force is a force that acts along the line joining the particle to the center of force (origin). Since the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear,the torque $\vec{\tau} = \vec{r} \times \vec{F}$ is equal to zero.
Since $\vec{\tau} = 0$,it follows from the relation $\frac{d\vec{L}}{dt} = 0$ that the angular momentum $\vec{L}$ remains constant over time.

(D) Let the square lamina have mass $M$ and side length $a$. The moment of inertia of a square lamina about an axis passing through its centre $O$ and perpendicular to its plane $(I_z)$ is given by $I_z = I_x + I_y$,where $I_x$ and $I_y$ are moments of inertia about the axes passing through the centre and parallel to the sides. By symmetry,$I_x = I_y = I_{EF} = I_{GH}$ (where $EF$ and $GH$ are axes passing through the midpoints of opposite sides). Thus,$I_z = 2 I_{EF}$.
Similarly,the moment of inertia about the diagonal axes $AC$ and $BD$ are equal by symmetry,so $I_{AC} = I_{BD}$. According to the perpendicular axes theorem,$I_z = I_{AC} + I_{BD} = 2 I_{AC}$.
Equating the two expressions for $I_z$,we get $2 I_{EF} = 2 I_{AC}$,which simplifies to $I_{AC} = I_{EF}$.

(D) According to $Mayer's$ relation,the difference between molar specific heats is given by $C_{p,m} - C_{v,m} = R$.
To find the specific heat per unit mass,we divide the molar specific heats by the molar mass $M$ of the gas.
For nitrogen gas $(N_2)$,the molar mass $M = 28 \ g/mol = 28 \times 10^{-3} \ kg/mol$.
Therefore,the specific heat difference is $C_p - C_v = \frac{C_{p,m}}{M} - \frac{C_{v,m}}{M} = \frac{R}{M}$.
Substituting $M = 28$,we get $C_p - C_v = \frac{R}{28}$.

(B) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is a state function and depends only on the initial and final states,not on the path taken.
For the path $iaf$:
$\Delta U = Q_{iaf} - W_{iaf}$
$\Delta U = 50 \, cal - 20 \, cal = 30 \, cal$.
Since the initial state $i$ and final state $f$ are the same for the path $ibf$,the change in internal energy $\Delta U$ remains the same.
For the path $ibf$:
$\Delta U = Q_{ibf} - W_{ibf}$
$30 \, cal = 36 \, cal - W_{ibf}$
$W_{ibf} = 36 \, cal - 30 \, cal = 6 \, cal$.

(C) The efficiency $(\eta)$ of a Carnot engine and the coefficient of performance $(\beta)$ of a refrigerator are related as:
$\beta = \frac{1 - \eta}{\eta}$
Given $\eta = 1/10$,we calculate the coefficient of performance:
$\beta = \frac{1 - 1/10}{1/10} = \frac{9/10}{1/10} = 9$
The coefficient of performance $(\beta)$ is also defined as the ratio of heat absorbed from the cold reservoir $(Q_2)$ to the work done $(W)$ on the system:
$\beta = \frac{Q_2}{W}$
Given $W = 10 \ J$ and $\beta = 9$,we have:
$9 = \frac{Q_2}{10 \ J}$
$Q_2 = 9 \times 10 \ J = 90 \ J$
Therefore,the energy absorbed from the reservoir at a lower temperature is $90 \ J$.

(A) The two springs are connected in parallel to the mass $m$.
The effective spring constant $k_{eff}$ is given by $k_{eff} = k_1 + k_2$.
The frequency of oscillation is $f = \frac{1}{2\pi} \sqrt{\frac{k_1 + k_2}{m}}$ ... $(i)$
When both $k_1$ and $k_2$ are made four times their original values,the new spring constants are $k_1' = 4k_1$ and $k_2' = 4k_2$.
The new effective spring constant is $k_{eff}' = 4k_1 + 4k_2 = 4(k_1 + k_2)$.
The new frequency of oscillation $f'$ is given by $f' = \frac{1}{2\pi} \sqrt{\frac{k_{eff}'}{m}} = \frac{1}{2\pi} \sqrt{\frac{4(k_1 + k_2)}{m}}$.
$f' = 2 \times \left( \frac{1}{2\pi} \sqrt{\frac{k_1 + k_2}{m}} \right) = 2f$.

(B) The instantaneous kinetic energy $K$ of a particle executing $S.H.M.$ is given by $K = \frac{1}{2} m v_{inst}^2 = \frac{1}{2} m \omega^2 a^2 \cos^2(\omega t + \phi)$.
At the equilibrium position,the displacement is $x = 0$,so $x = a \sin(\omega t)$,which implies $\omega t = 0$. At the extreme position,$x = a$,so $\omega t = \pi/2$.
The average kinetic energy over an interval is defined as $\langle K \rangle = \frac{1}{T'} \int_0^{T'} K dt$,where $T'$ is the time taken to move from equilibrium to the extreme position $(T' = T/4 = 1/(4v))$.
$\langle K \rangle = \frac{1}{T/4} \int_0^{T/4} \frac{1}{2} m \omega^2 a^2 \cos^2(\omega t) dt$
Using $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:
$\langle K \rangle = \frac{4}{T} \cdot \frac{1}{2} m \omega^2 a^2 \int_0^{T/4} \frac{1 + \cos(2\omega t)}{2} dt$
$\langle K \rangle = \frac{m \omega^2 a^2}{T} [t + \frac{\sin(2\omega t)}{2\omega}]_0^{T/4}$
Since $\omega = 2\pi v$ and $T = 1/v$,at $t = T/4$,$2\omega t = 2(2\pi v)(1/4v) = \pi$.
$\langle K \rangle = \frac{m \omega^2 a^2}{T} [T/4 + 0 - 0] = \frac{m \omega^2 a^2}{4} = \frac{m (2\pi v)^2 a^2}{4} = \pi^2 m a^2 v^2$.

(B) The displacement equation is given by $x = 2 \times 10^{-2} \cos(\pi t)$.
To find the velocity $v$, we differentiate the displacement with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} [2 \times 10^{-2} \cos(\pi t)] = -2 \times 10^{-2} \pi \sin(\pi t)$.
The speed is the magnitude of velocity, $|v| = |2 \times 10^{-2} \pi \sin(\pi t)|$.
For the speed to be maximum, the value of $|\sin(\pi t)|$ must be $1$.
This occurs when $\sin(\pi t) = 1$ or $\sin(\pi t) = -1$.
The first time this occurs is when $\sin(\pi t) = 1$, which corresponds to $\pi t = \frac{\pi}{2}$.
Solving for $t$, we get $t = \frac{1}{2} = 0.5 \text{ s}$.

(A) Given the displacement equation: $x = x_0 \cos(\omega t - \frac{\pi}{4})$.
Velocity $v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t - \frac{\pi}{4})$.
Acceleration $a = \frac{dv}{dt} = -x_0 \omega^2 \cos(\omega t - \frac{\pi}{4})$.
Using the trigonometric identity $\cos(\theta + \pi) = -\cos(\theta)$,we can rewrite the acceleration as:
$a = x_0 \omega^2 \cos(\omega t - \frac{\pi}{4} + \pi) = x_0 \omega^2 \cos(\omega t + \frac{3\pi}{4})$.
Comparing this with the given form $a = A \cos(\omega t + \delta)$,we find:
$A = x_0 \omega^2$ and $\delta = \frac{3\pi}{4}$.

(C) The sound level $L$ in decibels $(dB)$ is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the reference intensity.
The change in sound level is given by $\Delta L = L_1 - L_2 = 20 \ dB$.
Substituting the formula,we get $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_0} \right) - 10 \log_{10} \left( \frac{I_2}{I_0} \right) = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Given $\Delta L = 20$,we have $20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Dividing by $10$,we get $2 = \log_{10} \left( \frac{I_1}{I_2} \right)$.
Taking the antilog,we get $\frac{I_1}{I_2} = 10^2 = 100$.
Therefore,the intensity decreases by a factor of $100$.

(A) The electronic charge is a fundamental physical constant and is an intrinsic property of an electron.
It does not depend on the gravitational field or the location where the measurement is performed.
Millikan's oil drop experiment determines the value of the elementary charge $e$,which is approximately $1.602 \times 10^{-19} \ C$.
Since the charge of an electron remains constant regardless of the acceleration due to gravity ($g_E$ or $g_M$),the ratio of the electronic charge on the moon to the electronic charge on the earth is $1$.

(B) The initial kinetic energy is given by $KE = \frac{1}{2} m u^2$,where $u$ is the initial velocity.
At the highest point of the trajectory,the vertical component of velocity becomes zero,and the velocity of the ball is equal to its horizontal component.
$v_x = u \cos \theta = u \cos 60^{\circ} = u \times \frac{1}{2} = \frac{u}{2}$.
The kinetic energy at the highest point $(KE_{top})$ is given by:
$KE_{top} = \frac{1}{2} m v_x^2 = \frac{1}{2} m \left( \frac{u}{2} \right)^2$.
$KE_{top} = \frac{1}{2} m \left( \frac{u^2}{4} \right) = \frac{1}{4} \left( \frac{1}{2} m u^2 \right)$.
Since $KE = \frac{1}{2} m u^2$,we have $KE_{top} = \frac{KE}{4}$.

(B) Let the mass of the entire disc be $M$.
The mass per unit area is $\sigma = \frac{M}{\pi(2R)^2} = \frac{M}{4\pi R^2}$.
The mass of the removed circular disc of radius $R$ is $M_1 = \sigma \cdot \pi R^2 = \frac{M}{4\pi R^2} \cdot \pi R^2 = \frac{M}{4}$.
The mass of the remaining disc is $M_2 = M - M_1 = M - \frac{M}{4} = \frac{3M}{4}$.
Let the centre of the bigger disc be the origin $(0,0)$. The centre of mass of the removed disc is at $x_1 = R$ and the centre of mass of the remaining disc is at $x_2 = -\alpha R$.
Since the centre of mass of the original disc was at the origin,we have:
$M_1 x_1 + M_2 x_2 = 0$
$\frac{M}{4} \cdot R + \frac{3M}{4} \cdot (-\alpha R) = 0$
$\frac{M}{4} \cdot R = \frac{3M}{4} \cdot \alpha R$
$\alpha = \frac{1}{3}$.

(C) The distance of point $A(\sqrt{2}, \sqrt{2})$ from the origin is:
$OA = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \text{ units}$.
The distance of point $B(2, 0)$ from the origin is:
$OB = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \text{ units}$.
The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.
Since both points $A$ and $B$ are at the same distance $r = 2 \text{ units}$ from the charge $Q = 10^{-3} \mu C$ placed at the origin,the potentials at these points are equal:
$V_A = \frac{1}{4\pi\epsilon_0} \frac{Q}{OA}$ and $V_B = \frac{1}{4\pi\epsilon_0} \frac{Q}{OB}$.
Since $OA = OB = 2$,it follows that $V_A = V_B$.
Therefore,the potential difference between points $A$ and $B$ is:
$V_A - V_B = 0 \text{ V}$.

(A) The electric potential $V$ at the centre of the square is the algebraic sum of the potentials due to individual charges: $V = \sum \frac{kq}{r}$. Since the distance $r$ from each vertex to the centre is the same,$V = \frac{k}{r} (q_A + q_B + q_C + q_D)$. Interchanging the charges does not change the sum $(q_A + q_B + q_C + q_D)$,so $V$ remains unchanged.
The electric field $\vec E$ is a vector quantity. The resultant electric field at the centre is the vector sum of the fields due to individual charges. Initially,the charges are $q$ at $A, B$ and $-q$ at $C, D$. After interchanging $A$ with $D$ and $B$ with $C$,the charges at $A$ and $B$ become $-q$,and at $C$ and $D$ become $q$. This effectively reverses the direction of the resultant electric field vector at the centre. Thus,$\vec E$ changes.

(A) Given the potential function $V(x) = \frac{20}{x^2 - 4} \text{ volt}$.
The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$.
$E = -\frac{d}{dx} \left( 20(x^2 - 4)^{-1} \right) = -20 \cdot (-1)(x^2 - 4)^{-2} \cdot (2x) = \frac{40x}{(x^2 - 4)^2}$.
Now,substituting $x = 4 \mu m$ into the expression for $E$:
$E = \frac{40(4)}{(4^2 - 4)^2} = \frac{160}{(16 - 4)^2} = \frac{160}{12^2} = \frac{160}{144}$.
Simplifying the fraction by dividing both numerator and denominator by $16$,we get $E = \frac{10}{9} \text{ V/}\mu m$.
Since the result is positive,the electric field is in the $+ve \ x$ direction.

(A) Let the initial capacitance be $C = \frac{K \epsilon_0 A}{d}$. The charge on the capacitor is $Q = CV = \frac{K \epsilon_0 A V}{d}$.
When the dielectric is removed,the capacitance becomes $C_0 = \frac{\epsilon_0 A}{d} = \frac{C}{K}$.
Since the capacitor is disconnected from the battery (implied by the process of removing/reinserting),the charge $Q$ remains constant.
The initial potential energy is $U_i = \frac{Q^2}{2C}$.
When the dielectric is removed,the potential energy becomes $U_{removed} = \frac{Q^2}{2C_0} = \frac{Q^2}{2(C/K)} = K \frac{Q^2}{2C} = K U_i$.
When the dielectric is reinserted,the capacitance returns to $C$,and the potential energy returns to $U_f = \frac{Q^2}{2C} = U_i$.
The net work done by the system is the change in potential energy: $W = -(U_f - U_i) = -(U_i - U_i) = 0$.
Thus,the net work done by the system in this process is zero.

(D) We know that the resistance at temperature $t$ is given by the formula:
$R_{t} = R_{0}(1 + \alpha t)$
where $R_{t}$ is the resistance at $t\, ^\circ C$,$R_{0}$ is the resistance at $0\, ^\circ C$,and $\alpha$ is the temperature coefficient of resistance.
For $t = 50\, ^\circ C$:
$5 = R_{0}(1 + 50\alpha)$ ......$(i)$
For $t = 100\, ^\circ C$:
$6 = R_{0}(1 + 100\alpha)$ ......$(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{5}{6} = \frac{R_{0}(1 + 50\alpha)}{R_{0}(1 + 100\alpha)}$
$\frac{5}{6} = \frac{1 + 50\alpha}{1 + 100\alpha}$
Cross-multiplying:
$5(1 + 100\alpha) = 6(1 + 50\alpha)$
$5 + 500\alpha = 6 + 300\alpha$
$200\alpha = 1$
$\alpha = \frac{1}{200} = 0.005\, ^\circ C^{-1}$
Now,substitute $\alpha$ back into equation $(i)$ to find $R_{0}$:
$5 = R_{0}(1 + 50 \times 0.005)$
$5 = R_{0}(1 + 0.25)$
$5 = R_{0}(1.25)$
$R_{0} = \frac{5}{1.25} = 4\, \Omega$
Thus,the resistance at $0\, ^\circ C$ is $4\, \Omega$.

(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net current $I_{\text{enclosed}}$ passing through the surface bounded by the loop.
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$
For a thin-walled pipe carrying current $I$ along its length,any closed loop drawn inside the pipe encloses zero net current $(I_{\text{enclosed}} = 0)$.
Since the current is distributed only on the surface of the pipe,there is no current flowing through the interior region.
Therefore,$\oint \vec{B} \cdot d\vec{l} = 0$,which implies that the magnetic field $\vec{B}$ at any point inside the pipe is zero.

(B) For a long straight wire with a steady current $i$ uniformly distributed across its cross-section:
$1$. Magnetic field inside the wire at a distance $r < a$ is given by $B_{in} = \frac{\mu_0 i r}{2 \pi a^2}$.
At $r = a/2$, $B_1 = \frac{\mu_0 i (a/2)}{2 \pi a^2} = \frac{\mu_0 i}{4 \pi a}$.
$2$. Magnetic field outside the wire at a distance $r > a$ is given by $B_{out} = \frac{\mu_0 i}{2 \pi r}$.
At $r = 2a$, $B_2 = \frac{\mu_0 i}{2 \pi (2a)} = \frac{\mu_0 i}{4 \pi a}$.
$3$. The ratio of the magnetic field at $a/2$ and $2a$ is:
$\frac{B_1}{B_2} = \frac{\frac{\mu_0 i}{4 \pi a}}{\frac{\mu_0 i}{4 \pi a}} = 1$.

(B) The net Lorentz force on the particle is $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
Since the particle passes through the region without any change in velocity,the net force must be zero,so $\vec{E} + \vec{v} \times \vec{B} = 0$,which implies $\vec{E} = -(\vec{v} \times \vec{B}) = \vec{B} \times \vec{v}$.
Taking the cross product with $\vec{B}$ on both sides: $\vec{E} \times \vec{B} = (\vec{B} \times \vec{v}) \times \vec{B}$.
Using the vector triple product identity $(\vec{A} \times \vec{B}) \times \vec{C} = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{A}(\vec{B} \cdot \vec{C})$,we get $\vec{E} \times \vec{B} = \vec{v}(\vec{B} \cdot \vec{B}) - \vec{B}(\vec{v} \cdot \vec{B})$.
Since $\vec{v} \perp \vec{B}$,$\vec{v} \cdot \vec{B} = 0$,so $\vec{E} \times \vec{B} = \vec{v} B^2$.
Therefore,$\vec{v} = \frac{\vec{E} \times \vec{B}}{B^2}$.

(C) The magnetic field produced by a long straight current-carrying wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.
Since the wires $AOB$ and $COD$ are placed at right angles to each other,the magnetic fields $B_1$ and $B_2$ produced by them at a point $P$ (at distance $d$ from $O$ along the normal to the plane) will also be perpendicular to each other.
Thus,the magnitude of the resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.
Substituting the values,$B_1 = \frac{\mu_0 I_1}{2 \pi d}$ and $B_2 = \frac{\mu_0 I_2}{2 \pi d}$.
Therefore,$B = \sqrt{\left( \frac{\mu_0 I_1}{2 \pi d} \right)^2 + \left( \frac{\mu_0 I_2}{2 \pi d} \right)^2}$.
$B = \frac{\mu_0}{2 \pi d} \sqrt{I_1^2 + I_2^2} = \frac{\mu_0}{2 \pi d} (I_1^2 + I_2^2)^{1/2}$.

(C) The power of the combination of lenses in contact is given by the algebraic sum of their individual powers:
$P = P_{1} + P_{2}$
Given $P_{1} = -15 \ D$ and $P_{2} = +5 \ D$,we have:
$P = (-15 + 5) \ D = -10 \ D$
Since the power $P$ is related to the focal length $f$ (in meters) by the formula $P = \frac{1}{f}$,we can find the focal length as:
$f = \frac{1}{P} = \frac{1}{-10} \ m$
To convert the focal length into centimeters,we multiply by $100$:
$f = \left( \frac{1}{-10} \times 100 \right) \ cm = -10 \ cm$
Thus,the focal length of the combination is $-10 \ cm$.

(A) The energy $E$ of a photon with frequency $f$ is given by the Planck-Einstein relation: $E = hf$.
According to Einstein's mass-energy equivalence principle,the energy of a photon is also given by $E = mc^2$,where $m$ is the relativistic mass of the photon.
Equating the two expressions for energy: $mc^2 = hf$.
We know that the momentum $p$ of a particle is given by $p = mc$.
Rearranging the energy equation: $mc = \frac{hf}{c}$.
Therefore,the momentum of the photon is $p = \frac{hf}{c}$.

(B) The growth of current in an $LR$ circuit is given by the formula: $I = I_0(1 - e^{-\frac{R}{L}t})$.
Here,$I_0 = \frac{E}{R}$ is the maximum steady-state current.
Given: $L = 10 \ H$,$R = 5 \ \Omega$,$E = 5 \ V$,and $t = 2 \ s$.
First,calculate the maximum current: $I_0 = \frac{5 \ V}{5 \ \Omega} = 1 \ A$.
Now,substitute the values into the growth equation:
$I = 1 \times (1 - e^{-\frac{5}{10} \times 2})$
$I = 1 - e^{-1} \ A$.

(C) $KEY$ $CONCEPT$: The average power consumed in an $a.c.$ circuit is given by the formula: $P = E_{rms} I_{rms} \cos \phi$.
Given:
Voltage $E = E_o \sin \omega t$
Current $I = I_o \sin \left( \omega t - \frac{\pi}{2} \right)$
Comparing these with the standard forms $E = E_o \sin \omega t$ and $I = I_o \sin (\omega t - \phi)$,we find the phase difference $\phi = \frac{\pi}{2}$.
Substituting the value of $\phi$ into the power formula:
$P = E_{rms} I_{rms} \cos \left( \frac{\pi}{2} \right)$
Since $\cos \left( \frac{\pi}{2} \right) = 0$,the power consumption is:
$P = E_{rms} I_{rms} \times 0 = 0$.

(A) The intensity $I$ at a point on the screen in a Young's double slit experiment is given by the formula $I = I_0 \cos^2 \left( \frac{\phi}{2} \right)$,where $I_0$ is the maximum intensity and $\phi$ is the phase difference.
The relationship between phase difference $\phi$ and path difference $\Delta x$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x$.
Given the path difference $\Delta x = \frac{\lambda}{6}$,we calculate the phase difference:
$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
Now,substitute $\phi$ into the intensity formula:
$\frac{I}{I_0} = \cos^2 \left( \frac{\pi/3}{2} \right) = \cos^2 \left( \frac{\pi}{6} \right)$.
Since $\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$,we have:
$\frac{I}{I_0} = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4}$.

(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{initial} - E_{final} = h\nu$,where $\nu$ is the frequency.
For emission to occur,the electron must transition from a higher energy level to a lower energy level. Thus,options $(b)$ and $(d)$ are excluded as they represent absorption.
Comparing the energy differences for the remaining transitions:
For $n = 2$ to $n = 1$: $\Delta E_1 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - 0.25 \right) = 13.6 \times 0.75 = 10.2 \text{ eV}$.
For $n = 6$ to $n = 2$: $\Delta E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = 13.6 \left( 0.25 - 0.0277 \right) = 13.6 \times 0.2223 \approx 3.02 \text{ eV}$.
Since $\nu = \frac{\Delta E}{h}$,the transition with the largest energy difference will emit the photon with the highest frequency.
Comparing $\Delta E_1$ and $\Delta E_2$,we see that $\Delta E_1 > \Delta E_2$.
Therefore,the transition from $n = 2$ to $n = 1$ emits the photon of the highest frequency.

(A) Gamma $(\gamma)$ emission occurs when a nucleus in an excited state transitions to a lower energy state or the ground state. During this process,the nucleus emits a high-energy photon known as a $\gamma$-ray. Since $\gamma$-rays have zero charge and zero rest mass,the emission does not alter the atomic number $(Z)$ or the mass number $(A)$ of the nucleus. Therefore,the number of protons and neutrons remains unchanged.

(C) The nuclear binding energy of a nucleus is defined as the energy equivalent of the mass defect,given by the formula:
$B.E. = \Delta m c^2 = (\text{mass of nucleons} - \text{mass of nucleus}) c^2$
where $\Delta m$ is the mass defect.
For the oxygen isotope $_8O^{17}$:
Number of protons $(p)$ = $8$
Number of neutrons $(n)$ = $17 - 8 = 9$
Mass of nucleons = $8 M_p + 9 M_n$
Mass of the nucleus = $M_o$
Therefore,the binding energy is:
$B.E. = (8 M_p + 9 M_n - M_o) c^2$

(A) The circuit shown is a half-wave rectifier configuration.
When the input signal is at $+5 \ V$,the $p-n$ junction diode is forward-biased and acts as a closed switch. Thus,the output voltage across the load resistor $R_L$ is $+5 \ V$.
When the input signal is at $-5 \ V$,the $p-n$ junction diode is reverse-biased and acts as an open switch. Thus,no current flows through $R_L$,and the output voltage across $R_L$ is $0 \ V$.
Therefore,the output signal is a square pulse of $+5 \ V$ during the positive half-cycle and $0 \ V$ during the negative half-cycle,which matches the shape shown in option $A$.

(A) Carbon $(C)$,Silicon $(Si)$,and Germanium $(Ge)$ all belong to group $14$ of the periodic table and have $4$ valence electrons.
At room temperature,the energy band gap $(E_g)$ for Carbon (diamond) is approximately $5.4 \ eV$,which is very large,making it an insulator.
The energy band gap for Silicon is approximately $1.1 \ eV$ and for Germanium is approximately $0.7 \ eV$.
Because these band gaps are relatively small,thermal energy at room temperature is sufficient to excite a significant number of electrons from the valence band to the conduction band in $Si$ and $Ge$.
Therefore,$Si$ and $Ge$ act as semiconductors,while $C$ acts as an insulator due to its large band gap.

(B) Given that the half-life of $x$ is equal to the mean life of $y$:
$(t_{1/2})_x = (\tau)_y$
Since $(t_{1/2})_x = \frac{\ln 2}{\lambda_x}$ and $(\tau)_y = \frac{1}{\lambda_y}$,we have:
$\frac{\ln 2}{\lambda_x} = \frac{1}{\lambda_y} \Rightarrow \lambda_x = \lambda_y \ln 2 \approx 0.693 \lambda_y$.
This implies $\lambda_x < \lambda_y$.
Initially,the number of atoms is the same: $N_x = N_y = N_0$.
The decay rate (activity) is given by $A = \lambda N$.
Since $\lambda_x < \lambda_y$ and $N_x = N_y$,it follows that $A_x < A_y$.
Therefore,element $y$ will decay faster than element $x$.

(B) Let the capacitance of the capacitor be $C$ and the e.m.f. of the battery be $V$.
When the capacitor is fully charged,the potential difference across the plates is $V$.
The charge stored on the capacitor is $q = CV$.
The energy stored in the capacitor is $U = \frac{1}{2} CV^2 = \frac{1}{2} qV$.
The work done by the battery in supplying charge $q$ is $W = qV = CV^2$.
The ratio of the energy stored in the capacitor to the work done by the battery is $\frac{U}{W} = \frac{\frac{1}{2} qV}{qV} = \frac{1}{2}$.

(B) When a charged particle moves in a magnetic field perpendicular to its velocity,the magnetic force $F = q(v \times B)$ acts perpendicular to the velocity vector at every instant.
Since the force is always perpendicular to the velocity,the work done by the magnetic force is $W = F \cdot ds = 0$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done,so the kinetic energy remains constant.
However,because the direction of the velocity vector changes continuously as the particle moves in a circular path,the momentum $p = mv$ changes because momentum is a vector quantity.
Therefore,the momentum changes while the kinetic energy remains constant.