1. The de-Broglie wavelength is given by:
\[ \lambda = \frac{h}{\sqrt{2meV}}, \]
where \(h\) is Planck's constant, \(m\) is the mass of the electron, \(e\) is the charge of the electron, and \(V\) is the accelerating voltage.


2. The wavelength is inversely proportional to the square root of the voltage:
\[ \lambda \propto \frac{1}{\sqrt{V}}. \]


3. For \(V = 20 \, kV\), \(\lambda = \lambda_0\). For \(V = 40 \, kV\):
\[ \lambda = \lambda_0 \times \frac{\sqrt{20}}{\sqrt{40}} = \frac{\lambda_0}{\sqrt{2}}. \]


Hence, the de-Broglie wavelength is \(\frac{\lambda_0}{\sqrt{2}}\).


The de-Broglie wavelength is inversely proportional to the square root of the accelerating voltage. Doubling the voltage reduces the wavelength by a factor of \(\sqrt{2}\). Quick Tip: For higher accelerating voltages, the de-Broglie wavelength decreases, leading to better resolution in electron microscopes.

1. For equilibrium, sum of torques about point C = 0:

- Weight of the rod acts at its center of gravity, i.e., 0.5 m from C.

- Tension \(T\) in the cable acts at point B at an angle of \(30^\circ\).


2. Calculating torques:

- Torque due to the rod: \(10 \times 0.5 = 5 \, Nm\).

- Torque due to the hanging object: \(80 \times 1 = 80 \, Nm\).

- Torque due to tension: \(T \times \sin 30^\circ \times 1 = T \times 0.5 \, Nm\).


3. Equating torques:
\[ 80 + 5 = T \times 0.5. \]

\[ T = \frac{85}{0.5} = 300 \, N. \]


Hence, the tension in the cable is 300 N.


Equilibrium conditions require that the net torque and net force on the system are zero. Tension is Calculatingd by balancing the clockwise and counterclockwise torques about the pivot. Quick Tip: In torque calculations, always resolve forces into perpendicular components with respect to the lever arm.

1. Efficiency of a Carnot engine:
\[ \eta = 1 - \frac{T_{sink}}{T_{source}}. \]


2. For 50% efficiency:
\[ 0.5 = 1 - \frac{T_{sink}}{600}. \]

\[ T_{sink} = 300 \, K. \]


3. For 70% efficiency:
\[ 0.7 = 1 - \frac{300}{T_{source}}. \]

\[ \frac{300}{T_{source}} = 0.3. \]

\[ T_{source} = \frac{300}{0.3} = 1000 \, K. \]


Hence, the new temperature of the source is 1000 K.


The efficiency of a Carnot engine depends on the temperatures of the heat source and sink. Increasing efficiency requires increasing the temperature of the source while keeping the sink temperature constant. Quick Tip: The efficiency of a Carnot engine increases with a higher temperature difference between the source and the sink.

1. The time period of a pendulum is given by:
\[ T \propto \frac{1}{\sqrt{g}}. \]


2. At height \(R\) above the Earth's surface, the acceleration due to gravity is:
\[ g' = \frac{g}{(1+h/R)^2}, \quad h = R \implies g' = \frac{g}{(1+1)^2} = \frac{g}{4}. \]


3. The new time period \(T'\) becomes:
\[ T' = T \times \sqrt{\frac{g}{g'}} = T \times \sqrt{\frac{g}{g/4}} = T \times \sqrt{4} = 2T. \]


Hence, \(x = 2\).


The time period increases as the effective gravity decreases with height. At a height equal to Earth's radius, the time period doubles. Quick Tip: Time period of a pendulum depends inversely on the square root of gravity. Reduced gravity at higher altitudes increases the time period.

1. For SHM inside the Earth, the effective force is proportional to displacement:
\[ F = -\frac{G M r}{R^3}, \quad a = \frac{F}{m} = -\frac{G M}{R^3} r. \]


2. The time period of SHM is given by:
\[ T = 2\pi \sqrt{\frac{R^3}{G M}}. \]


3. Substituting \(g = \frac{G M}{R^2}\), the expression becomes:
\[ T = 2\pi \sqrt{\frac{R}{g}}. \]


4. Using \(R = 6400 \, km = 6.4 \times 10^6 \, m\) and \(g = 10 \, m/s^2\):
\[ T = 2\pi \sqrt{\frac{6.4 \times 10^6}{10}} = 2\pi \sqrt{6.4 \times 10^5}. \]


5. Simplifying:
\[ T \approx 2\pi \times 800 = 5026 \, seconds = 1 \, hour \, 24 \, minutes. \]


Hence, the time period is approximately 1 hour 24 minutes.


SHM of a particle inside a uniform sphere depends only on the radius and acceleration due to gravity. The time period is the same for any particle mass. Quick Tip: Remember that SHM in the Earth's tunnel is analogous to oscillations of a spring with the gravitational force acting as the restoring force.

1. Total distance traveled:
\[ d = x + x = 2x. \]


2. Total time taken:
\[ t = \frac{x}{V_1} + \frac{x}{V_2}. \]


3. Average speed:
\[ v_{avg} = \frac{Total distance}{Total time} = \frac{2x}{\frac{x}{V_1} + \frac{x}{V_2}}. \]


4. Simplifying:
\[ v_{avg} = \frac{2x}{x \left(\frac{1}{V_1} + \frac{1}{V_2}\right)} = \frac{2}{\frac{1}{V_1} + \frac{1}{V_2}}. \]

\[ v_{avg} = \frac{2 V_1 V_2}{V_1 + V_2}. \]


Hence, the average speed is \(\frac{2 V_1 V_2}{V_1 + V_2}\).


The average speed for equal distances depends on the harmonic mean of the two speeds. This is because the time taken varies inversely with speed. Quick Tip: For equal distances, always use the harmonic mean formula for average speed: \(v_{avg} = \frac{2 V_1 V_2}{V_1 + V_2}\).

1. The system is equivalent to two capacitors in series:

- Capacitor with dielectric (\(C_1\)): \(d_1 = 1 \, mm, \, \kappa = 5\).

- Capacitor without dielectric (\(C_2\)): \(d_2 = 1 \, mm\).


2. Individual capacitances:
\[ C_1 = \frac{\varepsilon_0 A \kappa}{d_1}, \quad C_2 = \frac{\varepsilon_0 A}{d_2}. \]


3. Substitute \(A = 40 \, cm^2 = 40 \times 10^{-4} \, m^2\):
\[ C_1 = \frac{\varepsilon_0 \times 40 \times 10^{-4} \times 5}{1 \times 10^{-3}} = 200 \varepsilon_0, \quad C_2 = \frac{\varepsilon_0 \times 40 \times 10^{-4}}{1 \times 10^{-3}} = 40 \varepsilon_0. \]


4. Combine in series:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}. \]

\[ C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{200 \varepsilon_0 \times 40 \varepsilon_0}{200 \varepsilon_0 + 40 \varepsilon_0} = \frac{8000}{240} = \frac{10}{3} \varepsilon_0. \]


Hence, the capacitance is \(\frac{10}{3} \varepsilon_0 \, F\).


For capacitors with multiple dielectrics, Calculating individual capacitances and combine them using series or parallel formulas. Quick Tip: For dielectrics in capacitors, use \(\kappa\) to modify the effective plate separation.

1. The root mean square velocity (\(v_{rms}\)) is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}}, \]
where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass.


2. From the formula, \(v_{rms} \propto \sqrt{T}\).


Hence, the root mean square velocity is proportional to the square root of temperature.


The root mean square velocity depends on the square root of the temperature, as higher temperatures lead to higher molecular speeds. Quick Tip: Always use the Kelvin scale for temperature when dealing with gas laws or molecular velocity calculations.

1. Surface tension (A):
\[ Surface tension = \frac{Force}{Length} = \frac{Kg m/s^2}{m} = Kg s^{-2}. \]

Matches with (I).


2. Pressure (B):
\[ Pressure = \frac{Force}{Area} = \frac{Kg m/s^2}{m^2} = Kg m^{-1} s^{-2}. \]

Matches with (II).


3. Viscosity (C):
\[ Viscosity = \frac{Force}{Velocity gradient} = \frac{Kg m/s^2}{m/s} = Kg m^{-1} s^{-1}. \]

Matches with (III).


4. Impulse (D):
\[ Impulse = Force \times Time = Kg m/s^2 \times s = Kg m s^{-1}. \]

Matches with (IV).


Hence, the correct match is: A-I, B-II, C-III, D-IV.


Dimensional analysis ensures physical consistency in equations and helps match quantities with their correct units. Quick Tip: Always derive units using fundamental definitions (e.g., \(Pressure = Force/Area\)).

1. The resonant frequency of an LC circuit is given by:
\[ \omega_0 = \frac{1}{\sqrt{LC}}. \]


2. When \(L \to 2L\) and \(C \to 8C\), the new resonant frequency is:
\[ \omega = \frac{1}{\sqrt{2L \cdot 8C}} = \frac{1}{\sqrt{16LC}} = \frac{1}{4} \omega_0. \]


Hence, the value of \(x\) is \(1/4\).


The resonant frequency of an LC oscillator decreases as the inductance and capacitance increase, since \(\omega_0 \propto 1/\sqrt{LC}\). Quick Tip: For LC circuits, increasing \(L\) or \(C\) reduces the frequency due to their inverse relation with \(\omega_0\).

1. Nuclear density is given by:
\[ \rho = \frac{Mass}{Volume} = \frac{A_u}{\frac{4}{3} \pi R^3}. \]


2. Substituting \(R = R_0 A^{1/3}\), nuclear density becomes:
\[ \rho = \frac{A_u}{\frac{4}{3} \pi (R_0 A^{1/3})^3} = \frac{A_u}{\frac{4}{3} \pi R_0^3 A}. \]


3. Simplifying:
\[ \rho = \frac{3 A_u}{4 \pi R_0^3}. \]


4. Since nuclear density is independent of mass number \(A\), the ratio is:
\[ \rho_{O} : \rho_{He} = 1 : 1. \]


Hence, the ratio is 1:1.

Nuclear density is constant for all nuclei because the nuclear force binds nucleons in a fixed volume, irrespective of \(A\). Quick Tip: For nuclear densities, remember that the mass number \(A\) cancels out in the expression, leading to uniform density.

1. The bandwidth of amplitude modulation is given by:
\[ B = 2 f_m, \]
where \(f_m\) is the frequency of the message signal.


2. Substituting \(f_m = 5 \, kHz\):
\[ B = 2 \times 5 = 10 \, kHz. \]


Hence, the bandwidth is 10 kHz.

The bandwidth of an AM signal is twice the frequency of the modulating (message) signal, covering both upper and lower sidebands. Quick Tip: In amplitude modulation, the bandwidth is directly proportional to the highest frequency component of the modulating signal.

1. The energy transport in an electromagnetic wave is given by the Poynting vector:
\[ \vec{S} = \vec{E} \times \vec{H}, \]
where \(\vec{E}\) is the electric field vector and \(\vec{H}\) is the magnetic field vector.


2. Given:
- Energy transport (\(\vec{S}\)) is in the negative \(z\)-direction: \(\vec{S} = -\hat{k}\).
- Electric field (\(\vec{E}\)) is in the positive \(y\)-direction: \(\vec{E} = +\hat{j}\).


3. Substituting into the cross-product:
\[ \vec{S} = \vec{E} \times \vec{H} \implies -\hat{k} = (+\hat{j}) \times \vec{H}. \]


4. Solving for \(\vec{H}\):
\[ \vec{H} = -\hat{i}. \]


Hence, the magnetic field is in the positive direction of \(x\).


In an electromagnetic wave, the directions of the electric field, magnetic field, and energy transport (Poynting vector) are mutually perpendicular, forming a right-handed coordinate system. Quick Tip: Use the right-hand rule to determine the direction of the magnetic field in electromagnetic waves: \(\vec{E} \times \vec{H} = \vec{S}\).

1. The position of bright fringes in Young's double-slit experiment is given by:
\[ y_n = \frac{n \lambda D}{d}, \]
where \(n\) is the fringe order, \(\lambda\) is the wavelength of light, \(D\) is the distance between the slits and screen, and \(d\) is the distance between the slits.


2. Substituting the given values:
- \(n = 5, \, \lambda = 600 \, nm = 600 \times 10^{-9} \, m, \, y_n = 5 \, cm = 5 \times 10^{-2} \, m, \, D = 1 \, m\).


3. Rearrange for \(d\):
\[ d = \frac{n \lambda D}{y_n}. \]


4. Substituting:
\[ d = \frac{5 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-2}} = 6 \times 10^{-6} \, m = 48 \, \mum. \]


Hence, the distance between the slits is 48 \(\mum\).


The fringe spacing depends on the wavelength, the slit-to-screen distance, and the slit separation. Accurate calculations require converting all quantities to SI units. Quick Tip: In Young's experiment, the fringe position increases with the slit-to-screen distance (\(D\)) and decreases with slit separation (\(d\)).

1. A:
- Using Biot-Savart's law, the magnetic field at point \(O\) is:
\[ B = \frac{\mu_0 I}{4\pi r} [\pi + 2]. \]


2. B:
- The magnetic field at \(O\) is:
\[ B = \frac{\mu_0 I}{4\pi r} [2\pi - 1]. \]


3. C:
- The magnetic field at \(O\) is:
\[ B = \frac{\mu_0 I}{4\pi r} [\pi - 1]. \]


4. D:
- The magnetic field at \(O\) is:
\[ B = \frac{\mu_0 I}{4\pi r} [2\pi + 1]. \]


Hence, the correct match is: A-III, B-IV, C-I, D-II.


The magnetic field at a point due to a current is Calculatingd using Biot-Savart's law or Ampère's circuital law. Quick Tip: For circular loops, the magnetic field depends on the geometry of the current and distance from the center.

1. Observing Assertion A:

Photodiodes are not used in forward bias for light intensity measurements. Instead, they are used in reverse bias to ensure that the photocurrent is proportional to the incident light intensity.

Hence, Assertion A is false.


2. Observing Reason R:

The given statement about the current in a p-n junction diode in forward bias being greater than in reverse bias is correct. The reverse bias current is minimal (leakage current) until breakdown voltage \(V_z\) is reached.

Hence, Reason R is true.


Hence, the final answer is A is false but R is true.


Photodiodes operate in reverse bias mode because the reverse current is directly proportional to light intensity, providing accurate measurement. Forward bias operation is unsuitable for this purpose. Quick Tip: Remember, photodiodes are always operated in reverse bias for light sensing applications due to their linear response to light intensity.

1. The magnetic intensity \(H\) inside a solenoid is given by:
\[ H = nI, \]
where \(n\) is the number of turns per unit length and \(I\) is the current.


2. Calculating \(n\):
\[ n = \frac{Total number of turns}{Length of solenoid} = \frac{1200}{2} = 600 \, turns/m. \]


3. Substituting \(n = 600 \, turns/m\) and \(I = 2 \, A\):
\[ H = nI = 600 \times 2 = 1200 \, A m^{-1}. \]


Hence, the magnetic intensity is \(\mathbf{1.2 \times 10^3 \, A m^{-1}}\).


The magnetic intensity depends on the number of turns per unit length and the current flowing through the solenoid. Quick Tip: For solenoids, remember \(H = nI\), where \(n\) is the turn density (\(turns/m\)).

1. Using Ohm's Law:
\[ R = \frac{V}{I} = \frac{3.4}{2} = 1.7 \, \Omega. \]


2. The resistance of a wire is given by:
\[ R = \rho \frac{L}{A}, \]
where \(\rho\) is resistivity, \(L\) is length, and \(A\) is cross-sectional area.


3. The volume of the wire is given by:
\[ V = \frac{Mass}{Density} = \frac{8.92 \times 10^{-3}}{8.92 \times 10^3} = 10^{-6} \, m^3. \]


4. The cross-sectional area is:
\[ A = \frac{V}{L}. \]


5. Rearrange the resistance formula to solve for \(L\):
\[ R = \rho \frac{L}{A} \implies L = \frac{R A}{\rho}. \]


6. Substituting values:
\[ A = \frac{10^{-6}}{L}. \]

Substituting \(A\) into the equation for \(L\):
\[ L = \sqrt{\frac{R \times 10^{-6}}{\rho}}. \]

Substituting \(R = 1.7 \, \Omega, \rho = 1.7 \times 10^{-8} \, \Omega \, m\):
\[ L = \sqrt{\frac{1.7 \times 10^{-6}}{1.7 \times 10^{-8}}}. \]

\[ L = 10 \, m. \]


Hence, the length of the wire is 10 m.


The length of the wire is determined by calculating its volume and cross-sectional area using the given mass, density, and resistance values. Quick Tip: Use \(R = \rho L/A\) for resistance-related calculations, and remember that volume is proportional to the product of cross-sectional area and length.

1. According to Newton's Law of Cooling:
\[ \frac{\Delta Q}{\Delta t} = -k (T - T_0), \]
where \(T_0\) is the room temperature and \(T\) is the average temperature of the body during the time interval.


2. For the first cooling phase (\(98^\circ C \to 86^\circ C\)):
\[ \Delta t = 2 \, min, \quad T_{avg} = \frac{98 + 86}{2} = 92^\circ C. \]

Substituting:
\[ k = \frac{\Delta Q}{\Delta t} = \frac{1}{2} \times (92 - 22) = \frac{70}{2}. \]


3. For the second cooling phase (\(75^\circ C \to 69^\circ C\)):
\[ T_{avg} = \frac{75 + 69}{2} = 72^\circ C. \]

Substituting:
\[ \Delta t = \frac{\Delta Q}{k} = \frac{1}{\frac{70}{2}} \times (72 - 22) = \frac{6}{5}. \]


Thus:
\[ \Delta t = 1.4 \, minutes. \]


Hence, the time taken is 1.4 minutes.


The rate of cooling depends on the difference between the object's temperature and the surrounding temperature. Calculations must use the average temperature for each interval. Quick Tip: Use Newton's Law of Cooling for temperature intervals by averaging the temperatures to simplify calculations.

1. The forces acting on the bob are:

- Tension (\(T\)) in the string.

- Centripetal force (\(T \sin \theta = \frac{mv^2}{R}\)).

- Vertical component (\(T \cos \theta = mg\)).


2. Dividing these equations:
\[ \tan \theta = \frac{T \sin \theta}{T \cos \theta} = \frac{\frac{mv^2}{R}}{mg}. \]


3. Simplify:
\[ \tan \theta = \frac{v^2}{gR}. \]


4. Substituting values (\(v = 20 \, m/s, \, R = 40 \, m, \, g = 10 \, m/s^2\)):
\[ \tan \theta = \frac{20^2}{10 \times 40} = 1. \]

\[ \theta = \tan^{-1}(1) = 45^\circ. \]


Hence, the angle made by the string is 45°.


The angle of the string depends on the balance of centripetal force and gravitational force. The tangent of the angle is the ratio of horizontal to vertical forces. Quick Tip: For circular motion problems, always Calculating centripetal force and compare it with gravitational force for equilibrium.

1. The critical angle is given by:
\[ \sin \theta_c = \frac{1}{\mu} \implies \sin \theta_c = \frac{1}{\sqrt{2}} \implies \theta_c = 45^\circ. \]


2. Using geometry, the lateral displacement is:
\[ x = t \sin(\theta_i - r) \sec r, \]
where \(t\) is the thickness, \(\theta_i = \theta_c = 45^\circ\), and \(\sin r = \frac{\sin 45^\circ}{\sqrt{2}} = 0.5\).


3. Substituting values:
\[ x = \sqrt{5} \times \sin(45^\circ - 30^\circ) \times \sec(30^\circ). \]

\[ x = \sqrt{5} \times \sin(15^\circ) \times \sec(30^\circ). \]

\[ x = \sqrt{5} \times 0.26 \times \frac{2}{\sqrt{3}} = \frac{\sqrt{5} \times 0.52}{\sqrt{3}} = 0.52 \, cm. \]


Hence, the lateral displacement is 0.52 cm.


Lateral displacement in a refractive medium depends on the thickness of the plate, the angle of incidence, and the refractive index. Quick Tip: For lateral displacement, always use trigonometric relations involving the refracted angle and critical angle for precise calculations.

1. Firstly observe the circuit:

The \(4 \, \Omega\) resistors are shorted as they are connected across the same potential. Hence, they can be removed.


2. The effective circuit becomes:
\[ R_{eq} = 3 + \frac{2 \parallel 12}{1} + 6, \]
where:
\[ 2 \parallel 12 = \frac{2 \times 12}{2 + 12} = \frac{24}{14} = 1.71 \, \Omega. \]


3. Substituting:
\[ R_{eq} = 3 + 1.71 + 6 = 10 \, \Omega. \]


Hence, the equivalent resistance is 10.00 \(\Omega\).


Simplify circuits by removing elements with no current flow. Combine series and parallel resistors step by step for accurate results. Quick Tip: In circuits, identify shorted components or elements with no current to simplify calculations.

1. From the graph:
\[ \tan \theta = \frac{\Delta l}{F}, \]
where \(\Delta l\) is extension and \(F\) is the force.


2. The formula for Young's modulus is:
\[ Y = \frac{F L}{A \Delta l}. \]


3. Substitute:
\[ Y = \frac{\tan \theta \cdot L}{A}. \]


4. Calculating the area:
\[ A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{4 \times 10^{-3}}{2}\right)^2 = 12.57 \times 10^{-6} \, m^2. \]


5. Substituting values:
\[ Y = \frac{62.8 \times 10^{-2}}{12.57 \times 10^{-6}} = 5 \times 10^{10} \, Nm^{-2}. \]


Hence, the value of \(x\) is 5.


Young's modulus is proportional to the ratio of stress to strain. Use the area of cross-section and extension formula for precise calculations. Quick Tip: Ensure unit conversions (e.g., cm to m) for consistency when calculating Young's modulus.

1. The force acting on the object is:
\[ F_x = F \cos \theta. \]


2. The equation of motion is:
\[ F \cos \theta = ma \implies m \frac{dv}{dx} v = F \cos \theta. \]


3. Rearrange and integrate:
\[ \int_0^v v \, dv = \int_0^x \frac{F}{m} \cos (kx) \, dx. \]

\[ \frac{1}{2} mv^2 = \frac{F}{k} \sin (kx). \]


4. The kinetic energy is:
\[ K.E. = \frac{1}{2} mv^2 = \frac{F}{k} \sin \theta. \]

Comparing with \(E = \frac{n}{k} \sin \theta\), \(n = 2\).


Hence, the value of \(n\) is 2.


The variation of force with position affects the kinetic energy. Use integration for non-constant forces to find the energy relation. Quick Tip: For variable force problems, relate displacement and velocity through work-energy principles.

1. Transition from second excited state (\(n=3\)) to first excited state (\(n=2\)):
\[ \frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right), \]
where \(R\) is the Rydberg constant.

Simplify:
\[ \frac{1}{\lambda_0} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \cdot \frac{5}{36}. \]


2. Transition from third excited state (\(n=4\)) to second orbit (\(n=2\)):
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right). \]
Simplify:
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4 - 1}{16} \right) = R \cdot \frac{3}{16}. \]


3. Wavelength ratio:
\[ \frac{\lambda_0}{\lambda} = \frac{\frac{5}{36}}{\frac{3}{16}} = \frac{5 \cdot 16}{36 \cdot 3} = \frac{80}{108} = \frac{20}{27}. \]
Thus:
\[ \lambda = \frac{20}{27} \lambda_0. \]


Hence, \(x = 27\).


The wavelength of radiation depends on the energy difference between the two levels. Simplify Rydberg's formula step by step to Calculating ratios. Quick Tip: For hydrogen transitions, use \(\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\), ensuring \(n_2 > n_1\).

1. Moment of inertia about the center of mass:
\[ I_{CM} = \frac{1}{2} mR^2. \]


2. Using the parallel axis theorem:
\[ I_{AB} = I_{CM} + m \left(\frac{2}{3} R \right)^2. \]

Substituting:
\[ I_{AB} = \frac{1}{2} mR^2 + m \left(\frac{4}{9} R^2 \right). \]

Simplify:
\[ I_{AB} = \frac{1}{2} mR^2 + \frac{4}{9} mR^2 = \frac{9}{18} mR^2 + \frac{8}{18} mR^2 = \frac{17}{18} mR^2. \]


3. Ratio:
\[ \frac{I_{AB}}{I_{CM}} = \frac{\frac{17}{18} mR^2}{\frac{1}{2} mR^2} = \frac{17}{9}. \]

Thus:
\[ x = 17. \]


Hence, the value of \(x\) is 17.


The moment of inertia about an axis parallel to the center of mass axis can be Calculatingd using the parallel axis theorem. Quick Tip: Always add \(md^2\) to the center of mass moment of inertia when using the parallel axis theorem.

1. The phase difference is given by:
\[ \Delta \phi = k \Delta x, \]
where \(k = \frac{2\pi}{\lambda}\) is the wave number.


2. Substituting \(\Delta \phi = \frac{\pi}{3}\) and \(\Delta x = 6 \, m\):
\[ \frac{\pi}{3} = \frac{2\pi}{\lambda} \cdot 6. \]
Simplify:
\[ \lambda = 36 \, m. \]


3. The wave velocity is:
\[ v = \lambda f = 36 \cdot 500 = 18000 \, m/s = 18 \, km/s. \]


Hence, the velocity is 18 km/s.


Wave velocity depends on the wavelength and frequency. Use the relation \(v = \lambda f\) to Calculating the speed of propagation. Quick Tip: For phase differences, use \(\Delta \phi = k \Delta x\), where \(k = \frac{2\pi}{\lambda}\).

1. Kinetic energy of the electron:
\[ \frac{1}{2} mv^2 = 5 \, eV. \]

Convert \(5 \, eV\) to joules:
\[ 5 \cdot 1.6 \times 10^{-19} = 8 \times 10^{-19} \, J. \]

Solve for \(v\):
\[ v_x = \sqrt{\frac{2 \cdot 8 \times 10^{-19}}{9.1 \times 10^{-31}}}. \]


2. Vertical displacement:
\[ \tan \theta = \frac{E_y}{E_x}. \]

Substituting:
\[ \theta = \tan^{-1}(1) = 45^\circ. \]


Hence, the angle of deviation is 45°.


The deviation of the path in a uniform electric field depends on the balance between horizontal and vertical forces on the electron. Quick Tip: Convert electron energy from eV to joules for calculations involving motion in an electric field.

1. The condition for maximum current amplitude in an LCR circuit is resonance, where:
\[ f = \frac{1}{2 \pi \sqrt{LC}}. \]


2. Rearrange to solve for \(L\):
\[ L = \frac{1}{(2 \pi f)^2 C}. \]


3. Substituting the values:
- \(f = 2000 \, Hz\),
- \(C = 62.5 \, nF = 62.5 \times 10^{-9} \, F\),

\[ L = \frac{1}{(2 \pi \cdot 2000)^2 \cdot 62.5 \times 10^{-9}}. \]


4. Simplify:
\[ L = \frac{1}{4 \pi^2 \cdot 4 \cdot 10^6 \cdot 62.5 \times 10^{-9}}. \]

Using \(\pi^2 = 10\):
\[ L = \frac{1}{4 \cdot 10 \cdot 10^6 \cdot 62.5 \times 10^{-9}}. \]

\[ L = \frac{1}{2.5 \times 10^{-2}} = 100 \, mH. \]


Hence, the inductance is 100 mH.


At resonance, the inductive reactance cancels out the capacitive reactance. Use \(f = \frac{1}{2 \pi \sqrt{LC}}\) to find the inductance. Quick Tip: For resonance in an LCR circuit, always solve for \(L\) or \(C\) using \(f = \frac{1}{2 \pi \sqrt{LC}}\) with proper unit conversions.

1. Cross product \(\vec{P} \times \vec{Q}\):
\[ \vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & \sqrt{3} & 2
4 & \sqrt{3} & 2.5 \end{vmatrix}. \]


2. Expanding the determinant:
\[ \vec{P} \times \vec{Q} = \hat{i} \begin{vmatrix} \sqrt{3} & 2
\sqrt{3} & 2.5 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2
4 & 2.5 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & \sqrt{3}
4 & \sqrt{3} \end{vmatrix}. \]


3. Solve each minor:
\[ \hat{i}: \begin{vmatrix} \sqrt{3} & 2
\sqrt{3} & 2.5 \end{vmatrix} = (\sqrt{3} \cdot 2.5 - \sqrt{3} \cdot 2) = \sqrt{3} \cdot 0.5 = 0.5 \sqrt{3}. \]

\[ \hat{j}: \begin{vmatrix} 3 & 2
4 & 2.5 \end{vmatrix} = (3 \cdot 2.5 - 4 \cdot 2) = 7.5 - 8 = -0.5. \]

\[ \hat{k}: \begin{vmatrix} 3 & \sqrt{3}
4 & \sqrt{3} \end{vmatrix} = (3 \cdot \sqrt{3} - 4 \cdot \sqrt{3}) = -\sqrt{3}. \]


4. Combine:
\[ \vec{P} \times \vec{Q} = \hat{i} \cdot (0.5 \sqrt{3}) - \hat{j} \cdot (-0.5) - \hat{k} \cdot \sqrt{3}. \]

\[ \vec{P} \times \vec{Q} = 0.5 \sqrt{3} \hat{i} + 0.5 \hat{j} - \sqrt{3} \hat{k}. \]


5. Find magnitude:
\[ |\vec{P} \times \vec{Q}| = \sqrt{\left(0.5 \sqrt{3}\right)^2 + (0.5)^2 + \left(-\sqrt{3}\right)^2}. \]

\[ |\vec{P} \times \vec{Q}| = \sqrt{\frac{3}{4} + \frac{1}{4} + 3} = \sqrt{4} = 2. \]


6. Unit vector:
\[ Unit vector = \frac{\vec{P} \times \vec{Q}}{|\vec{P} \times \vec{Q}|} = \frac{1}{2} \left(\sqrt{3} \hat{i} + \hat{j} - 2 \sqrt{3} \hat{k}\right). \]


Comparing with the given unit vector:
\[ \frac{1}{x} = \frac{1}{4}. \]

Thus:
\[ x = 4. \]


Final Answer: (4)


The cross product of two vectors is perpendicular to both, and its unit vector is obtained by dividing the cross product by its magnitude. Quick Tip: For cross products, carefully compute each determinant component and simplify to avoid errors.