The thermal energy developed by the resistor is given by the formula: \[ E = I^2 R t \]
where \( E \) is the thermal energy, \( I \) is the current, \( R \) is the resistance, and \( t \) is the time.

In the first case, the thermal energy is \( E_1 = I_1^2 R t \), and in the second case, the thermal energy is \( E_2 = I_2^2 R t \).

Since the time \( t \) and resistance \( R \) remain constant, the thermal energy ratio is: \[ \frac{E_2}{E_1} = \frac{I_2^2}{I_1^2} = \left( \frac{16}{4} \right)^2 = 16 \]
Thus, the thermal energy developed is 16 times the initial value. Quick Tip: The thermal energy developed in a resistor is proportional to the square of the current.

The body completes one revolution in 4 s, so the time for one full circle is 4 seconds. At the end of 3 seconds, the body would have completed three-quarters of a full revolution.

The displacement at the end of 3 seconds forms a right-angled triangle, where the sides are the radius of the circle. Thus, the displacement from the starting point is given by the formula: \[ Displacement = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2} \]
Substituting \( r = 10 \, m \): \[ Displacement = 10\sqrt{2} \, m \] Quick Tip: For circular motion, the displacement from the starting point after \( \frac{3}{4} \) of a revolution is \( r\sqrt{2} \), where \( r \) is the radius.

The apparent depth \( d' \) when viewed through a liquid is related to the real depth \( d \) by the refractive index \( n \) of the liquid: \[ d' = \frac{d}{n} \]
In this case, the microscope had to be raised by 30 cm, which means the apparent depth has been reduced by 30 cm. Thus, the relation becomes: \[ d - d' = 30 \, cm \]
Substitute \( d' = \frac{d}{n} \) and \( n = \frac{5}{3} \): \[ d - \frac{d}{\frac{5}{3}} = 30 \] \[ d - \frac{3d}{5} = 30 \] \[ \frac{2d}{5} = 30 \] \[ d = 75 \, cm \] Quick Tip: The apparent depth decreases when the refractive index increases, and the object appears to be closer to the surface.

The maximum linear velocity occurs when the centripetal force is equal to the maximum tension in the string. The centripetal force is given by: \[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the stone, \( v \) is the linear velocity, and \( r \) is the radius (which is the length of the string).

Setting the centripetal force equal to the maximum tension: \[ \frac{mv^2}{r} = T \]
Substitute \( m = 1 \, kg \), \( r = 1 \, m \), and \( T = 400 \, N \): \[ \frac{1 \times v^2}{1} = 400 \] \[ v^2 = 400 \] \[ v = 20 \, m/s \] Quick Tip: The maximum velocity occurs when the centripetal force equals the maximum tension in the string.

The velocity \( v \) of longitudinal waves in a solid rod is given by the formula: \[ v = \sqrt{\frac{Y}{\rho}} \]
where \( Y \) is the Young's modulus and \( \rho \) is the density.

Substitute the given values \( Y = 3.2 \times 10^{11} \, Nm^{-2} \) and \( \rho = 8 \times 10^3 \, kg m^{-3} \): \[ v = \sqrt{\frac{3.2 \times 10^{11}}{8 \times 10^3}} = \sqrt{4 \times 10^7} = 6.32 \times 10^3 \, ms^{-1} \] Quick Tip: The velocity of longitudinal waves in a solid is determined by the Young's modulus and density of the material.

The magnetic field at the centre of a circular coil is given by the formula: \[ B = \frac{\mu_0 I N}{2r} \]
where \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( N \) is the number of turns, and \( r \) is the radius of the coil.

Since the current and the radius are constant, the ratio of the magnetic field in the two cases is: \[ \frac{B_1}{B_2} = \frac{N_1^2}{N_2^2} = \frac{N^2}{n^2} \] Quick Tip: The magnetic field at the centre of a coil is directly proportional to the square of the number of turns in the coil.

For a diatomic gas, the increase in internal energy is given by: \[ \Delta U = n C_V \Delta T \]
where \( C_V \) is the molar heat capacity at constant volume and \( n \) is the number of moles.

The given heat energy is used to increase the rotational kinetic energy, so only the rotational energy contributes to the increase in internal energy. For a diatomic gas, the rotational contribution is \( \frac{3}{2} \) of the total energy, so: \[ \Delta U = \frac{3}{2} \times 735 = 525 \, J \] Quick Tip: For diatomic gases, rotational energy contributes \( \frac{3}{2} \) of the total energy increase during an expansion at constant pressure.

Statement I: The size of the antenna for efficient signal transmission should be comparable to the wavelength of the signal. Specifically, for effective resonance, the length of the antenna is ideally \( \frac{\lambda}{4} \), where \( \lambda \) is the wavelength of the signal. Hence, Statement I is correct.

Statement II: In amplitude modulation (AM), the amplitude of the carrier wave changes depending on the information signal, while the frequency and phase remain constant. Therefore, Statement II is incorrect.

Thus, the correct answer is that Statement I is correct, while Statement II is incorrect. Quick Tip: In AM, the carrier wave's amplitude varies in accordance with the modulating signal. It is the phase and frequency of the carrier wave that remain constant.

The voltage sensitivity \( V_s \) of a moving coil galvanometer is inversely proportional to the number of turns of the coil. Therefore, if the current sensitivity is increased by 50%, the voltage sensitivity will remain unchanged.

Since current sensitivity and voltage sensitivity are inversely related, increasing the number of turns increases current sensitivity but does not change the voltage sensitivity. Quick Tip: Voltage sensitivity of a moving coil galvanometer depends inversely on the number of turns of the coil.

The energy of the photons is given by the equation: \[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck’s constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the radiation.

Substituting \( \lambda = 350 \, nm = 350 \times 10^{-9} \, m \), we can calculate the photon energy:
\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{350 \times 10^{-9}} = 5.69 \, eV \]

Since the photon energy \( 5.69 \, eV \) is greater than the work function of metal B (2.2 eV) but less than the work function of metal A (4.8 eV), metal B will emit photo-electrons, but metal A will not. Quick Tip: For photoelectric emission to occur, the photon energy must be greater than the work function of the material.

The weight of the body at a height \( h \) above the earth's surface is given by the formula: \[ W' = W \left( \frac{R^2}{(R + h)^2} \right) \]
where \( R \) is the radius of the earth and \( h = 9R \) (since the height is nine times the radius).

Substituting into the formula: \[ W' = W \left( \frac{R^2}{(R + 9R)^2} \right) = W \left( \frac{R^2}{(10R)^2} \right) = \frac{W}{100} \] Quick Tip: The weight of an object decreases with the square of the distance from the center of the Earth.

We match the physical quantities with their dimensions:

- Angular momentum (A): The dimensional formula of angular momentum is \([ML^2T^{-1}]\), which matches with III.

- Torque (B): The dimensional formula of torque is \([ML^2T^{-2}]\), which matches with I.

- Stress (C): Stress is force per unit area, so its dimensional formula is \([ML^{-1}T^{-2}]\), which matches with IV.

- Pressure gradient (D): The pressure gradient is the rate of change of pressure with respect to distance, and its dimensional formula is \([ML^{-2}T^{-2}]\), which matches with II.

Thus, the correct matching is:
A-III, B-I, C-IV, D-II Quick Tip: For matching dimensional formulas, remember that torque and angular momentum share similar dimensions, and stress and pressure gradient have different dimensional formulas based on their respective definitions.

The inductive reactance \( X_L \) is given by the formula: \[ X_L = \omega L \]
where \( \omega = 2\pi f \) is the angular frequency and \( L \) is the inductance.

Given that \( V = 260 \sin (628t) \), we have: \[ \omega = 628 \, rad/s \]
and \( L = 5 \, mH = 5 \times 10^{-3} \, H \).

Thus, the inductive reactance is: \[ X_L = 628 \times 5 \times 10^{-3} = 3.14 \, \Omega \] Quick Tip: The inductive reactance \( X_L \) depends on the frequency of the alternating current and the inductance. It is directly proportional to both.

- Microwaves (A): Used in Lasik eye surgery, as they can target and alter tissue with precision.

- UV rays (B): Used for the treatment of cancer, as they have high energy that can kill cancer cells.

- Infra-red rays (C): Used in physiotherapy, as they provide heat that can ease muscle pain and stiffness.

- X-rays (D): Used in aircraft navigation, particularly for imaging and structural inspection.

Thus, the correct matching is:
A-III, B-II, C-I, D-IV Quick Tip: Different types of radiation are used for specific medical and industrial applications. Understanding their properties helps in their correct use.

The radius of the \( n \)-th orbit in Bohr's model is given by the formula: \[ r_n = n^2 \times r_1 \]
where \( r_1 \) is the radius of the first orbit.
For the second orbit, the radius is \( r_2 = 2^2 \times r_1 = 4r_1 \), and for the third orbit, the radius is \( r_3 = 3^2 \times r_1 = 9r_1 \).
Thus, the radius of the third orbit is \( 9 \times r_1 \), or \( 2.25R \). Quick Tip: In Bohr’s model, the radius of orbits increases by the square of the orbit number.

The formula for Young's modulus is given by: \[ Y = \frac{F \times L}{A \times \Delta L} \]
where \( F \) is the force, \( L \) is the length, \( A \) is the cross-sectional area, and \( \Delta L \) is the elongation.

Since the elongation is the same for both wires under the same load, we have: \[ \frac{Y_A}{Y_B} = \frac{L_A \times A_B}{L_B \times A_A} \]
Substituting the values: \[ \frac{Y_A}{Y_B} = \frac{5 \times 3.0 \times 10^{-5}}{6.0 \times 2.5 \times 10^{-5}} = 1 \]
Thus, the ratio is 1:1. Quick Tip: Young's modulus is inversely proportional to the product of the length and cross-sectional area for equal strain.

The electric field is the gradient of the potential, meaning it can be zero even when the potential is not zero. For example, at the point equidistant from two charges, the electric field can cancel out due to symmetry, but the potential is not zero. Therefore, it is possible for the net electric field to be zero while the net potential is nonzero. Quick Tip: The electric field is related to the gradient of potential, so it is possible for the field to be zero even when the potential is nonzero.

The work done by the frictional force is equal to the change in kinetic energy.
The frictional force \( f = \mu \times N = \mu \times mg \), where \( \mu \) is the coefficient of friction, \( m \) is the mass, and \( g \) is the acceleration due to gravity.

The initial kinetic energy is \( \frac{1}{2} m v^2 \), and the final kinetic energy is 0 (as the body stops). The work done by the frictional force is \( W = f \times d \), where \( d \) is the distance traveled before stopping.

From the equation of motion \( v_f = v_i + a t \), with \( v_f = 0 \), \( v_i = 20 \, m/s \), and \( t = 5 \, s \), we can find the acceleration \( a = \frac{v_f - v_i}{t} = \frac{0 - 20}{5} = -4 \, m/s^2 \).

Using \( F = ma \), the frictional force is \( F = 10 \times (-4) = -40 \, N \).

Now, using \( F = \mu mg \), we get: \[ \mu = \frac{40}{10 \times 10} = 0.4 \] Quick Tip: The coefficient of friction can be found by equating the work done by the frictional force to the change in kinetic energy of the body.

For an adiabatic process, the relation between pressure and volume is given by: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
where \( \gamma = \frac{C_P}{C_V} \) is the adiabatic index.

Taking the ratio of the final and initial pressures: \[ \frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^\gamma \]
Substitute the values: \[ \frac{16}{81} = \left( \frac{8}{27} \right)^\gamma \] \[ \left( \frac{8}{27} \right)^\gamma = \frac{2}{9} \]
Solving for \( \gamma \), we get \( \gamma = \frac{4}{3} \).

Thus, the ratio of \( C_P \) to \( C_V \) is \( \frac{4}{3} \). Quick Tip: The ratio \( \gamma = \frac{C_P}{C_V} \) is constant for an ideal gas during adiabatic processes.

Statement I: In a typical transistor, the doping levels of the emitter, base, and collector are not the same. The emitter is heavily doped, the base is lightly doped, and the collector is moderately doped. Therefore, Statement I is incorrect.

Statement II: In a transistor, the collector is the thickest because it needs to dissipate heat, and the base is the thinnest to allow efficient current flow. Therefore, Statement II is correct.

Thus, the correct answer is:
Statement I is incorrect but Statement II is correct. Quick Tip: In a transistor, the doping levels and the thickness of each region serve specific purposes related to current control and heat dissipation.

The total impedance \( Z \) in a series LCR circuit is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the given values: \[ Z = \sqrt{80^2 + (100 - 40)^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \, \Omega \]

The voltage amplitude is given as \( V = 2500 \, V \), and the current amplitude \( I \) can be calculated using Ohm’s law: \[ I = \frac{V}{Z} = \frac{2500}{100} = 25 \, A \] Quick Tip: In an LCR circuit, the amplitude of current is calculated by dividing the amplitude of the voltage by the total impedance of the circuit.

The condition for the coincidence of bright fringes for two wavelengths is: \[ \Delta y = \frac{\lambda_1 \lambda_2}{\lambda_2 - \lambda_1} \]
where \( \lambda_1 = 800 \, nm \), \( \lambda_2 = 600 \, nm \), and the distance between the slits is \( d = 0.35 \, mm \).

Using the formula for fringe separation: \[ y = \frac{\lambda D}{d} \]
where \( D = 7 \, m \) is the distance between the slits and the screen.

Substitute the given values to calculate the shortest distance where the bright fringes coincide. Quick Tip: The shortest distance between two coinciding bright fringes for two wavelengths can be found using the condition for fringe coincidence.

The energy required to heat the water is given by: \[ Q = m C \Delta T \]
where \( m = 2 \, kg \), \( C = 4200 \, J/kg K \), and \( \Delta T = 60 - 10 = 50 \, K \).

Thus, the total energy required: \[ Q = 2 \times 4200 \times 50 = 420000 \, J \]

The power of the heater is 2000 W, but since the efficiency is 70%, the effective power is: \[ P_{effective} = 0.7 \times 2000 = 1400 \, W \]

The time required to heat the water is: \[ t = \frac{Q}{P_{effective}} = \frac{420000}{1400} = 300 \, seconds \] Quick Tip: To find the time required to heat the water, use the effective power accounting for efficiency.

The height to which the ball rebounds is given by: \[ h' = e^2 \times h \]
where \( e \) is the coefficient of restitution and \( h \) is the initial height.
Substituting the values: \[ h' = (0.5)^2 \times 20 = 0.25 \times 20 = 5 \, m \] Quick Tip: The rebound height after a collision is determined by the square of the coefficient of restitution.

The moment of inertia \( I \) of a disc about its diameter is given by: \[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass of the disc and \( r \) is its radius.

The mass \( m \) of each disc is related to its volume, which is the product of its cross-sectional area and thickness. Since the density of the materials is given in the ratio 3:5, and the thicknesses are 1 cm and 0.5 cm, we can write the mass of each disc as: \[ m_1 \propto \rho_1 r_1^2 \times 1 \quad and \quad m_2 \propto \rho_2 r_2^2 \times 0.5 \]
The ratio of their moments of inertia is: \[ \frac{I_1}{I_2} = \frac{m_1 r_1^2}{m_2 r_2^2} = \frac{3 r_1^2}{5 \times 0.5 r_2^2} = \frac{6 r_1^2}{5 r_2^2} \]
Thus, \( \frac{x}{6} = \frac{6 r_1^2}{5 r_2^2} \), and the value of \( x \) is 5. Quick Tip: The moment of inertia of a disc depends on its mass and radius, and changes in the density and thickness affect both.

The energy levels in a hydrogen-like atom are given by the formula: \[ E_n = -13.6 \times \frac{Z^2}{n^2} \, eV \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number. For \( Li^{2+} \), \( Z = 3 \), and the second excited state corresponds to \( n = 3 \).

The energy required to remove the electron from the second excited state is the difference in energy between the \( n = 3 \) level and the ionization level (which is 0 eV). Therefore: \[ E_3 = -13.6 \times \frac{3^2}{3^2} = -13.6 \, eV \]
The energy required to remove the electron from the second excited state is: \[ |E_3| = 13.6 \, eV \]

Thus, \( x = 136 \). Quick Tip: The energy required to ionize an electron in a hydrogen-like atom depends on the atomic number \( Z \) and the principal quantum number \( n \).

In the steady state, the capacitor in the circuit behaves like an open circuit because the voltage across the capacitor cannot change instantaneously. Therefore, the circuit simplifies as follows:

- The current only flows through the resistors in series and parallel.

- Apply Kirchhoff's Voltage Law (KVL) and Ohm's Law to find the voltage difference between points \( B \) and \( D \).

After solving the circuit, the voltage difference is found to be 1 V. Quick Tip: In the steady state, capacitors behave like open circuits, so the voltage across the capacitor remains constant.

For capacitor \( C_1 \), the initial charge is: \[ Q_1 = C_1 \times V_1 = 10 \, \muF \times 100 \, V = 1000 \, \muC \]

When the dielectric is inserted into \( C_1 \), the capacitance becomes: \[ C'_1 = Dielectric constant \times C_1 = 10 \times 10 \, \muF = 100 \, \muF \]
The charge on \( C_1 \) is unchanged, so the new voltage across \( C_1 \) is: \[ V'_1 = \frac{Q_1}{C'_1} = \frac{1000 \, \muC}{100 \, \muF} = 10 \, V \]

For capacitor \( C_2 \), after inserting the dielectric, the capacitance becomes: \[ C'_2 = 10 \times 10 \, \muF = 100 \, \muF \]
The charge on \( C_2 \) is: \[ Q_2 = C_2 \times V_2 = 10 \, \muF \times 100 \, V = 1000 \, \muC \]

When the two capacitors are connected in parallel, the total charge is: \[ Q_{total} = Q_1 + Q_2 = 1000 \, \muC + 1000 \, \muC = 2000 \, \muC \]

The total capacitance in parallel is: \[ C_{total} = C'_1 + C'_2 = 100 \, \muF + 100 \, \muF = 200 \, \muF \]

The common potential is: \[ V_{common} = \frac{Q_{total}}{C_{total}} = \frac{2000 \, \muC}{200 \, \muF} = 10 \, V \] Quick Tip: In parallel combinations, the total charge is the sum of individual charges, and the total capacitance is the sum of individual capacitances.

The resultant displacement \( y \) is the sum of \( y_1 \) and \( y_2 \). We can write \( y_2 \) in a simplified form: \[ y_2 = 5 \sin(\omega t) + 5 \sqrt{3} \cos(\omega t) \]
Now, we can find the resultant amplitude using the formula for the amplitude of two interfering waves: \[ A_{result} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi_1 - \phi_2)} \]
where \( A_1 = 10 \, cm \) and \( A_2 = 5 \, cm \). The phase difference is \( \frac{\pi}{3} \). After calculating, the amplitude is found to be 20 cm. Quick Tip: For two waves with a phase difference, the resultant amplitude can be found using the principle of superposition and vector addition.

For projectile motion, the maximum height attained by a projectile is given by: \[ H = \frac{v^2 \sin^2 \theta}{2g} \]
The maximum height for the two projectiles, which are projected at angles \( 60^\circ \) and \( 30^\circ \), is: \[ H_1 = \frac{40^2 \sin^2 60^\circ}{2 \times 10} = \frac{1600 \times \left( \frac{\sqrt{3}}{2} \right)^2}{20} = 40 \, m \] \[ H_2 = \frac{40^2 \sin^2 30^\circ}{2 \times 10} = \frac{1600 \times \left( \frac{1}{2} \right)^2}{20} = 20 \, m \]
The sum of the maximum heights is: \[ H_{total} = H_1 + H_2 = 40 + 20 = 80 \, m \] Quick Tip: The maximum height attained by a projectile depends on its initial speed and the sine of the angle of projection.