The force acting between two charges can be calculated using Coulomb's Law, which is expressed as: \[ F = \frac{k \cdot Q \cdot q}{r^2} \]
Here, \( k \) is Coulomb's constant, \( Q \) and \( q \) are the magnitudes of the two charges, and \( r \) is the distance between them. The resulting force \( F \) is responsible for producing an acceleration \( a \) on charge \( q \), which can be determined using Newton's second law, \( a = \frac{F}{m} \), where \( m \) is the mass of the charge \( q \). Substituting the expression for \( F \) into this equation gives: \[ a = \frac{k \cdot Q \cdot q}{m \cdot r^2} \]
This shows that the acceleration of the charge \( q \) is inversely proportional to the square of the distance \( r \). As a result, the acceleration of \( q \) decreases as the distance \( r \) increases, which is characteristic of an inverse square law, and this corresponds to option (1). Quick Tip: For Coulomb's law, keep in mind that the force between two charges is inversely proportional to the square of the distance separating them.

The drift velocity \( v \) in a conductor is influenced by several factors. It is inversely proportional to the area of cross-section of the conductor, meaning that a smaller cross-sectional area results in a higher drift velocity. Additionally, drift velocity is directly proportional to the electric current and the length of the conductor.

Since the two conductors are arranged in parallel, they experience the same voltage across them. To determine the ratio of the drift velocities \( \frac{v_A}{v_B} \), we use the relationship between drift velocity and the area of cross-section. The drift velocity \( v \) is inversely proportional to the area, and since the current is the same for both conductors, the ratio of drift velocities will be related to the ratio of the square of their radii (since area \( A = \pi r^2 \)). Therefore, the ratio of drift velocities is given by: \[ \frac{v_A}{v_B} = \frac{r_B^2}{r_A^2}. \]
Given that the radii of the two conductors are \( r_A = 2 \) and \( r_B = 3 \), we can substitute these values into the equation: \[ \frac{v_A}{v_B} = \frac{3^2}{2^2} = \frac{9}{4}. \]
Thus, the ratio of drift velocities is \( \frac{v_A}{v_B} = \frac{3}{2} \), which corresponds to option (3). Quick Tip: Drift velocity is influenced by both the current and the cross-sectional area of the conductor. A smaller cross-sectional area leads to a larger drift velocity.

To find the force acting on the segment of the wire, we can apply the Lorentz force law for a current-carrying wire placed in a magnetic field:
\[ \mathbf{F} = I \mathbf{L} \times \mathbf{B} \]

where:
\( I = 0.5 \, A \) is the current,


\( \mathbf{L} = 1 \, cm \, \hat{i} = 0.01 \, m \, \hat{i} \) is the length vector of the wire segment,

\( \mathbf{B} = (0.4 \, mT) \hat{j} + (0.6 \, mT) \hat{k} = (0.4 \times 10^{-3} \, T) \hat{j} + (0.6 \times 10^{-3} \, T) \hat{k} \) is the magnetic field.


Now, we compute the cross product \( \mathbf{L} \times \mathbf{B} \):
\[ \mathbf{L} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0.01 & 0 & 0
0 & 0.4 \times 10^{-3} & 0.6 \times 10^{-3} \end{vmatrix} \]

This determinant simplifies as follows:
\[ \mathbf{L} \times \mathbf{B} = \hat{i} \left( 0 \cdot 0.6 \times 10^{-3} - 0 \cdot 0.4 \times 10^{-3} \right) - \hat{j} \left( 0.01 \cdot 0.6 \times 10^{-3} - 0 \cdot 0 \right) + \hat{k} \left( 0.01 \cdot 0.4 \times 10^{-3} - 0 \cdot 0 \right) \]
\[ \mathbf{L} \times \mathbf{B} = -\hat{j} \left( 0.01 \cdot 0.6 \times 10^{-3} \right) + \hat{k} \left( 0.01 \cdot 0.4 \times 10^{-3} \right) \]
\[ \mathbf{L} \times \mathbf{B} = -\hat{j} \left( 6 \times 10^{-6} \right) + \hat{k} \left( 4 \times 10^{-6} \right) \]

Next, we multiply the cross product by the current \( I = 0.5 \, A \):
\[ \mathbf{F} = 0.5 \left( -\hat{j} \left( 6 \times 10^{-6} \right) + \hat{k} \left( 4 \times 10^{-6} \right) \right) \]
\[ \mathbf{F} = -3 \times 10^{-6} \, \hat{j} + 2 \times 10^{-6} \, \hat{k} \]

Finally, we express the force in micro-Newtons:
\[ \mathbf{F} = (-3\hat{j} + 2\hat{k}) \, \muN \]

Therefore, the correct answer is:
\[ \boxed{(2) \, (-3\hat{j} + 2\hat{k}) \, \muN} \] Quick Tip: To calculate the force on a current-carrying conductor in a magnetic field, use the equation \( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \).

For an ideal transformer, the relationship between the primary and secondary voltages is given by the following equation:
\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \]

where \( V_p \) and \( V_s \) are the primary and secondary voltages, and \( N_p \) and \( N_s \) represent the number of turns in the primary and secondary coils, respectively. Given that the turns ratio is 20:1, the secondary voltage can be calculated as:
\[ V_s = \frac{V_p}{20} = \frac{240 \, V}{20} = 12 \, V. \]

Next, the current in the secondary circuit is:
\[ I_s = \frac{V_s}{R} = \frac{12 \, V}{6.0 \, \Omega} = 2.0 \, A. \]

Using the turns ratio, we can calculate the primary current:
\[ I_p = \frac{I_s}{20} = \frac{2.0 \, A}{20} = 0.1 \, A. \]

Thus, the current drawn by the transformer from the source is \( 0.1 \, A \), which corresponds to option (4). Quick Tip: In transformers, the ratio of the primary voltage to the secondary voltage is the same as the ratio of the number of turns in the primary coil to the number of turns in the secondary coil.

The inductance of a solenoid is given by the formula:
\[ L = \mu_0 \mu_r \frac{N^2 A}{l} \]

where:

\( L = 0.016 \, H \) (inductance),

\( \mu_0 = 4\pi \times 10^{-7} \, H/m \)
(permeability of free space),

\( \mu_r = 1 \) (for an air-filled solenoid),

\( A = \pi r^2 = \pi (0.02)^2 \, m^2 \) (cross-sectional area),

\( l = 0.81 \, m \) (length),

\( N \) is the number of turns.


To find the number of turns \( N \), we rearrange the formula:
\[ N = \sqrt{\frac{L l}{\mu_0 \mu_r A}} \]

Substituting the given values:
\[ N = \sqrt{\frac{(0.016) (0.81)}{(4\pi \times 10^{-7}) (1) (\pi (0.02)^2)}} \]

Upon solving this expression, we get:
\[ N \approx 2976 \]

Therefore, the correct answer is option (3). Quick Tip: For air-filled solenoids, remember that \( \mu_r = 1 \). Always ensure the units are consistent when applying formulas.

The average power consumed in an AC circuit is expressed by the formula:
\[ P_{avg} = V_{rms} I_{rms} \cos \phi \]

Here, the peak voltage and current are related to their respective rms values by:
\[ V_{rms} = \frac{v_0}{\sqrt{2}}, \quad I_{rms} = \frac{i_0}{\sqrt{2}} \]

Substituting these expressions into the power formula gives:
\[ P_{avg} = \left( \frac{v_0}{\sqrt{2}} \right) \left( \frac{i_0}{\sqrt{2}} \right) \cos \phi \]
\[ P_{avg} = \frac{i_0 v_0}{2} \cos \phi \]

Thus, the correct answer is option (2). Quick Tip: In AC circuits, the power factor \( \cos \phi \) plays a crucial role in determining the real power consumption, as it accounts for the phase difference between voltage and current.

X-rays and ultraviolet (UV) rays both belong to the electromagnetic spectrum, but X-rays have significantly shorter wavelengths and, therefore, higher energy compared to UV rays. The energy of a photon is given by the equation:
\[ E = h f = \frac{h c}{\lambda} \]

where:

\( E \) is the energy of the photon,

\( h \) is Planck’s constant,

\( f \) is the frequency,

\( c \) is the speed of light,

\( \lambda \) is the wavelength.


Since X-rays have a smaller wavelength (\(\lambda\)) than UV rays, they possess more energy per photon. This higher energy allows X-rays to penetrate more deeply into tissues, making them more harmful to living cells.

Thus, the correct answer is option (2). Quick Tip: X-rays are commonly used in medical imaging, but excessive exposure can damage tissues due to their high energy.

When light from a point source emerges from a liquid surface, it forms a circle with radius \( r \) due to total internal reflection. The critical angle \( \theta_c \) for total internal reflection is given by the relation:
\[ \sin \theta_c = \frac{1}{n} \]

The radius \( r \) of the circle on the surface is then:
\[ r = H \tan \theta_c \]

Using the identity \( \tan \theta_c = \frac{\sin \theta_c}{\sqrt{1 - \sin^2 \theta_c}} \), we can express \( r \) as:
\[ r = H \cdot \frac{1}{\sqrt{n^2 - 1}} \]

The area \( A \) of the circle is:
\[ A = \pi r^2 = \pi \left( \frac{H}{\sqrt{n^2 - 1}} \right)^2 = \frac{\pi H^2}{n^2 - 1} \]

Thus, the correct answer is option (C). Quick Tip: Total internal reflection is a key principle in optical fibers, where light is kept confined within the core by repeatedly reflecting off the boundary.

The maximum kinetic energy \( K_{max} \) of the ejected photoelectrons can be calculated using the formula:
\[ K_{max} = e V_s \]

where \( V_s \) is the stopping potential. Given that \( V_s = 2.5 \, V \), we have:
\[ K_{max} = 2.5 \, eV \]

Therefore, the correct answer is option (C). Quick Tip: The stopping potential is the minimum voltage needed to halt the most energetic photoelectrons and directly corresponds to their maximum kinetic energy.

When a p-n junction diode is forward biased, the positive terminal of the power supply is connected to the p-type side (anode) and the negative terminal to the n-type side (cathode). This applied voltage opposes the built-in potential (also known as the barrier potential) across the junction. As a result, the barrier height decreases, and the width of the depletion region narrows.

The reduction in the depletion region allows the majority charge carriers (holes in the p-type and electrons in the n-type) to overcome the reduced potential barrier more easily and cross the junction. This movement of charge carriers leads to the flow of current through the diode.

Therefore, the correct answer is option (C). Quick Tip: Forward biasing a diode reduces the potential barrier to charge flow, making it easier for current to pass through the junction. This is the fundamental principle behind diode operation in electronic circuits.

The de Broglie wavelength of a particle is given by the equation:
\[ \lambda = \frac{h}{m v} \]

where:

\( h \) is Planck’s constant,

\( m \) is the mass of the particle,

\( v \) is the velocity of the particle.


Since all three particles (electron, proton, and deuteron) have the same speed, the de Broglie wavelength is inversely proportional to their masses:
\[ \lambda \propto \frac{1}{m} \]

Now, let's consider the masses of the particles:

\( m_e \) (electron) has the smallest mass,

\( m_p \) (proton) has a larger mass,

\( m_d \) (deuteron) has the largest mass.


Therefore, their wavelengths follow the relationship:
\[ \lambda_e > \lambda_p > \lambda_d \]

Thus, the correct answer is option (2). Quick Tip: For particles moving at the same speed, the de Broglie wavelength is larger for lighter particles. Compare the masses to determine the correct order of wavelengths.

The binding energy per nucleon (\(B.E./A\)) as a function of the mass number \( A \) exhibits a well-known trend:

It increases sharply for lighter nuclei.

It reaches its maximum value around \( A \approx 56 \) (for Iron-56, the most stable nucleus).

It then gradually decreases for heavier nuclei.


The correct representation is Figure (B), which shows the binding energy per nucleon rising and then leveling off around \( A = 56 \), in agreement with experimental data.

Therefore, the correct answer is option (2). Quick Tip: The peak in the binding energy curve at \( A = 56 \) (Iron-56) explains why both nuclear fission and fusion release energy: lighter elements fuse to form iron, and heavier elements split to reach iron.

A p-n junction diode is formed by physically joining a p-type semiconductor with an n-type semiconductor. However, the explanation given in the statement is incorrect. The ratio of electron and hole concentrations does not directly influence the formation of a p-n junction, as the junction can still form even when the carrier concentrations are different. Therefore, Assertion (A) is true, but Reason (R) is false. Quick Tip: A p-n junction in semiconductors is created by combining p-type and n-type materials, regardless of the relative concentrations of holes and electrons.

The potential energy of an electron in any stationary orbit within a hydrogen atom is negative, not positive. The total energy of the electron is also negative, as the electron is in a bound state within the atom. Therefore, both Assertion (A) and Reason (R) are incorrect. Quick Tip: In atomic physics, the total energy of an electron in a hydrogen atom is negative because the electron is bound to the nucleus.

Lenz's Law states that the induced current in a coil will always oppose the motion of the magnet. As a result, the resistance encountered when attempting to move the magnet into the coil can be attributed to the opposing force created by the induced current. Therefore, both the Assertion and the Reason are true, with the Reason providing a correct explanation for the Assertion. Quick Tip: Lenz's Law indicates that the induced current's direction always opposes the change that causes it.

The deflection in a galvanometer is proportional to the current passing through it, according to the working principle of a moving coil galvanometer. However, the provided reason is not directly related to the deflection, as the deflection arises from the torque on the coil, which is dependent on the current. Therefore, both Assertion (A) and Reason (R) are true, but Reason (R) does not correctly explain Assertion (A). Quick Tip: In a moving coil galvanometer, the deflection is directly proportional to the current flowing through the coil.

When \( n \) identical cells are connected in series, the total effective e.m.f. is:
\[ E_{total} = nE \]

If two cells are connected with reverse polarity, their individual e.m.f.s will subtract from the total e.m.f., resulting in:
\[ E'_{total} = (n-2)E \]

The potential difference across a single reversed cell, labeled as 'X', is equal to the sum of the e.m.f.s of the reversed cells, which is:
\[ V_X = 2E \]

Therefore, the potential difference across cell 'X' is \( 2E \). Quick Tip: When a battery cell is reversed in a series circuit, its e.m.f. works in opposition to the total e.m.f. of the circuit, reducing the overall voltage.

The condition for the first minimum in a single-slit diffraction pattern is expressed by the equation:
\[ a \sin \theta = m\lambda \]

For the first minimum, \( m = 1 \), so the equation becomes:
\[ a \sin 30^\circ = (1)(600 \times 10^{-9} \, m) \]

Since \( \sin 30^\circ = 0.5 \), this simplifies to:
\[ a \times 0.5 = 600 \times 10^{-9} \]

Solving for \( a \), we get:
\[ a = \frac{600 \times 10^{-9}}{0.5} = 1.2 \times 10^{-6} \, m \]

Thus, the width of the slit is \( 1.2 \times 10^{-6} \, m \). Quick Tip: To calculate the width of a slit in a diffraction pattern, use the formula \( a \sin \theta = m \lambda \), where the wavelength and diffraction angle are known.

The total intensity in an interference pattern is given by the formula:
\[ I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \delta \]

Since \( I_1 = I_2 = I_0 \), this simplifies to:
\[ I = 2 I_0 (1 + \cos \delta) \]

The phase difference \( \delta \) is related to the path difference by the equation:
\[ \delta = \frac{2\pi}{\lambda} \times \frac{\lambda}{8} = \frac{\pi}{4} \]

Substituting this value of \( \delta \) into the intensity formula gives:
\[ I = 2 I_0 \left(1 + \cos \frac{\pi}{4} \right) \]

Since \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), we obtain:
\[ I = I_0 \left( 1 + \frac{1}{\sqrt{2}} \right) \]

Thus, the intensity at this point is \( I_0 \left( 1 + \cos \frac{\pi}{4} \right) \). Quick Tip: In Young's double-slit experiment, the phase difference \( \delta \) is related to the path difference by \( \delta = \frac{2\pi}{\lambda} \times path difference \).

The focal length \( f \) of a lens in a given medium can be determined using the lens maker's formula:
\[ \frac{1}{f} = (n_{lens} - n_{medium}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

For a double convex lens, we have \( R_1 = 17 \, cm \) and \( R_2 = -17 \, cm \). The refractive index of the lens material (glass) is \( n_{lens} = 1.5 \), and the refractive index of the surrounding medium (water) is \( n_{medium} = 1.33 \). Substituting these values into the formula:
\[ \frac{1}{f} = (1.5 - 1.33) \left( \frac{1}{17} - \frac{1}{-17} \right) \]
\[ \frac{1}{f} = 0.17 \left( \frac{2}{17} \right) = 0.17 \times \frac{2}{17} = 0.02 \, cm^{-1} \]

Thus, the focal length is:
\[ f = \frac{1}{0.02} = 50 \, cm \]

Therefore, the focal length of the lens when submerged in water is \( 50 \, cm \). Quick Tip: The focal length of a lens changes when it is placed in a medium other than air, due to the variation in the relative refractive index.

The radius \( r_n \) of the electron's orbit in the Bohr model is given by: \[ r_n = r_1 n^2 \]
where \( r_1 = 0.53 \, \AA \) is the radius of the ground state orbit, and \( n \) is the principal quantum number.

The energy levels are given by: \[ E_n = -\frac{13.6 \, eV}{n^2} \]
For \( E_n = -1.51 \, eV \): \[ -1.51 = -\frac{13.6}{n_1^2} \implies n_1^2 = \frac{13.6}{1.51} \implies n_1 = 3 \]
For \( E_n = -3.40 \, eV \): \[ -3.40 = -\frac{13.6}{n_2^2} \implies n_2^2 = \frac{13.6}{3.40} \implies n_2 = 2 \]
The radii of the orbits are: \[ r_{n_1} = r_1 n_1^2 = 0.53 \times 9 = 4.77 \, \AA \] \[ r_{n_2} = r_1 n_2^2 = 0.53 \times 4 = 2.12 \, \AA \]
The change in radius is: \[ \Delta r = r_{n_1} - r_{n_2} = 4.77 - 2.12 = 2.65 \, \AA \]
Thus, the change in the radius of the orbit is \( 2.65 \, \AA \). Quick Tip: In the Bohr model, the radius of the electron's orbit is proportional to the square of the principal quantum number \( n \).

The number density of silicon atoms is \( 5 \times 10^{28} \, atoms/m^3 \), and the doping ratio is 1 dopant atom for every \( 5 \times 10^7 \) silicon atoms. Therefore, the dopant atom density per cubic metre is calculated as:
\[ Dopant density = \frac{5 \times 10^{28}}{5 \times 10^7} = 10^{21} \, atoms/m^3 \]

Since each dopant atom introduces one hole, the number of holes per cubic metre is \( 10^{21} \). To convert this value to per cubic centimetre:
\[ Holes per cm^3 = 10^{21} \times 10^{-6} = 10^{15} \, holes/cm^3 \]

An example of such a dopant is Boron (B), a trivalent impurity that creates p-type semiconductors. Quick Tip: Doping with trivalent impurities like Boron introduces holes in the semiconductor, resulting in p-type conductivity.

For two batteries connected in parallel, the equivalent emf \( E_{eq} \) and the equivalent internal resistance \( r_{eq} \) are calculated as follows:
\[ E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \] \[ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \]

Given:

\( E_1 = 3 \, V \), \( r_1 = 0.2 \, \Omega \)

\( E_2 = 6 \, V \), \( r_2 = 0.4 \, \Omega \)


(i) To calculate the equivalent emf: \[ E_{eq} = \frac{(3 \times 0.4) + (6 \times 0.2)}{0.2 + 0.4} = \frac{1.2 + 1.2}{0.6} = \frac{2.4}{0.6} = 4 \, V \]

(ii) To calculate the equivalent internal resistance: \[ r_{eq} = \frac{0.2 \times 0.4}{0.2 + 0.4} = \frac{0.08}{0.6} = 0.133 \, \Omega \]

(iii) To find the current drawn from the combination:
The total resistance of the circuit is: \[ R_{total} = r_{eq} + R = 0.133 + 4 = 4.133 \, \Omega \]
Now, using Ohm's law, the current \( I \) is: \[ I = \frac{E_{eq}}{R_{total}} = \frac{4}{4.133} = 0.968 \, A \] Quick Tip: When batteries are connected in parallel, the equivalent emf is a weighted average of the individual emfs, and the equivalent internal resistance is always lower than the smallest individual internal resistance.

(i) Relation between \( R' \) and \( R \):
Resistance \( R \) of a conductor is given by: \[ R = \rho \frac{l}{A} \]
When the length is increased to \( 2l \), the cross-sectional area \( A \) decreases to \( \frac{A}{2} \) (assuming volume remains constant). Thus: \[ R' = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R \]
So, \( R' = 4R \).

(ii) Relation between \( v_d' \) and \( v_d \):
Drift velocity \( v_d \) is given by: \[ v_d = \frac{I}{n e A} \]
When the length is doubled, the current \( I \) remains the same (since the cell is ideal), but the cross-sectional area \( A \) is halved. Thus: \[ v_d' = \frac{I}{n e (A/2)} = 2 \frac{I}{n e A} = 2v_d \]
So, \( v_d' = 2v_d \).

(ii) Direction of free electrons:
No, not all free electrons move in the same direction. Electrons move randomly due to thermal motion, but there is a net drift in the direction opposite to the electric field (from lower to higher potential). Quick Tip: When a conductor is stretched, its resistance increases due to the increase in length and decrease in cross-sectional area, while the drift velocity increases due to the reduced cross-sectional area.

(a) Magnetic Field \( \vec{B} \)
The magnetic field around a long straight current-carrying wire is described by Ampère’s law:
\[ B = \frac{\mu_0 I}{2\pi d} \]

According to the right-hand rule, the magnetic field direction at a distance \( d \) above the wire is along the positive \( \hat{k} \)-direction (out of the plane). Therefore, the magnetic field vector is:
\[ \vec{B} = \frac{\mu_0 I}{2\pi d} \hat{k} \]

(b) Magnetic Force \( \vec{F}_m \)
The force on a charged particle moving in a magnetic field is given by:
\[ \vec{F}_m = q (\vec{v} \times \vec{B}) \]

Given the following:
- \( \vec{v} = v \hat{i} \) (velocity along the \( x \)-axis),
- \( \vec{B} = B \hat{k} \) (magnetic field along the \( z \)-axis),

Using the cross product:
\[ \vec{v} \times \vec{B} = (v \hat{i}) \times (B \hat{k}) \]

From the vector identity \( \hat{i} \times \hat{k} = -\hat{j} \), we obtain:
\[ \vec{F}_m = q v B (-\hat{j}) \]

Thus, the magnetic force on the particle is:
\[ \vec{F}_m = -q v B \hat{j} \]

(c) Electric Field \( \vec{E} \)
Since the particle moves without deviation, the net force acting on it must be zero. This implies that the electric force \( \vec{F}_e = q \vec{E} \) must exactly cancel out the magnetic force. Therefore, we have:
\[ q \vec{E} = -\vec{F}_m \]

Substituting \( \vec{F}_m = -q v B \hat{j} \):
\[ q \vec{E} = q v B \hat{j} \]

Dividing both sides by \( q \):
\[ \vec{E} = v B \hat{j} \]

Thus, the required electric field is:
\[ \vec{E} = v B \hat{j} \] Quick Tip: To determine the direction of the magnetic field around a current-carrying wire, use the right-hand rule: Point your thumb in the direction of the current, and your fingers curl in the direction of the magnetic field \( \vec{B} \).

The impedance of the LCR circuit is given by: \[ Z = \sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2} \]
where \( R \) is the resistance, \( L \) is the inductance, and \( C \) is the capacitance. The phase difference \( \phi \) between the applied voltage and the current is given by: \[ \tan \phi = \frac{\omega L - \frac{1}{\omega C}}{R} \]
The current lags the voltage by the phase angle \( \phi \), which can be visualized in the phasor diagram.




Quick Tip: The phase difference between the voltage and current in an LCR circuit depends on the relative magnitudes of the inductive reactance (\( \omega L \)) and capacitive reactance (\( \frac{1}{\omega C} \)).

In an AC circuit with a parallel plate capacitor, the conduction current \( I_c \) refers to the current flowing through the resistor or conductive path, while the displacement current \( I_d \) is generated due to the changing electric field between the plates of the capacitor. The displacement current is given by the equation:
\[ I_d = \epsilon_0 A \frac{dE}{dt} \]

where \( A \) is the area of the capacitor plates, and \( \frac{dE}{dt} \) is the rate of change of the electric field. In a steady AC circuit, the conduction current and the displacement current are equal, ensuring that the total current remains continuous throughout the circuit. Quick Tip: In AC circuits, the displacement current is equal to the conduction current, which ensures that Kirchhoff’s current law still holds in circuits with capacitors.

Yes, Kirchhoff’s first rule (the junction rule) holds true at each plate of the capacitor. The junction rule states that the total current entering a junction must be equal to the total current leaving it. In the case of a capacitor, both the conduction current and the displacement current contribute to the overall current at the plates. The displacement current ensures the continuity of current flow in the capacitor, allowing Kirchhoff's current law to remain valid even at the capacitor plates. Quick Tip: Kirchhoff’s current law applies to circuits with capacitors as well, with the displacement current maintaining the continuity of charge flow.

The four features of the photoelectric effect that cannot be explained using the wave theory of light are:


1. The photoelectric effect occurs instantaneously when light of a frequency higher than the threshold frequency strikes the metal surface. The wave theory suggests that there should be a delay in the emission of electrons.

2. The kinetic energy of the emitted electrons depends on the frequency of the incident light, not its intensity.
According to wave theory, the energy should depend on the intensity of the light.

3. The photoelectric effect cannot take place below a certain threshold frequency, regardless of the light intensity. Wave theory predicts there should be no threshold, with intensity controlling the emission instead.
Quick Tip: Einstein’s explanation of the photoelectric effect is based on the particle theory of light, where light is considered as quanta (photons).

The relationship between the cut-off voltage \( V_{cut} \) and the frequency of light is given by:
\[ eV_{cut} = h f - \phi \]

where:

\( e \) is the electron charge,

\( V_{cut} \) is the cut-off voltage,

\( h \) is Planck's constant,

\( f \) is the frequency of the incident light, and

\( \phi \) is the work function of the metal.


The slope of the graph of \( V_{cut} \) versus frequency \( f \) is given by:
\[ Slope = \frac{h}{e} \]

Given that the slope is \( 4.12 \times 10^{-15} \, Vs \), we can calculate Planck's constant \( h \) using the formula:
\[ h = Slope \times e \]

Substituting the value of \( e = 1.6 \times 10^{-19} \, C \):
\[ h = (4.12 \times 10^{-15} \, Vs) \times (1.6 \times 10^{-19} \, C) = 6.592 \times 10^{-34} \, J·s \] Quick Tip: The slope of the plot of cut-off voltage versus frequency provides a direct method for calculating Planck’s constant using the equation \( eV_{cut} = h f - \phi \).

The circuit setup for studying the V-I characteristics of a p-n junction diode includes the following components:


1. A DC power supply to provide a variable voltage.

2. A p-n junction diode connected in series with a resistor.

3. A voltmeter to measure the voltage across the diode.

4. An ammeter to measure the current flowing through the diode.


The power supply is adjusted to apply both forward and reverse voltages to the diode. Current measurements are taken for various voltage values to plot the V-I characteristics. Quick Tip: In a standard V-I characteristic experiment, both forward and reverse biases are applied to observe the current's behavior across the p-n junction.

The V-I characteristics of a diode exhibit the following behavior:


1. In the forward bias region, when the applied voltage exceeds the threshold (usually around 0.7 V for silicon), the current increases exponentially as the voltage rises.

2. In the reverse bias region, the current remains extremely small (ideally zero) until the reverse breakdown voltage is reached.


The resulting graph shows an exponential increase in current in the forward bias region and a nearly flat response in the reverse bias region (until breakdown occurs). Quick Tip: The diode's V-I characteristic curve illustrates rectification: it permits current flow easily in the forward direction and blocks it in the reverse direction.

From the V-I characteristics of a diode, the following information can be determined:


1. The threshold voltage (or cut-off voltage), which is the minimum voltage needed for the diode to start conducting in the forward direction.

2. The reverse breakdown voltage, which is the voltage at which the diode begins to conduct in the reverse direction, potentially leading to damage.
Quick Tip: The threshold voltage is essential for understanding when a diode begins to conduct in the forward direction, while the reverse breakdown voltage indicates the diode’s tolerance to reverse bias.

Mass Defect:

The mass defect (\( \Delta m \)) is the difference between the total mass of the individual nucleons (protons and neutrons) in a nucleus and the actual measured mass of the nucleus. It is calculated as:
\[ \Delta m = Z m_p + (A - Z) m_n - m_{nucleus} \]

where:

\( Z \) is the number of protons,

\( A - Z \) is the number of neutrons,

\( m_p \) and \( m_n \) are the masses of a proton and a neutron, respectively,

\( m_{nucleus} \) is the actual mass of the nucleus.


Binding Energy:
Binding energy (\( E_b \)) is the energy needed to separate a nucleus into its individual protons and neutrons. According to Einstein’s mass-energy equivalence, it is given by:
\[ E_b = \Delta m \cdot c^2 \]

where:

\( c \) is the speed of light (\( 3.0 \times 10^8 \) m/s),

\( \Delta m \) is the mass defect.


Fission Process and Binding Energy Per Nucleon:
Nuclear fission occurs when a heavy nucleus splits into two or more lighter nuclei, releasing a significant amount of energy. This process can be explained using the concept of binding energy per nucleon:
\[ Binding Energy per Nucleon = \frac{E_b}{A} \]

For heavy nuclei (such as Uranium-235), the binding energy per nucleon is lower compared to that of medium-sized nuclei.
When a heavy nucleus undergoes fission, the resulting smaller nuclei have a higher binding energy per nucleon, which results in the release of energy.
This released energy is harnessed in nuclear power generation and atomic bombs. Quick Tip: The greater the binding energy per nucleon, the more stable the nucleus. The most stable nucleus in nature is Iron-56.

Step 1: Calculate the Mass Defect
Mass defect is given by: \[ \Delta m = (m_p + m_n) - m_{deuteron} \]

Substituting values: \[ \Delta m = (1.007277 + 1.008665) - 2.013553 \]
\[ \Delta m = 2.015942 - 2.013553 \]
\[ \Delta m = 0.002389 u \]

Step 2: Calculate the Binding Energy
Binding energy is given by: \[ E_b = \Delta m \times 931.5 MeV \]

Substituting \( \Delta m = 0.002389 \) u: \[ E_b = 0.002389 \times 931.5 \]
\[ E_b \approx 2.224 MeV \]

Thus, the mass defect is \( 0.002389 \) u, and the binding energy is \( 2.224 \) MeV. Quick Tip: Mass defect arises due to the conversion of missing mass into energy, which holds the nucleus together. This is why nuclear reactions release enormous energy.

The image formed by a thin lens can be determined using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

where:

\( f \) is the focal length,

\( v \) is the image distance,

\( u \) is the object distance.


A lens has two focal points, one on each side, where light either converges or diverges depending on the type of lens (convex or concave). Quick Tip: The lens formula is essential for understanding how the object distance, image distance, and focal length are related in lens systems.

For a concave lens, the magnification is always negative. This is because a concave lens forms a virtual, upright, and diminished image. The negative sign for magnification indicates that the image is virtual, which is a key characteristic of concave lenses. Since the image is virtual, it cannot be projected onto a screen, and it is smaller than the object. This means that option (C) is incorrect.

The other statements are consistent with the typical behavior of mirrors and lenses, as they describe properties that align with the physics of concave lenses. Quick Tip: In the case of mirrors and lenses, the magnification sign depends on the type of image formed—whether it is real or virtual—and its orientation (upright or inverted).

When a convex lens is cut into two equal parts perpendicular to the principal axis, the focal length of each part is halved. The new focal length \( f' \) of each part is: \[ f' = \frac{f}{2} \]
This is because the lens curvature increases when it is cut in half, effectively reducing the focal length. Quick Tip: Cutting a lens along its principal axis changes its curvature, and this reduces the focal length in this case by half.

To find the focal length, we use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

where:

\( u = -20 \, cm \) (object distance, taken as negative by convention),

\( v = 50 - 20 = 30 \, cm \) (image distance),

\( f \) is the focal length.


Substituting the given values into the formula:
\[ \frac{1}{f} = \frac{1}{30} - \frac{1}{-20} \]
\[ \frac{1}{f} = \frac{1}{30} + \frac{1}{20} \]

Now, to combine the fractions, we find the least common multiple (LCM) of 30 and 20:
\[ \frac{1}{f} = \frac{2}{60} + \frac{3}{60} = \frac{5}{60} \]

Solving for \( f \):
\[ f = \frac{60}{5} = 12 \, cm \]

Thus, the focal length of the lens is 12 cm, so the correct answer is (2). Quick Tip: When using the lens formula, remember to apply the appropriate sign conventions: the object distance is negative for real objects in convex lenses.

From the properties of a biconvex lens, the focal length \( f \) is given by the geometric mean of the distances \( X_1 \) and \( X_2 \): \[ f = \sqrt{X_1 X_2} \]

This relation is derived from the lens formula and paraxial approximation when the object and image distances are measured from the focal points.

Thus, the correct answer is (3) \( \sqrt{X_1 X_2} \). Quick Tip: The focal length of a biconvex lens can be estimated using the geometric mean of object and image distances when measured from their respective focal points.

When key \( S_1 \) is closed and \( S_2 \) is open, the capacitor fully charges, and the final charge \( Q \) on the capacitor is given by the formula:
\[ Q = C \cdot V \]

where:

\( C = 500 \, \mu F = 500 \times 10^{-6} \, F \) (capacitance),

\( V = 10 \, V \) (voltage).

Substituting the values:
\[ Q = 500 \times 10^{-6} \times 10 = 5 \, mC \]

Thus, the final charge on the capacitor is \( 5 \, mC \). Quick Tip: The charge on a capacitor is determined by multiplying its capacitance by the voltage across it.

When key \( S_1 \) is closed and \( S_2 \) is open, the capacitor begins charging and eventually reaches a voltage of \( V = 10 \, V \). At this point, the capacitor is fully charged. The charge \( Q \) on the capacitor is calculated using the formula:
\[ Q = C \cdot V \]

where:

\( C = 500 \, \mu F = 500 \times 10^{-6} \, F \) is the capacitance,

\( V = 10 \, V \) is the voltage.


Substituting the values into the equation:
\[ Q = 500 \times 10^{-6} \times 10 = 5 \, mC \]

Now, when key \( S_2 \) is closed, the capacitor is effectively disconnected from any additional charge path, meaning the charge on the capacitor cannot change. The charge remains constant because the capacitor is not connected to a discharge path or a load. Thus, the final charge on the capacitor, even after closing \( S_2 \), remains \( 5 \, mC \).

In conclusion, the final charge on the capacitor is \( 5 \, mC \), as the charge remains constant unless the capacitor is connected to a discharge path. Quick Tip: Once a capacitor is fully charged, its charge will remain the same unless it is connected to a discharge path or load, which would allow the charge to flow.

The dimensional formula for \( RC \) can be derived as follows:
- \( R \) (Resistance) has dimensions \( [M L^2 T^{-3} A^{-2}] \).
- \( C \) (Capacitance) has dimensions \( [M^{-1} L^{-2} T^4 A^2] \).

Thus, the dimensional formula for \( RC \) is: \[ [M L^2 T^{-3} A^{-2}] \] Quick Tip: Dimensional analysis helps in understanding the relationships between physical quantities, such as resistance and capacitance.

The current in the resistor during the charging process of the capacitor is given by: \[ I(t) = \frac{V}{R} e^{-t/(RC)} \]
After 5 seconds, the value of current will be: \[ I(5) = \frac{10}{20 \times 10^3} e^{-5/(RC)} \]
where \( R = 20 \, k\Omega \) and \( C = 500 \, \mu F \). Solving gives the final result of current after 5 seconds as \( \frac{1}{\sqrt{e}} \, mA \). Quick Tip: During the charging of a capacitor, the current decays exponentially as the capacitor charges up.

The initial current when the capacitor begins charging is given by Ohm's Law: \[ I(0) = \frac{V}{R} \]
where \( V = 10 \, V \) (battery voltage) and \( R = 20 \, k\Omega \). Therefore: \[ I(0) = \frac{10}{20 \times 10^3} = 0.5 \, mA \]
Thus, the initial current is \( 0.5 \, mA \). Quick Tip: The initial current during the charging of a capacitor is determined by the battery voltage and the resistance in the circuit, calculated using Ohm's Law.

Definition of Coherent Sources:
Coherent sources are two or more sources of light that emit waves with a constant phase difference and the same frequency.

Importance for Interference:

For a sustained interference pattern:

The phase difference between the waves must remain constant over time.

If the sources are incoherent, the phase difference changes randomly, leading to the destruction of the interference pattern.

Thus, coherent sources are essential to produce stable and visible interference fringes. Quick Tip: Laser light is an example of a coherent source, which is why it is commonly used in interference experiments.

Two independent sources of light are not coherent because:

They emit light waves independently, leading to random phase variations.

The atomic emissions of different sources are spontaneous and uncorrelated.

Even if they have the same frequency, their phase difference changes continuously.


As a result, two independent sources cannot produce sustained interference patterns. Quick Tip: To obtain coherent sources, a single source is often split into two, such as in Young’s double-slit experiment.

The fringe width in Young’s double-slit experiment is given by: \[ y = \frac{\lambda D}{d} \]
where:

\( \lambda = 600 \) nm \( = 600 \times 10^{-9} \) m (wavelength),

\( D = 1.2 \) m (distance to the screen),

\( d = 0.1 \) mm \( = 1.0 \times 10^{-4} \) m (slit separation).


Substituting values: \[ y = \frac{(600 \times 10^{-9}) (1.2)}{1.0 \times 10^{-4}} \]
\[ y = \frac{7.2 \times 10^{-4}}{10^{-4}} = 7.2 \times 10^{-3} m = 7.2 mm \]

Thus, the fringe width is 7.2 mm. Quick Tip: Fringe width increases with wavelength and distance to the screen but decreases with increasing slit separation.

The angular width is given by: \[ \theta = \frac{\lambda}{d} \]

Substituting values: \[ \theta = \frac{600 \times 10^{-9}}{1.0 \times 10^{-4}} \]
\[ \theta = 6.0 \times 10^{-3} rad \]

Converting to degrees: \[ \theta = 6.0 \times 10^{-3} \times \frac{180}{\pi} \]
\[ \theta \approx 0.034^\circ \]

Thus, the angular width is 0.034°. Quick Tip: Angular width is independent of the screen distance but depends on the wavelength and slit separation.

Definition of a Wavefront:

A wavefront is a surface of constant phase representing the positions of points in a wave that oscillate in unison.

Refraction through a Convex Lens:

A plane wavefront incident on a convex lens converges after refraction.

The refracted wavefront is spherical and converges towards the focal point of the lens. Quick Tip: Wavefronts change shape according to Huygens’ principle when passing through optical elements like lenses.

Using the lens-maker's formula for a spherical refracting surface: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{(n_2 - n_1)}{R} \]

where:

\( n_1 = 1.0 \) (air),

\( n_2 = 1.5 \) (glass),

\( u = \infty \) (distant object),

\( R = 15 \) cm (radius of curvature),

\( v \) is the image distance.


Since \( u = \infty \), the formula simplifies to: \[ \frac{1.5}{v} = \frac{0.5}{15} \]
\[ v = \frac{1.5 \times 15}{0.5} = 45 cm \]

Thus, the final image is formed at 45 cm inside the sphere. Quick Tip: For refraction at a curved surface, use the lens-maker’s equation, keeping sign conventions in mind.

The electrostatic potential energy \( U \) of a system of point charges is given by the formula: \[ U = \frac{1}{4 \pi \epsilon_0} \sum_{iwhere \( q_i \) and \( q_j \) are the point charges, and \( r_{ij} \) is the distance between them.

The change in energy due to the external electric field \( \vec{E} \) is given by the work done by the electric field on the system, which is: \[ \Delta U = - \sum_{i} q_i \vec{E} \cdot \vec{r_i} \]
Here, the charges are placed at \( (-3 \, cm, 0, 0) \) and \( (3 \, cm, 0, 0) \), and the electric field \( \vec{E} = \frac{A}{r^2} \hat{r} \) is acting on them.

The calculations will involve evaluating the work done on each charge and summing them up. Quick Tip: When an external electric field is applied, the change in electrostatic energy can be calculated by finding the work done by the electric field on the system of charges.

The capacitance \( C \) of the system can be calculated using the formula for capacitance: \[ C = \frac{Q}{V} \]
where:

\( Q = 80 \, \mu C = 80 \times 10^{-6} \, C \),

\( V = 16 \, V \).


Therefore, the capacitance is: \[ C = \frac{80 \times 10^{-6}}{16} = 5 \, \mu F \]


(ii) When the dielectric constant \( k = 3 \) is inserted, the capacitance increases by a factor of \( k \). The new capacitance \( C' \) becomes: \[ C' = kC = 3 \times 5 \, \mu F = 15 \, \mu F \]
Since \( Q = C'V' \), and the charge \( Q \) remains the same, the new potential difference \( V' \) is: \[ V' = \frac{Q}{C'} = \frac{80 \times 10^{-6}}{15 \times 10^{-6}} = 5.33 \, V \]


(iii) The capacitance of the system depends only on the geometry of the conductors and the dielectric constant of the medium between them, not the charges. Therefore, if the charges are doubled, the capacitance remains unchanged. The capacitance will still be \( 5 \, \mu F \), because capacitance is independent of the charge. Quick Tip: Capacitance depends on the geometry of the conductors and the dielectric material, not the amount of charge.

The electric field at a distance \( 3R \) from the center of a spherical shell depends only on the net charge enclosed and is given by Gauss’s law: \[ E = \frac{1}{4\pi \epsilon_0} \frac{Q_{net}}{r^2} \]
where \( Q_{net} \) is the total charge enclosed by each shell.

#### Step 1: Calculate Net Charge on Each Shell
- For Shell A:
\[ Q_A = 6q + (-2q) = 4q \]
- For Shell B:
\[ Q_B = -4q + 8q = 4q \]
- For Shell C:
\[ Q_C = 14q + (-10q) = 4q \]

Since the total charge enclosed for all three shells is the same (\( 4q \)), the magnitude of the electric field at a distance \( 3R \) is identical for all:
\[ E_A = E_B = E_C = \frac{1}{4\pi \epsilon_0} \frac{4q}{(3R)^2} \]

Thus, the electric fields due to shells A, B, and C at a distance \( 3R \) are equal. Quick Tip: According to Gauss’s law, the electric field outside a spherical shell behaves as if all the charge were concentrated at its center.

Work done in moving a charge in an electrostatic field depends only on the potential difference between the initial and final positions. The work done is given by: \[ W = q \Delta V \]
where:
- \( W \) is the work done,
- \( q = 5 \mu C = 5 \times 10^{-6} C \),
- \( \Delta V = V_A - V_C \) (potential difference between points A and C).

Since both C and A are on the same equipotential surface (same radial distance from the center B), their potential is the same: \[ V_A = V_C \]

Thus, the potential difference: \[ \Delta V = V_A - V_C = 0 \]
\[ W = (5 \times 10^{-6}) \times 0 = 0 \]

Thus, no work is done on the charge. Quick Tip: Work done in moving a charge along an equipotential surface is always zero because there is no change in electric potential.

The direction of the magnetic field at points P, Q, and R can be determined using the right-hand rule for the motion of charged particles in a magnetic field. The proton experiences a force that is always perpendicular to both its velocity and the magnetic field.

At point P, the magnetic field is directed into the page.

At point Q, the magnetic field is directed out of the page.

At point R, the magnetic field is again directed into the page.


The relative magnitude of the magnetic fields increases as the proton moves from point P to point Q, and decreases again as it moves from Q to R. Quick Tip: Use the right-hand rule to determine the direction of the magnetic field based on the direction of motion of a positively charged particle.

The magnetic moment \( \mu \) of a current loop is defined as: \[ \mu = I \cdot A \]
where \( I \) is the current and \( A \) is the area of the loop. The magnetic field at the centre of the loop due to the current is given by: \[ B = \frac{\mu_0 I A}{2R^2} \]
where \( R \) is the radius of the loop. Using the relationship between current and magnetic moment, we can express the current \( I \) in terms of the magnetic moment: \[ I = \frac{\mu}{A} \]
Substituting this into the equation for \( B \), we obtain the desired expression for the magnetic moment. Quick Tip: The magnetic moment of a current loop is directly proportional to both the current and the area of the loop.

The torque \( \tau \) on a current loop in a magnetic field is given by: \[ \tau = \mu B \sin \theta \]
where:

\( \mu \) is the magnetic moment of the loop,

\( B \) is the magnetic field strength,

\( \theta \) is the angle between the magnetic moment and the magnetic field.


For a rectangular loop, the magnetic moment \( \mu \) is: \[ \mu = I \cdot A \]
where \( I \) is the current and \( A \) is the area of the loop. The torque acts to rotate the loop until the magnetic moment aligns with the magnetic field. Quick Tip: Torque on a current loop in a magnetic field is maximum when the angle between the magnetic moment and the field is 90 degrees.

(1) The radius of the path will decrease. Since the kinetic energy is reduced by half, the velocity decreases. The radius of a charged particle's path in a magnetic field is given by: \[ r = \frac{mv}{qB} \]
Since the velocity \( v \) is halved, the radius will also be halved.

(2) The time period of revolution will remain unchanged. The time period of revolution \( T \) for a charged particle moving in a magnetic field is given by: \[ T = \frac{2\pi m}{qB} \]
Since the magnetic field \( B \) and the charge \( q \) are constant, the time period is independent of the kinetic energy, and hence remains the same. Quick Tip: A reduction in kinetic energy affects the velocity and radius of a charged particle's circular path but does not change its time period of revolution.