CBSE Class 12 Physics Question Paper 2024 (Set 1- 55/2/1) with Answer Key

The acceleration \( a \) of each particle is given by Newton’s second law: \[ F = ma \quad and \quad F = qE \quad \Rightarrow \quad a = \frac{qE}{m}. \]
Thus, the time taken for each particle to travel the same distance is given by the kinematic equation: \[ d = \frac{1}{2} a t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2dm}{qE}}. \]
The ratio of times \( t_p \) and \( t_q \) for particles P and Q is: \[ \frac{t_p}{t_Q} = \sqrt{\frac{m_p}{m_Q}}. \] Quick Tip: For this problem, remember that the time for a particle to travel a distance depends on the mass and charge, and use the kinematic equations effectively.

The drift velocity \( v_d \) is related to the potential difference \( V \) by the equation: \[ v_d \propto V. \]
Thus, when the potential difference is halved, the drift velocity will also be halved: \[ v_d \rightarrow \frac{v_d}{2}. \] Quick Tip: Drift velocity is directly proportional to the potential difference. Halving the potential difference will halve the drift velocity.

The magnetic moment \( M \) of a current-carrying loop is given by: \[ M = I A, \]
where \( I \) is the current and \( A \) is the area of the loop. The area of a circular loop is: \[ A = \pi r^2. \]
The circumference of the loop is \( 2\pi r = 4.4 \, m \), so: \[ r = \frac{4.4}{2\pi} = 0.7 \, m. \]
Now, calculating the area: \[ A = \pi (0.7)^2 = 1.54 \, m^2. \]
Thus, the magnetic moment is: \[ M = 1.0 \times 1.54 = 1.54 \, Am^2. \] Quick Tip: The magnetic moment of a circular loop is the product of the current and the area. Use the circumference to find the radius of the loop.

The magnetic flux \( \Phi_B \) linked with the coil is given by: \[ \Phi_B = B A \cos \theta. \]
Where:

\( B \) is the magnetic field,

\( A \) is the area of the coil,

\( \theta \) is the angle between the normal to the surface of the coil and the magnetic field.


The area of the coil is: \[ A = \pi r^2 = \pi (0.1)^2 = 0.0314 \, m^2. \]
The magnetic field \( B \) is the dot product of the vector \( \vec{B} \) and the unit vector normal to the surface of the coil: \[ \vec{B} \cdot \hat{n} = (1.0 \hat{i} + 0.5 \hat{j}) \cdot (0.6 \hat{i} + 0.8 \hat{j}) = 1.0 \times 0.6 + 0.5 \times 0.8 = 0.6 + 0.4 = 1.0 \, mT. \]
Thus, the magnetic flux is: \[ \Phi_B = 1.0 \times 0.0314 = 0.0314 \, Wb = 31.4 \, \mu Wb. \] Quick Tip: For calculating magnetic flux, remember to use the dot product to find the effective magnetic field component normal to the coil's surface.

In an ideal transformer:
The magnetic flux remains the same in both the primary and secondary coils because the transformer operates based on mutual induction, where the same magnetic flux links both coils.

The power is conserved in an ideal transformer, meaning the input power (in the primary coil) is equal to the output power (in the secondary coil), ignoring losses.


Thus, the quantities that remain the same are: \[ \boxed{Magnetic flux and Power both}. \] Quick Tip: In an ideal transformer, the power is conserved, and the magnetic flux linking the primary and secondary coils remains unchanged.

When a resistor and an ideal inductor are connected in series, the total voltage across both components is the supply voltage \( 100 \sqrt{2} \, V \), which is the RMS value.

The voltmeter reading across the resistor and the inductor is the same because the voltage drop across each component will be equal. This happens because the total impedance of the circuit is such that both the resistor and the inductor have the same voltage drop.
Step by Step Solution:

Step 1:
Let the voltage across the resistor be \( V_R \) and the voltage across the inductor be \( V_L \). Given that \( V_R = V_L \).

Step 2:
The total voltage \( V_{total} \) is given by the vector sum of \( V_R \) and \( V_L \). Since they are in quadrature, we use the Pythagorean theorem: \[ V_{total} = \sqrt{V_R^2 + V_L^2} \]

Step 3:
Substitute \( V_R = V_L \) into the equation: \[ V_{total} = \sqrt{V_R^2 + V_R^2} = \sqrt{2V_R^2} = V_R \sqrt{2} \]

Step 4:
Given that \( V_{total} = 100\sqrt{2} \, V \), we can solve for \( V_R \): \[ 100\sqrt{2} = V_R \sqrt{2} \quad \Rightarrow \quad V_R = 100 \, V \]


Thus, the reading of the voltmeter is: \[ \boxed{100 \, V}. \] Quick Tip: In a series L-R circuit, the voltage across each component is the same if they share the total supply voltage.

Electromagnetic waves are classified according to their wavelengths. A wavelength of 10 nm corresponds to gamma rays, which are the highest frequency and shortest wavelength electromagnetic waves. This is much shorter than the wavelength for ultraviolet rays or X-rays. Quick Tip: Electromagnetic waves with wavelengths smaller than 100 nm are typically considered gamma rays.

The cut-off wavelength \( \lambda_{cut} \) can be found using the photoelectric equation: \[ E_{photon} = \phi = \frac{hc}{\lambda_{cut}}, \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \phi \) is the work function. Rearranging to solve for \( \lambda_{cut} \): \[ \lambda_{cut} = \frac{hc}{\phi}. \]
Substitute \( h = 6.626 \times 10^{-34} \, J\cdots \), \( c = 3 \times 10^8 \, m/s \), and \( \phi = 3.315 \, eV = 3.315 \times 1.602 \times 10^{-19} \, J \): \[ \lambda_{cut} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{3.315 \times 1.602 \times 10^{-19}} \approx 375 \, nm. \] Quick Tip: The cut-off wavelength is inversely proportional to the work function. Higher work functions result in shorter cut-off wavelengths.

The wavelengths of radiation corresponding to atomic transitions are related by the energy difference between the levels. According to the Rydberg formula for transitions: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \]
where \( R \) is the Rydberg constant and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states, respectively. For the transitions described, the relation between the wavelengths is: \[ \frac{1}{\lambda_1} + \frac{1}{\lambda_2} = \frac{1}{\lambda_3}. \] Quick Tip: For transitions between energy levels, use the Rydberg formula to relate wavelengths to energy differences.

To determine the relationship between the distance \( d \) at which the alpha particle stops and its initial kinetic energy \( K \), we can use the principles of energy conservation and the Coulomb force.

1. Initial Kinetic Energy: The alpha particle has an initial kinetic energy \( K \).

2. Potential Energy at Distance \( d \): When the alpha particle stops, all its kinetic energy is converted into electrostatic potential energy. The potential energy \( U \) between the alpha particle (charge \( +2e \)) and the gold nucleus (charge \( +79e \)) at a distance \( d \) is given by:
\[ U = \frac{1}{4\pi\epsilon_0} \cdot \frac{(2e)(79e)}{d} \]
where \( \epsilon_0 \) is the permittivity of free space.

3. Energy Conservation: At the point where the alpha particle stops, its kinetic energy is zero, and its potential energy equals the initial kinetic energy:
\[ K = \frac{1}{4\pi\epsilon_0} \cdot \frac{158e^2}{d} \]

4. Solving for \( d \): Rearrange the equation to solve for \( d \):
\[ d = \frac{1}{4\pi\epsilon_0} \cdot \frac{158e^2}{K} \]
This shows that \( d \) is inversely proportional to \( K \):
\[ d \propto \frac{1}{K} \]

Therefore, the correct answer is:
\[ \boxed{C} \] Quick Tip: The distance at which a charged particle stops due to Coulomb's interaction is inversely proportional to the square root of its kinetic energy.

An n-type semiconductor is formed by doping an intrinsic semiconductor with a group V element (such as phosphorus), which has five valence electrons. These extra electrons contribute to conductivity. In this case, phosphorus (P) is used for doping silicon to form an n-type semiconductor. Quick Tip: For n-type semiconductors, use elements from group V (like phosphorus) to provide extra electrons.

In reverse biasing, the external voltage opposes the potential barrier of the p-n junction, which causes the depletion region to widen. This happens because the external voltage increases the barrier height, preventing current from flowing (except for a very small leakage current). Quick Tip: In reverse bias, the barrier height increases and the depletion region widens, leading to the prevention of current flow.

When the intensity of incident radiation increases (for a constant frequency), the number of photons hitting the surface increases. This results in the emission of more photoelectrons per second, thereby increasing the photocurrent. This explanation aligns perfectly with the assertion. Quick Tip: The intensity of radiation affects the number of photoelectrons emitted, which in turn increases the photocurrent.

Lenz’s law is a consequence of the conservation of energy, as it ensures that the induced current always flows in a direction that opposes the change in magnetic flux. The statement that there is no power loss in an ideal inductor is also true because an ideal inductor does not dissipate energy; instead, it stores energy in the magnetic field. However, this does not directly explain Lenz’s law, making option (2) the correct choice. Quick Tip: Lenz’s law follows the principle of conservation of energy, while an ideal inductor stores energy without dissipation, making both statements true but unrelated in explanation.

The assertion is correct because when an electron and a proton enter the magnetic field with the same momentum, they experience the same force due to their charge and follow a circular path with the same radius. The radius of the circular path is given by the formula:
\[ r = \frac{p}{qB} \]

Since both have the same momentum and the formula depends on \( p \), the radii of their circular paths will be the same. Thus, both the assertion and the reason are true, and the reason correctly explains the assertion. Quick Tip: For a charged particle in a magnetic field, the radius of the circular motion depends on its momentum, charge, and magnetic field strength.

The assertion is correct because the magnifying power of a compound microscope is considered negative, as the final image is inverted relative to the object. However, the reason is incorrect because the final image formed is not erect but rather inverted with respect to the object. Therefore, the assertion is true, but the reason is false. Quick Tip: A compound microscope produces an inverted image, and its magnifying power is conventionally taken as negative.

Resistivity \( \rho \) is the property of a material that quantifies how strongly it resists the flow of electric current. It is given by the formula: \[ \rho = \frac{m}{ne^2\tau}, \]
where:
 \( m \) is the mass of the electron,
 \( e \) is the charge of the electron,
 \( n \) is the number density of free electrons in the conductor,
 \( \tau \) is the relaxation time.

(a) Resistivity is inversely proportional to the number density of free electrons: \[ \rho \propto \frac{1}{n}. \]
(b) Resistivity is directly proportional to the relaxation time \( \tau \): \[ \rho \propto \frac{1}{\tau}. \] Quick Tip: Resistivity depends on the material properties such as the number of free electrons and their ability to move (relaxation time).

Let the equations of the two waves be: \[ x_1 = a \cos(\omega t), \] \[ x_2 = a \cos(\omega t + \phi), \]
where \( a \) is the amplitude, \( \omega \) is the frequency, and \( \phi \) is the phase difference.

The resultant displacement \( x \) is: \[ x = x_1 + x_2 = a \cos(\omega t) + a \cos(\omega t + \phi) = a ( \cos(\omega t) + \cos(\omega t + \phi) ). \]
Using the trigonometric identity: \[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right), \]
we get: \[ x = 2a \cos\left(\frac{\phi}{2}\right) \cos\left(\omega t + \frac{\phi}{2}\right). \]
The intensity \( I \) is proportional to the square of the amplitude: \[ I = K (Amplitude)^2 = K \left(2a \cos\left(\frac{\phi}{2}\right)\right)^2 = 4K a^2 \cos^2\left(\frac{\phi}{2}\right). \]
Let \( I_0 = Ka^2 \) be the intensity of each incident wave. Thus, the resultant intensity is: \[ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right). \] Quick Tip: The intensity of two superimposed waves depends on the square of the cosine of half the phase difference.

(i) As the source slit is moved closer to the plane of the slits, the sharpness of the interference pattern decreases. This happens because the light from different parts of the source overlaps and the fringes begin to blur, eventually disappearing if the slit is moved too close. The fringe sharpness is affected by the ratio \( \frac{s}{d} \), where \( s \) is the distance between the source and the slits, and \( d \) is the separation between the slits. If \( s \) decreases, the sharpness of the interference pattern decreases.

(ii) As the separation between the two slits increases, the fringe spacing \( \beta \) decreases. The fringe spacing is given by: \[ \beta = \frac{\lambda D}{d}, \]
where \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( d \) is the slit separation. As \( d \) increases, \( \beta \) decreases, and the fringes become closer together. Quick Tip: The sharpness of interference fringes depends on the relative distances between the source, slits, and screen. Adjusting these distances changes the fringe spacing.

The lens maker’s formula is: \[ \frac{1}{f} = \left(\frac{n_g}{n_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

For the convex lens in air: \[ \frac{1}{f_a} = \left(\frac{n_g}{n_a} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

For the convex lens in liquid: \[ \frac{1}{f_l} = \left(\frac{n_g}{n_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Taking the ratio: \[ \frac{f_l}{f_a} = \frac{\frac{1}{n_g - 1}}{\frac{1}{n_g - n_l}} \]

Substituting values: \[ \frac{f_l}{f_a} = \frac{1.52 - 1}{1.52 - 1.65} \]
\[ = \frac{0.52}{-0.13} = -6.6 \]

Thus: \[ f_l = -6.6 \times 15 = -99 cm. \]

Nature of the Lens:
Since the focal length is negative, the lens behaves as a diverging (concave) lens in the liquid. Quick Tip: Immersing a lens in a liquid with a higher refractive index decreases the focal length and may change the nature of the lens.

The binding energy of a nucleus is given by the formula: \[ B.E. = \left( Zm_p + (A - Z)m_n - M_N \right) \times 931.5 \, MeV, \]
where:
 \( Z \) is the atomic number,
 \( A \) is the mass number,
 \( m_p \) is the mass of a proton,
 \( m_n \) is the mass of a neutron,
 \( M_N \) is the nuclear mass.

For \( {}^{12}_6C \), \( Z = 6 \), \( A = 12 \), and the given masses are: \[ m_p = 1.007825 \, u, \, m_n = 1.008665 \, u, \, M_N = 12.000000 \, u. \]
Thus, the binding energy is: \[ B.E. = \left( 6 \times 1.007825 + 6 \times 1.008665 - 12.000000 \right) \times 931.5 \, MeV. \]
After calculation: \[ B.E. = 92.16 \, MeV. \] Quick Tip: The binding energy is a measure of the energy required to disassemble the nucleus into its constituent protons and neutrons.

(i) When a semiconductor is doped with a trivalent impurity (like boron), an acceptor level is created just above the valence band. This decreases the energy gap because the electrons in the valence band can now easily jump to the acceptor level, making the material more conductive.

(ii) When a semiconductor is doped with a pentavalent impurity (like phosphorus), a donor level is created just below the conduction band. This also decreases the energy gap, as electrons can easily be excited from the donor level into the conduction band, increasing conductivity. Quick Tip: Doping with trivalent impurities creates acceptor levels, while doping with pentavalent impurities creates donor levels, both affecting the energy gap.

By applying Kirchhoff’s laws to the circuit, we can determine the currents in the different branches.

First, we use Kirchhoff’s current law at point B: \[ I_1 + I_2 = I_3 \]
Next, applying Kirchhoff’s loop rule for the closed loop AGFBA: \[ 3 + 2I_3 - 6 + 4I_2 + 2I_3 = 0 \]
Simplifying: \[ I_2 + I_3 = \frac{3}{4} \]
We substitute this in another equation based on another loop, BFDCB: \[ -4I_2 + 6 + 2I_1 - 6 + 2I_1 = 0 \]
Which simplifies to: \[ I_2 = I_1 \]
By solving the equations, we find: \[ I_1 = \frac{1}{4} \, A, \, I_2 = \frac{1}{4} \, A, \, I_3 = \frac{1}{2} \, A. \] Quick Tip: Kirchhoff’s laws are helpful for analyzing circuits with multiple batteries and resistors to determine currents and voltages.

(a) The speed \( v \) of an electromagnetic wave in a medium depends on:
The permittivity \( \epsilon \) of the medium.
The magnetic permeability \( \mu \) of the medium.
The speed of EM waves is given by: \[ v = \frac{1}{\sqrt{\mu \epsilon}}. \]

(b) Electromagnetic waves are produced by oscillating or accelerated charges, such as electrons in an antenna. These oscillating charges generate both electric and magnetic fields that propagate through space as an electromagnetic wave.

(c) The electric and magnetic fields for an electromagnetic wave propagating along the z-axis are perpendicular to each other and the direction of propagation. The electric field oscillates along the x-axis and the magnetic field along the y-axis, as shown in the diagram:




\[ Electric Field (E) \quad and \quad Magnetic Field (B) \quad in the direction of propagation (z-axis). \] Quick Tip: The speed of electromagnetic waves in a medium depends on the permittivity and permeability of the medium.

The induced emf \( \varepsilon \) is given by Faraday’s law of induction: \[ \varepsilon = -N \frac{d\phi}{dt}. \]
Here, \( N \) is the number of turns in the coil, and \( \phi \) is the magnetic flux. The flux is related to the magnetic field \( B \) and the area \( A \) of the coil.

The magnetic flux changes due to the change in current in the solenoid, causing an induced emf. The induced emf is: \[ \varepsilon = -N \mu_0 n \pi r^2 \frac{dI}{dt}. \]
Given values:
\( N = 100 \),

\( n = 250 \, turns/cm \),

\( r = 1.6 \, cm \),

\( dI = 1.5 \, A \),

\( dt = 25 \, ms \).


Substituting the values: \[ \varepsilon = 0.1536 \, V. \]
The current induced in the coil is: \[ I = \frac{\varepsilon}{R} = \frac{0.1536}{5} = 0.03 \, A. \] Quick Tip: Induced current depends on the rate of change of magnetic flux and the resistance of the coil.

The force between two parallel conductors carrying currents is due to the magnetic field produced by one conductor acting on the other. The force is repulsive if the currents are in opposite directions and attractive if the currents are in the same direction.

The magnitude of the force per unit length between two conductors is given by Ampère's force law: \[ F = \frac{\mu_0 I_1 I_2}{2 \pi d}, \]
where:
 \( I_1 \) and \( I_2 \) are the currents in the two conductors,
 \( d \) is the distance between them,
 \( \mu_0 \) is the permeability of free space.

One ampere is defined as the current that, when flowing through two parallel conductors one meter apart, produces a force of \( 2 \times 10^{-7} \, N/m \) on each conductor. Quick Tip: The force between two conductors depends on the current in each conductor and the distance between them.

The torque on a current loop in a magnetic field is given by the expression: \[ \tau = I A B \sin \theta, \]
where:
 \( I \) is the current in the loop,
 \( A \) is the area of the loop,
 \( B \) is the magnetic field strength,
 \( \theta \) is the angle between the normal to the loop and the magnetic field.

The forces on arms BC and DA are equal and opposite and cancel each other out, while the forces on arms AB and CD form a couple, leading to a torque.
\[ \tau = I A B \sin \theta. \] Quick Tip: The torque on a current loop in a magnetic field depends on the current, area of the loop, and the angle between the magnetic field and the loop's normal.

From Bohr’s second postulate, the angular momentum of the electron in the \( n \)-th orbit is quantized: \[ m v r = \frac{nh}{2 \pi}, \]
where \( m \) is the mass of the electron, \( v \) is its velocity, \( r \) is the radius of the orbit, and \( n \) is the principal quantum number.

Also, from Coulomb's law and centripetal force, we have: \[ \frac{m v^2}{r} = \frac{e^2}{4 \pi \epsilon_0 r^2}, \]
where \( e \) is the charge of the electron and \( \epsilon_0 \) is the permittivity of free space.

By solving these equations, we get the expression for the radius of the \( n \)-th orbit: \[ r = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}. \]
For \( n = 1 \), this is the Bohr radius \( a_0 \): \[ a_0 = \frac{6.63 \times 10^{-34} \times 8.854 \times 10^{-12}}{3.14 \times 9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^2} = 5.29 \times 10^{-11} \, m = 0.53 \, Å. \] Quick Tip: Bohr’s radius gives the size of the smallest orbit for an electron in a hydrogen atom.

(a)
The de Broglie wavelength is given by:
\[ \lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2m}} \times \frac{1}{\sqrt{K}} \]

Comparing with the equation of a straight line (\( y = mx + c \)), the slope of the line is:
\[ slope = \frac{h}{\sqrt{2m}} \]

Thus, the slope of the line represents \( \frac{h}{\sqrt{2m}} \), which is inversely proportional to \( \sqrt{m} \).


(b)
From the equation:
\[ slope \propto \frac{1}{\sqrt{m}} \]

Since the slope of \( m_2 \) is greater than that of \( m_1 \), it implies that \( m_2 \) is lighter, and \( m_1 \) is heavier.


(c)
No, this graph is not valid for a photon. The given equation for momentum:
\[ p = \sqrt{2mK} \]

is not applicable to a photon, as a photon has zero rest mass and its momentum is given by:
\[ p = \frac{h}{\lambda} \] Quick Tip: For massive particles, the de Broglie wavelength follows \( \lambda = \frac{h}{\sqrt{2mK}} \), while for photons, momentum is given by \( p = \frac{h}{\lambda} \), making the given relation inapplicable.

A full-wave rectifier uses two diodes arranged in a bridge configuration. The AC input is applied to the transformer, and the diodes rectify the current to produce a pulsating DC output.

Working:
1. During the positive half-cycle of the input AC voltage, the current flows through the diode D1, and diode D2 is reverse biased and does not conduct. Therefore, current flows through the load resistor \( R_L \) in one direction.
2. During the negative half-cycle, diode D1 is reverse biased and does not conduct, but diode D2 becomes forward biased and allows current to flow through the load resistor in the same direction.

This results in a unidirectional output current.
\[ Input waveform \quad and \quad Output waveform \]




The input AC voltage alternates in both positive and negative cycles, while the output remains in one direction, providing a pulsating DC. Quick Tip: A full-wave rectifier produces an output that is smoother and more consistent than half-wave rectification by using both positive and negative cycles of the input signal.

The incorrect statement is: (B) The internal resistance of a cell decreases with the decrease in temperature of the electrolyte. The internal resistance actually increases as the temperature of the electrolyte decreases. Lower temperatures hinder the movement of ions, which increases the resistance.

For cells connected in parallel, the equivalent emf \( \varepsilon_{eq} \) is given by the formula: \[ \varepsilon_{eq} = \frac{r_1 \varepsilon_2 + r_2 \varepsilon_1}{r_1 + r_2}, \]
where \( \varepsilon_1 \) and \( \varepsilon_2 \) are the emf values of the two cells, and \( r_1 \) and \( r_2 \) are their respective internal resistances.

Substituting the given values: \[ \varepsilon_{eq} = \frac{0.1 \times 6.0 + 0.4 \times 2.0}{0.1 + 0.4} = \frac{0.6 + 0.8}{0.5} = 2.8 \, V. \]

Thus, the correct answer is: \[ \boxed{(B)} \quad 2.8 \, V. \] Quick Tip: When cells are connected in parallel, the equivalent emf depends on the internal resistances and emfs of the cells.

The electromotive force (emf) of the cell is given by the sum of the potential differences between the positive and negative electrodes relative to the electrolyte. Since both \( V_+ \) and \( V_- \) are positive, their sum is also positive, making option (A) correct.
 

The internal resistance \( r \) is given by:
\[ V = E - Ir \]

Substituting the given values:
\[ 4 = 6 - (2 \times r) \]
\[ 2r = 2 \]
\[ r = 1.0 \Omega \] Quick Tip: The internal resistance of a cell affects the terminal voltage when a current is drawn. For parallel-connected identical cells, the total emf remains the same, but the internal resistance is reduced.

The angle of incidence at the critical angle is the angle at which the refracted ray grazes the interface, i.e., it refracts along the boundary. At this point, the angle of refraction is \(90^\circ\), so by the law of reflection, the angle of reflection must also be \(90^\circ\).

Thus, the correct answer is:
\[
\boxed{\text{(D) } 90^\circ}.
\]

When light passes from one medium to another, the wavelength of the light changes. The relationship between the wavelengths in the two media is given by:
\[
\lambda_2 = \lambda_1 \frac{v_2}{v_1} = \lambda_1 \frac{n_1}{n_2},
\]
where \( \lambda_1 \) is the wavelength in the first medium (water), and \( \lambda_2 \) is the wavelength in the second medium (air).

Given:
 \( \lambda_1 = 600 \, \text{nm} \),
 \( n_1 = \frac{4}{3} \) (water),
  \( n_2 = 1 \) (air).

Thus:
\[
\lambda_2 = 600 \times \frac{4}{3} = 800 \, \text{nm}.
\]

Thus, the correct answer is:
\[
\boxed{\text{(D) } 800 \, \text{nm}}.
\]

given
1. Medium A is denser than medium B.
2. The incident ray PQ in medium A makes an angle of \( 30^\circ \) with the horizontal.
3. The refracted ray in medium B is parallel to the interface AB.
4. We need to find the refractive index of medium B with respect to medium A (\( \mu_{BA} \)).

Step 1: Understand the geometry and angles
 The incident ray PQ makes an angle of \( 30^\circ \) with the horizontal. This means the angle of incidence (\( i \)) with respect to the normal is:
  \[
  i = 90^\circ - 30^\circ = 60^\circ
  \]
The refracted ray is parallel to the interface AB. This means the angle of refraction (\( r \)) is \( 90^\circ \) (since the refracted ray is parallel to the interface).

Step 2: Apply Snell's Law
Snell's Law states:
\[
\mu_A \sin i = \mu_B \sin r
\]
where:
 \( \mu_A \) = refractive index of medium A,
 \( \mu_B \) = refractive index of medium B,
 \( i \) = angle of incidence,
 \( r \) = angle of refraction.

From the problem:
 \( i = 60^\circ \),
 \( r = 90^\circ \).

Substitute these values into Snell's Law:
\[
\mu_A \sin 60^\circ = \mu_B \sin 90^\circ
\]

We know:
 \( \sin 60^\circ = \frac{\sqrt{3}}{2} \),
 \( \sin 90^\circ = 1 \).

So:
\[
\mu_A \cdot \frac{\sqrt{3}}{2} = \mu_B \cdot 1
\]

Step 3: Find the refractive index of medium B with respect to medium A
The refractive index of medium B with respect to medium A is defined as:
\[
\mu_{BA} = \frac{\mu_B}{\mu_A}
\]

From the equation in Step 2:
\[
\mu_B = \mu_A \cdot \frac{\sqrt{3}}{2}
\]

Divide both sides by \( \mu_A \):
\[
\mu_{BA} = \frac{\mu_B}{\mu_A} = \frac{\sqrt{3}}{2}
\]

The refractive index of medium B with respect to medium A is:
\[
\boxed{\frac{\sqrt{3}}{2}}
\]

Thus, the correct option is (1).

Using Snell’s law for the critical angle \( \theta_c \): \[ n_A \sin \theta_c = n_B \sin 90^\circ = 1. \]
Hence, \[ \sin \theta_c = \frac{v_A}{v_B} = \frac{2 \times 10^8}{2.5 \times 10^8} = \frac{4}{5}. \]
Thus, the critical angle is: \[ \theta_c = \sin^{-1} \left( \frac{4}{5} \right). \]

Thus, the correct answer is: \[ \boxed{(B) \sin^{-1} \frac{4}{5}}. \]

For the given triangular prism, the angle \( \theta \) is related to the refractive index of the material of the prism. Using Snell's law and geometrical principles, the relationship is given by: \[ \boxed{(A) \sin^{-1} \sqrt{n^2 - 1}}. \] Quick Tip: In cases involving refraction and critical angles, remember to use Snell's Law, \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), to calculate refractive indices or angles accurately.

(a) (i) Expression for electric potential: 
The electric potential due to a dipole is the sum of potentials due to the charges \( +q \) and \( -q \) separated by distance \( 2a \):

\[
V = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r_1} - \frac{q}{r_2} \right) 
\]

Using the geometry of the dipole, we have:

\[
r_1^2 = r^2 + a^2 - 2a r \cos\theta, \quad r_2^2 = r^2 + a^2 + 2a r \cos\theta
\]

For \( r \gg a \), using binomial expansion and retaining first-order terms:

\[
\frac{1}{r_1} \approx \frac{1}{r} \left(1 - \frac{a \cos\theta}{r} \right), \quad \frac{1}{r_2} \approx \frac{1}{r} \left(1 + \frac{a \cos\theta}{r} \right)
\]

Substituting these into the potential equation:

\[
V = \frac{q}{4\pi\epsilon_0} \left( \frac{2a \cos\theta}{r^2} \right)
\]

Since dipole moment \( p = 2qa \), we get:

\[
V = \frac{p \cos\theta}{4\pi\epsilon_0 r^2}
\]

Potential energy of the system:

\[
U = \frac{k q q_2}{a} + \frac{k q_2 q_3}{a} + \frac{k q_3 q}{a}
\]

Substituting given values:

\[
U = \frac{k(q)(2q)}{a} + \frac{k(2q)(nq)}{a} + \frac{k(q)(nq)}{a} = 0
\]

\[
\frac{2q^2}{a} + \frac{2nq^2}{a} + \frac{nq^2}{a} = 0
\]

\[
2 + 2n + n = 0 \quad \Rightarrow \quad 3n = -2
\]

\[
n = -\frac{2}{3}
\]

Gauss’s Law states:

\[
\oint \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0}
\]

For a uniformly charged infinite sheet, the electric field is given by:

\[
E = \frac{\sigma}{2\epsilon_0}
\]
 

The electric field due to a long straight wire is given by:



\[ E = \frac{\lambda}{2\pi\epsilon_0 r} \]

For the two wires:
\[ E_1 = \frac{\lambda_1}{2\pi\epsilon_0 r_1}, \quad E_2 = \frac{\lambda_2}{2\pi\epsilon_0 r_2} \]

Substituting the given values:
\[ E_1 = \frac{10 \times 10^{-6}}{2\pi\epsilon_0 (10 \times 10^{-2})} (-\hat{j}) \]
\[ E_2 = \frac{20 \times 10^{-6}}{2\pi\epsilon_0 (20 \times 10^{-2})} (-\hat{j}) \]

Net electric field:
\[ E_{net} = \frac{10 \times 10^{-6}}{2\pi\epsilon_0} \left(\frac{1}{0.1} + \frac{2}{0.2} \right) (-\hat{j}) \]
\[ E_{net} = 3.6 \times 10^6 (-\hat{j}) N/C \]

Force on the electron:
\[ F_{net} = qE_{net} \]
\[ F_{net} = (-1.6 \times 10^{-19}) \times (3.6 \times 10^6) N \]
\[ F_{net} = 5.76 \times 10^{-13} N (\hat{j}) \] Quick Tip: The electric potential due to a dipole follows an inverse square law, whereas for a charged infinite plane, the electric field remains uniform and independent of distance.

The motion of a charged particle in a magnetic field can be analyzed into two components of velocity:
1. A component \( v_{\perp} \) perpendicular to the magnetic field, causing circular motion.
2. A component \( v_{\parallel} \) parallel to the magnetic field, causing linear motion along the direction of the field.

The total velocity \( v \) is the vector sum of these two components: \[ v_{\perp} = v \sin \theta, \quad v_{\parallel} = v \cos \theta. \]
The perpendicular component causes circular motion, and the parallel component causes linear motion along the magnetic field direction, resulting in a helical path.

The magnetic force provides the centripetal force for the circular motion, which gives the radius \( r \) of the circular path: \[ \frac{mv_{\perp}^2}{r} = q v_{\perp} B. \]
Solving for \( r \): \[ r = \frac{mv_{\perp}}{qB}. \]

The time period \( T \) for one complete revolution is the circumference of the circle divided by the velocity: \[ T = \frac{2\pi r}{v_{\perp}} = \frac{2\pi m}{qB}. \]

The frequency of revolution \( \nu \) is the reciprocal of the time period: \[ \nu = \frac{1}{T} = \frac{qB}{2\pi m}. \]

Thus, the particle follows a helical path with a frequency of revolution \( \nu = \frac{qB}{2\pi m} \).


Quick Tip: A charged particle moving in a magnetic field will follow a helical path due to the combination of circular motion caused by the perpendicular velocity component and linear motion caused by the parallel velocity component.

We are given the radius \( r = 2 \, Å = 2 \times 10^{-10} \, m \) and the frequency of revolution \( \nu = 8 \times 10^{14} \, rev/s \).

The magnetic moment \( m \) of a moving charge is given by: \[ m = I \cdot A, \]
where \( I \) is the current due to the motion of the electron, and \( A \) is the area of the circular orbit traced by the electron.

The current \( I \) is related to the charge and the frequency of revolution: \[ I = q \cdot \nu. \]
Substituting the known values for charge of the electron \( q = 1.6 \times 10^{-19} \, C \) and frequency \( \nu = 8 \times 10^{14} \, rev/s \): \[ I = 1.6 \times 10^{-19} \cdot 8 \times 10^{14} = 1.28 \times 10^{-4} \, A. \]

The area \( A \) of the circular orbit is: \[ A = \pi r^2 = \pi (2 \times 10^{-10})^2 = 1.26 \times 10^{-19} \, m^2. \]

Thus, the magnetic moment is: \[ m = I \cdot A = (1.28 \times 10^{-4}) \cdot (1.26 \times 10^{-19}) = 1.61 \times 10^{-23} \, Am^2. \]

The magnetic moment associated with the orbital motion of the electron is \( 1.61 \times 10^{-23} \, Am^2 \).


Quick Tip: The magnetic moment of a particle moving in a circular path is proportional to the current generated by the motion and the area enclosed by the path.

The current sensitivity of a galvanometer is defined as the deflection produced per unit current passing through it. It is given by the formula: \[ I_s = \frac{\theta}{I} = \frac{NBA}{K}, \]
where \( \theta \) is the angular deflection, \( I \) is the current, \( N \) is the number of turns of the coil, \( B \) is the magnetic field strength, \( A \) is the area of the coil, and \( K \) is the torsional constant.

The current sensitivity can be increased by:

Increasing the number of turns in the coil.
Increasing the area of the coil in the magnetic field.
Decreasing the torsional constant \( K \) (which means making the suspension more flexible).


Increasing the current sensitivity by changing the coil's properties may also affect the resistance of the galvanometer, which in turn can alter its voltage sensitivity. As the current sensitivity increases, the voltage sensitivity may decrease because the resistance increases, which can reduce the voltage required for full-scale deflection.


The current sensitivity is directly proportional to the number of turns and the area of the coil, but increasing the current sensitivity does not necessarily lead to an increase in voltage sensitivity. Voltage sensitivity is related to the resistance of the galvanometer, and increasing the number of turns to increase current sensitivity will also increase the resistance, thereby decreasing the voltage sensitivity. Quick Tip: The current sensitivity can be increased by increasing the number of turns, the area of the coil, or decreasing the torsional constant. However, increasing the current sensitivity may not always increase voltage sensitivity.

To convert the galvanometer into a voltmeter, we need to connect a series resistance \( R_s \) with the galvanometer. The value of \( R_s \) can be calculated using the following relation: \[ V = I_G (R + R_s), \]
where \( V \) is the full-scale deflection voltage (100 V), \( I_G \) is the current for full-scale deflection (20 mA), and \( R \) is the resistance of the galvanometer (15 Ω).

Rearranging to solve for \( R_s \): \[ R_s = \frac{V}{I_G} - R = \frac{100}{20 \times 10^{-3}} - 15 = 5000 - 15 = 4985 \, \Omega. \]

Thus, the required series resistance is 4985 Ω.

By connecting a 4985 Ω resistor in series with the galvanometer, it can be converted into a voltmeter with a full-scale range of 0 to 100 V. Quick Tip: To convert a galvanometer into a voltmeter, a high resistance is connected in series. The value of the resistance is calculated to achieve the desired voltage range for full-scale deflection.

The interference pattern obtained in Young's double-slit experiment has equally spaced bright bands, with constant intensity. In contrast, the diffraction pattern from a single slit shows maxima and minima, where the maxima become weaker on either side of the central maximum.

The graph depicts maxima and minima due to constructive and destructive interference, with maxima at integer multiples of \( \lambda \) and minima at odd multiples of \( \lambda / 2 \). Quick Tip: The intensity distribution for a double-slit interference pattern follows a sinusoidal curve, with alternating bright and dark bands representing constructive and destructive interference.

We know that the intensity at a point in Young's double-slit experiment is given by: \[ I = I_0 \cos^2 \left( \frac{\pi \Delta}{\lambda} \right), \]
where \( I_0 \) is the maximum intensity and \( \Delta \) is the path difference.

For path difference \( \Delta = \lambda \), the intensity is \( K \). Thus, we have: \[ K = I_0 \cos^2 \left( \pi \right) = I_0. \]

Now, for the path difference \( \Delta = \lambda / 6 \), we calculate the intensity: \[ I = I_0 \cos^2 \left( \frac{\pi}{6} \right) = I_0 \times \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} I_0. \]

Since \( K = I_0 \), the intensity at a point where the path difference is \( \lambda / 6 \) is: \[ I = \frac{3}{4} K. \] Quick Tip: When the path difference is not an integer multiple of \( \lambda \), use the formula for intensity and calculate using the cosine squared function for accurate results.

In a compound microscope, the objective lens forms a real, inverted, and diminished image at the focal plane of the eyepiece. The eyepiece acts as a magnifier to form a virtual, erect, and magnified image at the least distance of distinct vision.

The magnifying power \( M \) of the compound microscope is given by the product of the magnifying powers of the objective lens \( M_o \) and the eyepiece lens \( M_e \): \[ M = M_o \times M_e. \]

The magnifying power of the objective lens is given by: \[ M_o = \frac{v_o}{u_o}, \]
where \( v_o \) is the image distance and \( u_o \) is the object distance for the objective lens. Since the image is formed at the focal length of the objective lens \( f_o \), we have: \[ v_o = f_o. \]

For the eyepiece, the magnifying power is given by: \[ M_e = \frac{D}{f_e}, \]
where \( D \) is the least distance of distinct vision and \( f_e \) is the focal length of the eyepiece.

Thus, the total magnifying power is: \[ M = \frac{f_o}{u_o} \times \frac{D}{f_e}. \] Quick Tip: For a compound microscope, the magnifying power is the product of the magnifying powers of the objective and the eyepiece. The objective magnifies the object, and the eyepiece further magnifies the image formed.

In a telescope, the magnifying power \( M \) is given by the ratio of the focal lengths of the objective and the eyepiece: \[ M = \frac{f_o}{f_e}. \]
Here, \( f_o = 100 \, cm \) and \( f_e = 5 \, cm \), so the magnifying power is: \[ M = \frac{100}{5} = 20. \]

Thus, the magnifying power of the telescope is 20. Quick Tip: The magnifying power of a telescope is simply the ratio of the focal length of the objective lens to the focal length of the eyepiece when the final image is formed at infinity.