CBSE Class 12 Physics Question Paper 2024 (Set 1 - 55/1/1) Available: Download Solution PDF with Answer Key

The electric field at the center of a uniformly charged circular ring is zero. This is because the contributions to the electric field from all elements of the ring cancel out due to symmetry. The electric field due to a small charge element \( dq \) on the ring at the center is: \[ dE = \frac{k dq}{R^2}, \]
where \( k = \frac{1}{4\pi\varepsilon_0} \). However, the vector components of the electric field from opposite elements of the ring cancel out due to symmetry, resulting in a net electric field of: \[ 0. \] Quick Tip: For symmetric charge distributions, evaluate the net electric field by summing the contributions from all elements. Symmetry often results in cancellation of components at specific points.

In a parallel configuration, the total capacitance is the sum of individual capacitances: \[ C_{total} = n \cdot C = 10 \cdot 1 \mu F = 10 \mu F. \]
The energy stored in a capacitor is given by: \[ U = \frac{1}{2} C V^2. \]
Substitute the values: \[ U = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot (100)^2. \]
Simplify: \[ U = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot 10^4 = 10^{-2} J. \]
Thus, the total energy stored is: \[ 10^{-2} \, J. \] Quick Tip: For capacitors in parallel, the total capacitance is the sum of individual capacitances. Use \( U = \frac{1}{2} C V^2 \) to calculate the energy stored.

The circuit consists of two batteries and three resistors. To calculate the potential difference between points A and B, use Kirchhoff’s Voltage Law (KVL).

Step 1: Analyze the Loop
Starting from point A and moving clockwise: \[ 12 - 1 \cdot I - 0.5 \cdot I - 6 = 0, \]
where \( I \) is the current. Combine the terms: \[ 12 - 6 - 1.5I = 0 \implies 6 = 1.5I \implies I = \frac{6}{1.5} = 4 \, A. \]

Step 2: Calculate the Potential Difference
The potential difference across the \( 0.5 \, \Omega \) resistor is: \[ V = I \cdot R = 4 \cdot 0.5 = 2 \, V. \]
The potential at point A is \( 12 \, V \), and the potential at point B is \( 6 + 2 = 8 \, V \). The potential difference is: \[ V_{AB} = 12 - 8 = 9 \, V. \]
Thus, the potential difference between A and B is: \[ 9 \, V. \] Quick Tip: For circuits, always apply Kirchhoff’s Voltage Law systematically. The total potential difference across a loop is the sum of the EMFs minus the voltage drops across the resistors.

The magnetic field exerts a force on each segment of the current-carrying loop. Due to the interaction between the magnetic field and the current, the magnetic forces tend to pull the loop inward, reducing its area. This behavior is due to the Lorentz force acting on the current elements, which generates a net torque that compresses the loop.

Thus, the loop tends to: \[ shrink. \] Quick Tip: For a current-carrying loop in a uniform magnetic field, consider the forces and torques acting on each segment. The loop generally shrinks due to inward forces generated by the magnetic interaction.

The magnetic force \( \vec{F} \) on a current-carrying wire is given by: \[ \vec{F} = I (\vec{L} \times \vec{B}), \]
where:

\( I = 1.0 \, A \) is the current,
\( \vec{L} = (10 \, cm) \hat{j} = (0.1 \, m) \hat{j} \) is the length vector of the wire,
\( \vec{B} = (5 \, mT) \hat{j} - (8 \, mT) \hat{k} = (5 \times 10^{-3}) \hat{j} - (8 \times 10^{-3}) \hat{k} \).


Step 1: Compute \( \vec{L} \times \vec{B} \)
Using the determinant method: \[ \vec{L} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0 & 0.1 & 0
0 & 5 \times 10^{-3} & -8 \times 10^{-3} \end{vmatrix}. \]
Expand the determinant: \[ \vec{L} \times \vec{B} = \hat{i} \big( 0.1 \cdot (-8 \times 10^{-3}) - 0 \big) - \hat{j} (\dots) + \hat{k} (\dots). \]
Simplify: \[ \vec{L} \times \vec{B} = \hat{i} (-0.8 \times 10^{-3}) = -(0.8 \times 10^{-3}) \hat{i}. \]

Step 2: Calculate \( \vec{F} \) \[ \vec{F} = I (\vec{L} \times \vec{B}) = 1.0 \cdot -(0.8 \times 10^{-3}) \hat{i}. \]
Simplify: \[ \vec{F} = -(0.8 \, mN) \hat{i}. \]
Thus, the force on the wire is: \[ -(0.8 \, mN) \hat{i}. \] Quick Tip: To calculate the magnetic force on a wire, use \( \vec{F} = I (\vec{L} \times \vec{B}) \). Carefully expand the cross product, keeping the direction of vectors in mind.

To convert a galvanometer into an ammeter, a shunt resistance \( S \) is connected in parallel with the galvanometer. The current through the galvanometer is given as \( I_g = 0.001 \, A \) (0.1% of 1 A). The total current through the ammeter is \( I = 1 \, A \).

The current through the shunt is: \[ I_s = I - I_g = 1 - 0.001 = 0.999 \, A. \]
The potential difference across the galvanometer and the shunt is the same: \[ I_g G = I_s S. \]
Substitute \( I_s \) and rearrange for \( S \): \[ S = \frac{I_g G}{I_s} = \frac{(0.001)G}{0.999}. \]
Simplify: \[ S = \frac{G}{999}. \]
The total resistance of the ammeter is the parallel combination of \( G \) and \( S \): \[ R_{ammeter} = \frac{G S}{G + S}. \]
Substitute \( S = \frac{G}{999} \): \[ R_{ammeter} = \frac{G \cdot \frac{G}{999}}{G + \frac{G}{999}} = \frac{G^2}{999G + G} = \frac{G}{1000}. \]
Thus, the resistance of the ammeter is: \[ \frac{G}{1000} \, \Omega. \] Quick Tip: To convert a galvanometer into an ammeter, calculate the shunt resistance \( S \) using \( S = \frac{I_g G}{I_s} \). The total resistance of the ammeter is the effective parallel resistance of \( G \) and \( S \).

The capacitive reactance \( X_C \) is given by: \[ X_C = \frac{1}{\omega C}, \]
where \( \omega \) is the angular frequency, and \( C \) is the capacitance. Initially, the reactance is: \[ X = \frac{1}{\omega C}. \]
When the capacitance is doubled \( C' = 2C \) and the frequency is tripled \( \omega' = 3\omega \), the new reactance becomes: \[ X'_C = \frac{1}{\omega' C'} = \frac{1}{(3\omega)(2C)} = \frac{1}{6\omega C}. \]
Compare with the initial reactance: \[ X'_C = \frac{X}{6}. \]
Thus, the new reactance is: \[ \frac{X}{6}. \] Quick Tip: For capacitive reactance, use \( X_C = \frac{1}{\omega C} \). Changes in capacitance and frequency inversely affect the reactance proportionally.

Displacement current exists in regions where there is a time-varying electric field. In Region I, the electric field is: \[ E_x = E_0 \sin(kz - \omega t). \]
This electric field varies with time due to the presence of the term \( -\omega t \). The displacement current density \( J_d \) is given by: \[ J_d = \varepsilon_0 \frac{\partial E_x}{\partial t}. \]
Differentiating \( E_x \) with respect to time: \[ \frac{\partial E_x}{\partial t} = -\omega E_0 \cos(kz - \omega t). \]
Thus, a displacement current exists in Region I. In the other regions (II, III, IV), the electric field does not vary with time, so there is no displacement current.

Therefore, the displacement current exists in: \[ Region I. \] Quick Tip: Displacement current arises from time-varying electric fields. Look for time-dependent terms in the electric field expressions to identify such regions.

The Balmer series corresponds to transitions where the final energy level is \( n_f = 2 \). The second spectral line occurs when the transition happens from \( n_i = 4 \) to \( n_f = 2 \).

Using the formula for the wavelength of emitted radiation: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right), \]
where:

\( R_H \) is the Rydberg constant,
\( n_f = 2 \) and \( n_i = 4 \) for the second spectral line.

Substitute \( n_f = 2 \) and \( n_i = 4 \): \[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right). \]
Simplify: \[ \frac{1}{\lambda} = R_H \cdot \frac{4 - 1}{16} = R_H \cdot \frac{3}{16}. \]
Thus, the correct transition for the second line in the Balmer series is: \[ \boxed{n_f = 2 and n_i = 4}. \] Quick Tip: For the Balmer series, transitions end at \( n_f = 2 \). Spectral lines are numbered based on the initial energy level \( n_i \) (e.g., \( n_i = 3 \) for the first line, \( n_i = 4 \) for the second line).

When germanium (Ge), a group-IV element, is doped with arsenic (As), a group-V element, it creates an n-type semiconductor. Arsenic contributes extra electrons (conduction electrons) to the material because it has five valence electrons compared to four in germanium. These extra electrons become free to conduct electricity, thus increasing the number of conduction electrons.

Thus, the effect of doping Ge with As is: \[ The number of conduction electrons increases. \] Quick Tip: Doping a semiconductor with a group-V element increases the number of conduction electrons, resulting in an \textit{n-type semiconductor.

The maximum kinetic energy of emitted electrons is given by Einstein's photoelectric equation: \[ K_{max} = h\nu - \phi, \]
where:

\( h\nu \) is the photon energy,
\( \phi \) is the work function of the metal.

For beam \( A \): \[ K_{max,A} = 3.3 - 2.3 = 1.0 \, eV. \]
For beam \( B \): \[ K_{max,B} = 11.3 - 2.3 = 9.0 \, eV. \]
The maximum speed of the emitted electrons is related to the kinetic energy by: \[ K_{max} = \frac{1}{2} mv_{max}^2. \]
Thus: \[ v_{max} \propto \sqrt{K_{max}}. \]
The ratio of speeds is: \[ \frac{v_{max,A}}{v_{max,B}} = \sqrt{\frac{K_{max,A}}{K_{max,B}}}. \]
Substitute the values: \[ \frac{v_{max,A}}{v_{max,B}} = \sqrt{\frac{1.0}{9.0}} = \frac{1}{3}. \]
Thus, the ratio of maximum speeds is: \[ \boxed{\frac{1}{3}}. \] Quick Tip: For photoelectric problems, use \( K_{max} = h\nu - \phi \) to calculate the kinetic energy of emitted electrons. Relate it to speed using \( v_{max} \propto \sqrt{K_{max}} \).

The de Broglie wavelength \( \lambda \) of a particle is given by: \[ \lambda = \frac{h}{p}, \]
where:

\( h \) is Planck's constant,
\( p \) is the momentum of the particle.

If two particles have the same de Broglie wavelength, then their momenta must be the same, because \( \lambda \) is inversely proportional to \( p \). However, their masses may differ, and thus their speeds and energies can be different.

For a moving electron and a moving proton, having the same wavelength implies: \[ \boxed{p_{electron} = p_{proton}}. \]
Thus, the correct answer is: \[ Momentum. \] Quick Tip: The de Broglie wavelength \( \lambda \) depends only on the momentum \( p \) of the particle. If two particles have the same \( \lambda \), they must have the same momentum, regardless of their masses or speeds.

In the photoelectric effect, the kinetic energy of the emitted photoelectrons depends on the frequency (or wavelength) of the incident light, not its intensity. Hence, Assertion (A) is false. The photoelectric current depends on the intensity of the incident light and not its wavelength, so Reason (R) is also false.

Thus, the correct answer is: \[ Assertion (A) is false and Reason (R) is also false. \] Quick Tip: In the photoelectric effect, the intensity of light affects the photoelectric current, while the frequency (or wavelength) of light determines the kinetic energy of the emitted electrons.

Mutual inductance between two coils depends on the extent of magnetic flux linkage between them. When the coils are wound on each other, the magnetic flux linkage is maximum, and hence the mutual inductance is also maximum. Therefore, both the Assertion (A) and the Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Thus, the correct answer is: \[ Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). \] Quick Tip: Mutual inductance is maximum when the coils are wound on each other because the magnetic flux linkage is maximized in this configuration.

Two long parallel wires carrying current in the same direction attract each other, while wires carrying current in opposite directions repel each other. When two wires are connected in series to the same battery, the current through them flows in the same direction, causing them to attract rather than repel. Hence, Assertion (A) is false. Additionally, Reason (R) is also false because the wires connected in series do not carry current in opposite directions.

Thus, the correct answer is: \[ Assertion (A) is false and Reason (R) is also false. \] Quick Tip: Two parallel wires carrying current in the same direction attract each other, while those carrying current in opposite directions repel each other.

Plane and convex mirrors always produce virtual images of objects, regardless of the position of the object. A virtual image cannot act as an object for producing a real image. Hence, both the Assertion (A) and the Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Thus, the correct answer is: \[ Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). \] Quick Tip: Plane and convex mirrors always produce virtual images. For a real image to form, the object must be real or a virtual image must act as a real object.

The resistance of a material varies with temperature according to the formula: \[ R_T = R_0(1 + \alpha \Delta T), \]
where:

\( R_T \) is the resistance at temperature \( T \),
\( R_0 \) is the resistance at the reference temperature \( T_0 \),
\( \alpha \) is the temperature coefficient of resistance,
\( \Delta T = T - T_0 \) is the temperature difference.


Given that \( R_T = 2R_0 \), \( \alpha = 4.0 \times 10^{-3}\,K^{-1} \), and \( T_0 = 20^\circC \), substitute these values: \[ 2R_0 = R_0(1 + \alpha \Delta T). \]
Cancel \( R_0 \) on both sides: \[ 2 = 1 + \alpha \Delta T. \]
Rearrange to find \( \Delta T \): \[ \Delta T = \frac{2 - 1}{\alpha} = \frac{1}{4.0 \times 10^{-3}} = 250\,K. \]
Calculate the temperature \( T \): \[ T = T_0 + \Delta T = 20 + 250 = 270^\circC. \]

Thus, the temperature at which the resistance of the wire is twice its resistance at \( 20^\circC \) is: \[ 270^\circC. \] Quick Tip: To calculate the temperature at which resistance changes, always use the formula \( R_T = R_0(1 + \alpha \Delta T) \) and solve for \( \Delta T \).

The wavelength \( \lambda \) of light is related to its speed and frequency by: \[ \lambda = \frac{v}{f}. \]

For light in air: \[ v_{air} = c = 3.0 \times 10^8\,m/s. \]
The wavelength in air is: \[ \lambda_{air} = \frac{v_{air}}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 600\,nm. \]
(i) The wavelength of the reflected light remains the same as in air: \[ \lambda_{reflected} = 600\,nm. \]

(ii) For refracted light in the medium: \[ v_{medium} = \frac{v_{air}}{n} = \frac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8\,m/s. \]
The wavelength in the medium is: \[ \lambda_{refracted} = \frac{v_{medium}}{f} = \frac{2.0 \times 10^8}{5.0 \times 10^{14}} = 400\,nm. \]

Thus: \[ \lambda_{refracted} = 400\,nm. \] Quick Tip: The wavelength of reflected light does not change as it remains in the same medium. The wavelength of refracted light changes and is inversely proportional to the refractive index of the medium.

The focal length \( f \) of a plano-convex lens is related to its refractive index \( n \) and radius of curvature \( R \) by the lens maker's formula: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right). \]
Simplify: \[ \frac{1}{f} = (n - 1) \frac{1}{R}. \]
Rearrange to find \( R \): \[ R = (n - 1)f. \]
Substitute \( n = 1.4 \) and \( f = 16\,cm \): \[ R = \frac{16}{1.4 - 1} = \frac{16}{0.4} = 40\,cm. \]

Thus, the radius of the curved surface of the lens is: \[ 40\,cm. \] Quick Tip: For plano-convex lenses, use the lens maker's formula and remember that one radius of curvature is \( \infty \) for the plane surface.

The mirror formula is: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u}, \]
where:

\( f \) is the focal length,
\( v \) is the image distance,
\( u \) is the object distance.


The focal length \( f \) is related to the radius of curvature \( R \) by: \[ f = \frac{R}{2} = \frac{40}{2} = 20\,cm. \]

Given \( u = -30\,cm \) (negative because the object is in front of the mirror), substitute into the mirror formula: \[ \frac{1}{20} = \frac{1}{v} - \frac{1}{30}. \]

Rearrange to solve for \( \frac{1}{v} \): \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{30}. \]

Simplify: \[ \frac{1}{v} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60}. \]

Thus: \[ v = \frac{60}{5} = 12\,cm. \]

(i) The position of the image is: \[ v = 12\,cm (real and inverted). \]

The magnification \( M \) is given by: \[ M = -\frac{v}{u}. \]

Substitute \( v = 12\,cm \) and \( u = -30\,cm \): \[ M = -\frac{12}{-30} = \frac{12}{30} = 0.4. \]

(ii) The magnification of the image is: \[ M = 0.4 (diminished). \] Quick Tip: For mirrors, use the mirror formula \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \) and apply the sign convention carefully for object and image distances.

The de Broglie wavelength of the neutron is given by: \[ \lambda_n = \frac{h}{p_n}, \]
where \( p_n \) is the momentum of the neutron. The kinetic energy \( E \) of the neutron is: \[ E = \frac{p_n^2}{2m}. \]

Solve for \( p_n \): \[ p_n = \sqrt{2mE}. \]

Substitute \( p_n \) into the expression for \( \lambda_n \): \[ \lambda_n = \frac{h}{\sqrt{2mE}}. \]

The wavelength of the photon \( \lambda_p \) is related to its energy by: \[ \lambda_p = \frac{hc}{E}. \]

The ratio of wavelengths is: \[ \frac{\lambda_n}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}. \]

Simplify: \[ \frac{\lambda_n}{\lambda_p} = \frac{E}{\sqrt{2mE} \cdot c}. \]

Further simplification: \[ \frac{\lambda_n}{\lambda_p} = \frac{\sqrt{E}}{\sqrt{2m} \cdot c}. \]

Thus, the ratio of the wavelengths is: \[ \frac{\lambda_n}{\lambda_p} = \frac{\sqrt{E}}{\sqrt{2m} \cdot c}. \] Quick Tip: For de Broglie wavelength problems, use \( \lambda = \frac{h}{p} \) and carefully relate the momentum or energy for each particle or photon.

Gallium arsenide (GaAs) exhibits negative resistance in certain regions of its current-voltage (I-V) characteristic curve. The graph typically shows a linear region (Ohm's law), followed by a region of negative slope (negative resistance), and then a saturation region.

The regions can be identified as follows:

(a) The region where resistance is negative corresponds to the portion of the I-V curve with a negative slope (decreasing current with increasing voltage).
(b) The region where Ohm's law is obeyed corresponds to the initial linear part of the I-V curve.


A qualitative sketch of the I-V graph is shown below:

Quick Tip: For semiconductors like GaAs, identify negative resistance regions by looking for sections of the I-V curve with a negative slope.

The electric flux \(\Phi\) through a surface is given by: \[ \Phi = \int \vec{E} \cdot \vec{dA}, \]
where:

\(\vec{E}\) is the electric field,
\(\vec{dA}\) is the area vector perpendicular to the surface.

(a) The flux through the cube:
The electric field varies as \(\vec{E} = 500x\hat{i}\). Only the two faces of the cube perpendicular to the \(x\)-axis contribute to the flux. Let the cube extend from \(x = 0\) to \(x = 0.1\,\mathrm{m}\).

At \(x = 0\): The electric field is \(\vec{E} = 500 \times 0 = 0\,\mathrm{N/C}\).
At \(x = 0.1\,\mathrm{m}\): The electric field is \(\vec{E} = 500 \times 0.1 = 50\,\mathrm{N/C}\).


The area of each face of the cube is: \[ A = (0.1)^2 = 0.01\,\mathrm{m}^2. \]

The flux through the face at \(x = 0\) is: \[ \Phi_1 = \vec{E} \cdot A = 0 \times 0.01 = 0\,\mathrm{Nm}^2/\mathrm{C}. \]

The flux through the face at \(x = 0.1\,\mathrm{m}\) is: \[ \Phi_2 = \vec{E} \cdot A = 50 \times 0.01 = 0.5\,\mathrm{Nm}^2/\mathrm{C}. \]

The net flux through the cube is: \[ \Phi = \Phi_2 - \Phi_1 = 0.5 - 0 = 0.5\,\mathrm{Nm}^2/\mathrm{C}. \]

Thus, the flux passing through the cube is: \[ \Phi = 0.5\,\mathrm{Nm}^2/\mathrm{C}. \]


(b) The charge within the cube:
Using Gauss's law: \[ \Phi = \frac{q_{enclosed}}{\epsilon_0}, \]
where \(\epsilon_0 = 8.854 \times 10^{-12}\,\mathrm{C}^2/\mathrm{Nm}^2\) is the permittivity of free space.

Rearrange to find \(q_{enclosed}\): \[ q_{enclosed} = \Phi \cdot \epsilon_0. \]

Substitute the values: \[ q_{enclosed} = 0.5 \cdot 8.854 \times 10^{-12} = 4.427 \times 10^{-12}\,\mathrm{C}. \]

Thus, the charge within the cube is: \[ q_{enclosed} = 4.43 \times 10^{-12}\,\mathrm{C}. \] Quick Tip: For calculating flux through a cube with a non-uniform field, consider the electric field at each relevant face and use Gauss’s law to find the enclosed charge.

Definition of Current Density:
The current density \(\vec{j}\) is defined as the current per unit cross-sectional area. Mathematically: \[ \vec{j} = \frac{\vec{I}}{A}, \]
where \(\vec{I}\) is the current and \(A\) is the cross-sectional area. Current density is a vector quantity because it has both magnitude and direction.

Derivation:
The current \(I\) through a conductor is given by: \[ I = nqAv_d, \]
where:

\(n\) is the number of charge carriers per unit volume,
\(q\) is the charge of each carrier,
\(A\) is the cross-sectional area,
\(v_d\) is the drift velocity of the charge carriers.


Substitute \(v_d\) using \(v_d = \mu E\), where \(\mu\) is the mobility of charge carriers and \(E\) is the electric field: \[ I = nqA(\mu E). \]

The current density \(\vec{j}\) is: \[ \vec{j} = \frac{I}{A} = nq(\mu E). \]

Using \(\mu = \frac{e\tau}{m}\), where \(\tau\) is the relaxation time and \(m\) is the mass of the charge carriers, substitute \(\mu\): \[ \vec{j} = nq \left( \frac{e\tau}{m} \right) E. \]

Simplify: \[ \vec{j} = \frac{ne^2 \tau}{m} E. \]

Thus, \(\alpha = \frac{ne^2 \tau}{m}\), and: \[ \vec{j} = \alpha \vec{E}. \] Quick Tip: Current density \(\vec{j}\) is always a vector quantity. It is proportional to the electric field \(\vec{E}\), with proportionality constant \(\alpha = \frac{ne^2 \tau}{m}\).

Definition of Wheatstone Bridge:
A Wheatstone bridge is a circuit used to measure unknown resistances. It consists of four resistors arranged in a diamond shape, with a galvanometer connected between two opposite nodes.

Condition for Balance:
Let the resistances of the four arms of the Wheatstone bridge be \(R_1\), \(R_2\), \(R_3\), and \(R_4\). The bridge is balanced when no current flows through the galvanometer. This occurs when: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4}. \]

Derivation:
At balance, the potentials at the two points connected to the galvanometer are equal. Using Ohm's law: \[ \frac{V_{AB}}{R_1} = \frac{V_{AD}}{R_2}, \quad \frac{V_{AB}}{R_3} = \frac{V_{AD}}{R_4}. \]

Simplify: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4}. \]

Thus, the condition for the Wheatstone bridge to be balanced is: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4}. \] Quick Tip: For a balanced Wheatstone bridge, ensure \(\frac{R_1}{R_2} = \frac{R_3}{R_4}\). This eliminates current through the galvanometer.

(a) The speed of the proton:
The kinetic energy of the proton is: \[ K = \frac{1}{2} mv^2. \]

Solve for \(v\): \[ v = \sqrt{\frac{2K}{m}}. \]

Substitute \(K = 1.3384 \times 10^{-14} \, \mathrm{J}\) and \(m = \frac{q}{q/m} = \frac{1.6 \times 10^{-19}}{1.0 \times 10^8} = 1.6 \times 10^{-27} \, \mathrm{kg}\): \[ v = \sqrt{\frac{2 \cdot 1.3384 \times 10^{-14}}{1.6 \times 10^{-27}}} = \sqrt{1.672 \times 10^{13}} = 1.29 \times 10^6 \, \mathrm{m/s}. \]

Thus: \[ v = 1.29 \times 10^6 \, \mathrm{m/s}. \]



(b) The magnitude of acceleration of the proton:
The force acting on the proton in the magnetic field is: \[ F = qvB, \]
where \(q = 1.6 \times 10^{-19} \, \mathrm{C}\) and \(B = 2.0 \, \mathrm{mT} = 2.0 \times 10^{-3} \, \mathrm{T}\).

Substitute: \[ F = (1.6 \times 10^{-19})(1.29 \times 10^6)(2.0 \times 10^{-3}) = 4.13 \times 10^{-16} \, \mathrm{N}. \]

The acceleration is: \[ a = \frac{F}{m} = \frac{4.13 \times 10^{-16}}{1.6 \times 10^{-27}} = 2.58 \times 10^{11} \, \mathrm{m/s^2}. \]

Thus: \[ a = 2.58 \times 10^{11} \, \mathrm{m/s^2}. \]



(c) The radius of the circular path:
The radius is: \[ r = \frac{mv}{qB}. \]

Substitute: \[ r = \frac{(1.6 \times 10^{-27})(1.29 \times 10^6)}{(1.6 \times 10^{-19})(2.0 \times 10^{-3})}. \]

Simplify: \[ r = \frac{2.064 \times 10^{-21}}{3.2 \times 10^{-22}} = 6.45 \, \mathrm{m}. \]

Thus: \[ r = 6.45 \, \mathrm{m}. \] Quick Tip: For a charged particle in a magnetic field, use \(F = qvB\) to calculate force, \(a = \frac{F}{m}\) for acceleration, and \(r = \frac{mv}{qB}\) for the radius of the path.

The current in the circuit is: \[ i = i_m \sin(\omega t - \phi), \]
where \(\phi\) is the phase angle between voltage and current.

The instantaneous power is: \[ p(t) = v \cdot i = v_m \sin \omega t \cdot i_m \sin(\omega t - \phi). \]

Using the trigonometric identity: \[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)], \]
we get: \[ p(t) = \frac{v_m i_m}{2} [\cos \phi - \cos(2\omega t - \phi)]. \]

The average power dissipated over one cycle is: \[ P_{avg} = \frac{1}{T} \int_0^T p(t) \, dt = \frac{v_m i_m}{2} \cos \phi. \]

Substituting \(i_m = \frac{v_m}{Z}\), where \(Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}\), we get: \[ P_{avg} = \frac{v_m^2}{2Z} \cos \phi = \frac{v_m^2 R}{2Z^2}. \]

At resonance, \(\omega L = \frac{1}{\omega C}\), and the resonant frequency is: \[ \omega_0 = \frac{1}{\sqrt{LC}}. \] Quick Tip: The average power dissipated in an AC circuit depends only on the resistive part of the impedance, while the resonant frequency minimizes the overall impedance of the circuit.

1. Radio Waves: Large antenna arrays are used for radio wave transmission because the wavelength of radio waves is in meters, requiring similarly sized systems to efficiently radiate or receive these waves.
2. Microwaves: Microwaves have a wavelength on the order of centimeters, which matches the size of cavities or waveguides used for their propagation in microwave ovens or radar systems. Quick Tip: For efficient radiation or absorption, the system’s size should match the wavelength of the electromagnetic wave.

(i) Long-distance radio broadcasts use short-wave bands because these bands undergo ionospheric reflection, allowing them to travel large distances beyond the line of sight.


(ii) Optical and radio telescopes can be built on the ground because these waves can penetrate the Earth’s atmosphere. However, X-rays are absorbed by the atmosphere, making it necessary to place X-ray telescopes on satellites orbiting the Earth. Quick Tip: The Earth’s atmosphere selectively allows certain frequencies (like optical and radio waves) while blocking others (like X-rays and gamma rays).

1. Drawbacks of Rutherford’s Model:

The model could not explain the stability of the atom. Accelerating electrons should emit radiation and eventually collapse into the nucleus.
It could not explain the discrete spectral lines observed in atomic emission spectra.


2. Bohr’s Solution:

Bohr proposed that electrons revolve in specific quantized orbits where they do not radiate energy.
Energy is only emitted or absorbed when an electron transitions between these orbits.


3. Unequal Spacing of Orbits:
The radius of the \(n\)-th orbit is proportional to \(n^2\), and the energy of the \(n\)-th orbit is given by: \[ E_n = -\frac{13.6}{n^2} \, \mathrm{eV}. \]
The energy difference between successive orbits decreases as \(n\) increases, leading to unequally spaced orbits. Quick Tip: Bohr’s model successfully explained atomic stability and spectral lines by introducing quantized energy levels.

(a) Properties of a Nucleus:

The nucleus contains protons and neutrons, collectively called nucleons.
The nucleus has a positive charge due to the presence of protons.


(b) Nuclear Density:
The nucleus is much denser than the atom because most of the mass of an atom is concentrated in its tiny nucleus, whereas the atom’s size includes the much larger electron cloud.


(c) Density of Nuclear Matter:
The volume of a nucleus is proportional to \(A\), where \(A\) is the mass number. The radius \(R\) of a nucleus is given by: \[ R = R_0 A^{1/3}. \]

The density of the nucleus is: \[ \rho = \frac{Mass of Nucleus}{Volume of Nucleus} = \frac{m_N A}{\frac{4}{3}\pi R^3} = \frac{m_N}{\frac{4}{3}\pi R_0^3}. \]

Since \(R_0\) and \(m_N\) are constants, the density is the same for all nuclei. Quick Tip: The density of nuclear matter is constant because the nucleus’s radius increases with \(A^{1/3}\), while its mass is proportional to \(A\).

The lens maker’s formula is given by: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a double-convex lens, both surfaces have the same radius of curvature \( R \), and: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{-1}{R} \right) \] \[ \frac{1}{f} = (n - 1) \cdot \frac{2}{R} \]
The power \( P \) of the lens is: \[ P = \frac{1}{f} = \frac{2(n - 1)}{R} \] Quick Tip: For a double-convex lens, use the lens maker’s formula with the radii of curvature appropriately signed. The total power is proportional to the refractive index and inversely proportional to the radius.

When a lens is cut into two equal parts perpendicular to its principal axis, the focal length of each part remains unchanged. Since the power of a lens depends only on the focal length and is given by: \[ P = \frac{1}{f} \] The power of each part remains the same as the original lens. Thus, the power of one part is \( P \).

Quick Tip: Cutting a lens perpendicular to its principal axis does not change its focal length or power.

When two lenses are kept in contact, the total power of the combination is the algebraic sum of their individual powers. Since both parts have power \( P \), the total power is: \[ P_{\text{total}} = P + P = 2P \]

Quick Tip: For lenses in contact, the combined power is the sum of their individual powers.