UP Board Class 12 Physics Question Paper 2024 (Code 348 FY) Available- Download Solution PDF with Answer Key

Step 1: Resistivity (\( \rho \)) is a material property and does not change with stretching.
\[ Resistivity remains constant \]
\[ \boxed{\rho = constant} \] Quick Tip: Resistivity depends only on the material, not on dimensions.

Step 1: The formula for minimum deviation is:
\[ \delta_{\min} = 2 \theta - A \]

Step 2: Given \( A = 60^\circ \) and \( \mu = \sqrt{3} \), we solve for \( \delta_{\min} \).
\[ \boxed{\delta_{\min} = 60^\circ} \] Quick Tip: In an equilateral prism with refractive index \( \sqrt{3} \), the minimum deviation angle equals the prism angle.

Step 1: The dimensional formula of inductance \( L \) is \( [ML^2T^{-2}A^{-2}] \) and for capacitance \( C \) is \( [M^{-1}L^{-2}T^{4}A^{2}] \).

Step 2: Their product gives:
\[ \sqrt{LC} \propto T \]
\[ \boxed{Unit: Second} \] Quick Tip: The product of inductance and capacitance represents the square of time dimension.

Step 1: The work-function \( W \) and threshold wavelength \( \lambda \) are related as:
\[ W \propto \frac{1}{\lambda} \]

Step 2: If the work-function increases \( n \) times:
\[ \lambda' = \frac{\lambda_0}{n} \]
\[ \boxed{\lambda_0 / n} \] Quick Tip: Threshold wavelength is inversely proportional to the work-function.

Step 1: By balancing atomic number and mass number:
\[ 4 + A = (A+3) + 1 \]

Step 2: This indicates that the emitted particle is a proton.
\[ \boxed{Proton} \] Quick Tip: Use conservation of mass and atomic number to identify unknown particles.

Step 1: The shunt resistance is calculated as:
\[ S = \frac{I_G R}{I - I_G} \]

Step 2: Given 10% current passes through the galvanometer, the correct answer is:
\[ \boxed{R/11} \] Quick Tip: Shunt resistance allows a controlled fraction of current to bypass the galvanometer.

The magnetic dipole moment \( \mu \) is given by:
\[ \mu = I \cdot A \]

where \( I \) is the current and \( A \) is the area of the loop.

The unit of magnetic dipole moment is:
\[ Ampere-meter^2 \, (A \cdot m^2) \] Quick Tip: Magnetic dipole moment quantifies the strength and orientation of a magnetic source.

Microwaves are electromagnetic waves with frequencies between \( 1 \) GHz and \( 300 \) GHz.

Use: Microwaves are commonly used in radar systems for detecting objects at a distance.
\[ \boxed{Used in radar systems} \] Quick Tip: Microwaves are widely used in communication and cooking applications.

The angle of polarisation (\( \theta_p \)) is the angle of incidence at which light is completely polarised upon reflection.
\[ \theta_p = \tan^{-1} (\mu) \] Quick Tip: At Brewster's angle, reflected light is completely plane polarised.

The diode is reverse biased.

Step 1: In the given circuit, the negative terminal (\(-5V\)) is connected to the \( p \)-side, and the positive terminal (\(-3V\)) is connected to the \( n \)-side.

Step 2: Since the \( p \)-side is at a lower potential, the junction is reverse biased.
\[ \boxed{Reverse biased} \] Quick Tip: In reverse bias, the depletion region widens, and current flow is minimal.

A coil has an inductance of 1 henry if a current change of \( 1 \) ampere per second induces an electromotive force of \( 1 \) volt.
\[ \boxed{1 \, H = 1 \, V \cdot s / A} \] Quick Tip: Inductance measures the opposition to changes in current flow in a circuit.

Current density (\( J \)) is defined as the current flowing per unit cross-sectional area.
\[ J = \frac{I}{A} \]

The unit of current density is:
\[ Ampere per square meter (A/m^2) \] Quick Tip: Current density indicates how much electric current flows through a given area.

The drift velocity \( v_d \) is related to the applied potential difference \( V \) by the formula:
\[ v_d = \frac{e E \tau}{m} \]

Since \( E = \frac{V}{L} \), the relationship becomes:
\[ v_d = \frac{e \tau}{m} \cdot \frac{V}{L} \] Quick Tip: Drift velocity increases linearly with the applied potential difference.

The de-Broglie wavelength is given by:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

For equal kinetic energy:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}} \]

Since the mass of an alpha particle is four times that of a proton:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p}{m_p}} = 2 \]
\[ \boxed{2} \] Quick Tip: For particles with the same kinetic energy, de-Broglie wavelength is inversely proportional to the square root of mass.

Mass defect: The difference between the sum of the masses of individual nucleons and the actual mass of the nucleus.
\[ \Delta m = (Z m_p + N m_n) - M_{nucleus} \]

Binding energy: The energy required to separate the nucleus into its individual nucleons.
\[ E_b = \Delta m \cdot c^2 \]
\[ \boxed{E_b = \Delta m \cdot c^2} \] Quick Tip: Mass defect leads to the release of binding energy according to Einstein's equation.

Using Ohm's Law:
\[ V_{total} = 1.5 \, V - 0.5 \, V = 1.0 \, V \]
\[ I = \frac{V}{R} \]
\[ 5 \times 10^{-3} = \frac{1}{R} \]
\[ R = \frac{1}{5 \times 10^{-3}} = 200 \, \Omega \]
\[ \boxed{200 \, \Omega} \] Quick Tip: Subtract the diode voltage drop before applying Ohm's Law to find resistance.

Displacement Current: It is the current that appears in the region where an electric field is changing with time, given by:
\[ I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]

Step 1: The magnitude of the magnetic field is obtained using the relation:
\[ B = \frac{E}{c} \]

Step 2: Substituting values:
\[ B = \frac{6}{3 \times 10^8} \]
\[ B = 2 \times 10^{-8} \, T \]

Step 3: The direction is along the \( \hat{k} \)-axis due to cross-product rules.
\[ \boxed{B = 2 \times 10^{-8} T, along \hat{k}} \] Quick Tip: The magnetic field is perpendicular to both the electric field and propagation direction.

Using the formulas for parallel and series combinations:
\[ C_1 + C_2 = 20 \]
\[ \frac{C_1 C_2}{C_1 + C_2} = 4.8 \]

Solving for the ratio:
\[ C_1 = 16 \, \muF, \quad C_2 = 4 \, \muF \]
\[ \boxed{C_1 : C_2 = 4 : 1} \] Quick Tip: In series, reciprocal capacitances add; in parallel, capacitances sum directly.

(i) Attractive: When the currents flow in the same direction.

(ii) Repulsive: When the currents flow in opposite directions.
\[ \boxed{Attractive if same direction, repulsive if opposite direction.} \] Quick Tip: Parallel currents attract, anti-parallel currents repel due to the interaction of magnetic fields.

A resonant circuit is an electrical circuit in which resonance occurs at a particular frequency.

Resonance Condition:
\[ \omega L = \frac{1}{\omega C} \]

Resonant Frequency:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]
\[ \boxed{f = \frac{1}{2\pi \sqrt{LC}}} \] Quick Tip: At resonance, impedance is minimum and current is maximum in an L-C-R circuit.

Einstein’s Photoelectric Equation:
\[ K_{\max} = h\nu - \phi \]

Step 1: Using given data:
\[ k_1 = 1 - 0.5 = 0.5 \, eV \]
\[ k_2 = 2.5 - 0.5 = 2.0 \, eV \]
\[ \frac{k_1}{k_2} = \frac{0.5}{2.0} = \frac{1}{4} \]
\[ \boxed{\frac{k_1}{k_2} = \frac{1}{4}} \]

Step 2: Since kinetic energy is proportional to the square of velocity:
\[ \frac{v_1}{v_2} = \sqrt{\frac{k_1}{k_2}} \]
\[ = \sqrt{\frac{1}{4}} = \frac{1}{2} \]
\[ \boxed{\frac{v_1}{v_2} = \frac{1}{2}} \] Quick Tip: Kinetic energy is proportional to the square of velocity in photoelectric effect.

Torque on a current-carrying loop:

The torque \( \tau \) on a current loop in a uniform magnetic field is given by:
\[ \tau = \vec{m} \times \vec{B} \]

where \( \vec{m} \) is the magnetic dipole moment and \( \vec{B} \) is the magnetic field.

Magnetic dipole moment:
\[ m = N I A \]

where:
- \( N \) is the number of turns,
- \( I \) is the current,
- \( A \) is the area of the loop.
\[ \boxed{\tau = N I A B \sin \theta} \] Quick Tip: The torque is maximum when the plane of the loop is perpendicular to the magnetic field.

(i) Just at the moment switch \( S \) is pressed:

At \( t = 0 \), the inductor opposes any change in current, so the initial current through the inductor is zero.
\[ i_2 = 0, \quad i_1 = \frac{10}{5} = 2 A \]
\[ \boxed{i = i_1 = 2 A} \]

(ii) After a long time:

The inductor acts as a short circuit, and the total resistance becomes:
\[ R_{eq} = 5 + 10 = 15 \, \Omega \]
\[ i = \frac{10}{15} = \frac{2}{3} A \]

Current division:
\[ i_1 = \frac{2}{3} \times \frac{10}{15} = \frac{2}{3} A, \quad i_2 = \frac{2}{3} \times \frac{5}{15} = \frac{1}{3} A \]
\[ \boxed{i_1 = \frac{2}{3} A, \quad i_2 = \frac{1}{3} A} \] Quick Tip: In an RL circuit, the inductor initially resists current flow and allows full current flow after a long time.

(i) RMS Value of Source Voltage

The given voltage is in peak form:
\[ V_{peak} = 282 V \]
\[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \]
\[ = \frac{282}{\sqrt{2}} = 199.5 V \]
\[ \boxed{199.5 V} \] Quick Tip: RMS voltage is obtained by dividing peak voltage by \( \sqrt{2} \).

Kirchhoff's First Law (KCL - Junction Rule):

The sum of currents entering a junction is equal to the sum of currents leaving the junction.
\[ \sum I_{in} = \sum I_{out} \]

Kirchhoff's Second Law (KVL - Loop Rule):

The sum of all voltages around any closed loop in a circuit is zero.
\[ \sum V = 0 \]

Solution for Ammeter Reading:

Applying Kirchhoff's loop rule to the given circuit:
\[ \frac{5}{4 + 2} = \frac{5}{6} \, A \]

The voltage across the parallel resistors is:
\[ V = IR = \frac{5}{6} \times 2 = \frac{10}{6} V \]

Current through the 6Ω resistor:
\[ I = \frac{10}{6} \div 6 = \frac{5}{18} \, A \]
\[ \boxed{Ammeter reading = \frac{5}{18} A} \] Quick Tip: Use Kirchhoff's loop rule to calculate voltages and junction rule for current distribution.

According to Huygens' wave theory:


Each point on a wavefront acts as a secondary source of wavelets.
When light travels from a denser to a rarer medium, the speed of the wave increases.
The wavefront bends away from the normal due to the increase in speed.


The refractive index is given by:
\[ n = \frac{speed in rarer medium}{speed in denser medium} \]
\[ \boxed{\sin i = n \sin r} \] Quick Tip: When light travels to a rarer medium, it bends away from the normal.

Step 1: Applying Snell's law at face AB:
\[ n_{air} \sin 45^\circ = n_{prism} \sin r \]

Since \( r = 90^\circ \), the critical angle condition applies:
\[ \sin 45^\circ = \frac{1}{n} \]
\[ n = \frac{1}{\sin 45^\circ} \]
\[ n = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]
\[ \boxed{n = 1.414} \] Quick Tip: At the critical angle, light refracts tangentially along the interface.

Step 1: Consider an electric dipole consisting of charges \( +q \) and \( -q \) separated by a distance \( 2a \). The dipole moment is:
\[ \vec{p} = q \cdot 2a \]

Step 2: The electric field at an equatorial point (perpendicular bisector) of the dipole at a distance \( r \) from the center is given by the formula:
\[ E_{equatorial} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{(r^2 + a^2)^{3/2}} \]

Step 3: For large distances (\( r \gg a \)):
\[ E_{equatorial} \approx \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{r^3} \]
\[ \boxed{E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p}{r^3}} \] Quick Tip: The electric field at the equatorial position is directed opposite to the dipole moment.

Step 1: Gauss' law states that the total electric flux \( \Phi \) through a closed surface is equal to the charge enclosed divided by the permittivity of free space:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon_0} \]

Step 2: Consider a point charge \( q \) at the center of a spherical surface of radius \( r \):
\[ E \cdot 4 \pi r^2 = \frac{q}{\varepsilon_0} \]

Step 3: Solving for the electric field:
\[ E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \]
\[ \boxed{E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}} \] Quick Tip: Gauss' law simplifies calculations for symmetric charge distributions.

Conditions for Interference of Light:


The sources must be coherent (having a constant phase difference).
The sources should emit light waves of the same frequency and wavelength.
The sources must have equal or nearly equal intensity for clear visibility.
The superposition of waves should occur in the same medium.


Step 1: For coherent sources:
\[ I_{max} = (\sqrt{I} + \sqrt{4I})^2 \]
\[ = (1 + 2)^2 I = 9I \]
\[ \boxed{9I} \]

Step 2: For non-coherent sources, the intensities add directly:
\[ I_{total} = I + 4I = 5I \]
\[ \boxed{5I} \] Quick Tip: In coherent interference, amplitudes add; in non-coherent interference, intensities add.

Step 1: Using the lens formula for the convex lens:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-20} \]
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20} \]
\[ v = \frac{20}{3} \approx 6.67 cm \]

Step 2: The image acts as an object for the concave lens. Using the lens formula again:
\[ \frac{1}{-10} = \frac{1}{v'} - \frac{1}{30 - 6.67} \]
\[ \frac{1}{-10} = \frac{1}{v'} - \frac{1}{23.33} \]
\[ v' = -17.5 cm \]
\[ \boxed{Image distance = 17.5 cm behind concave lens.} \]

Step 3: For combined lenses in contact, the effective power is:
\[ P_{total} = P_1 + P_2 \]
\[ = \frac{100}{10} + \frac{100}{-10} \]
\[ = 10 - 10 = 0 \, D \]
\[ \boxed{Total power = 0 D (acts like a plane glass).} \] Quick Tip: When convex and concave lenses of equal focal length are combined, they cancel each other out.

Bohr's Postulates for Hydrogen Atom:

Electrons revolve in fixed orbits without emitting energy.
Only certain discrete orbits are allowed where angular momentum is quantized.
Electrons emit or absorb energy when transitioning between orbits.


Step 1: Binding Energy
\[ E_{binding} = 13.6 \, eV \]
\[ \boxed{E_{binding} = 13.6 eV} \]

Step 2: Angular Momentum

According to Bohr's quantization condition:
\[ m v r = n \hbar \]

For ground state \( n = 1 \):
\[ L = \hbar = \frac{h}{2\pi} \]
\[ \boxed{L = \frac{h}{2\pi}} \]

Step 3: Total Energy

Total energy of an electron in the ground state:
\[ E = -\frac{13.6}{n^2} \, eV \]

For \( n = 1 \):
\[ \boxed{E = -13.6 eV} \] Quick Tip: Bohr's model explains discrete energy levels and quantization of angular momentum.

Step 1: Definition of Nuclear Fission and Fusion

Nuclear Fission: The splitting of a heavy nucleus into two lighter nuclei with the release of energy.
\[ ^{235}U + n \rightarrow ^{141}Ba + ^{92}Kr + 3n + energy \]

Nuclear Fusion: The combining of two light nuclei to form a heavier nucleus with the release of energy.
\[ ^{2}H + ^{3}H \rightarrow ^{4}He + n + energy \]

Step 2: Why Energy is Released?

Energy is released in both processes due to the mass defect, which is converted into energy according to Einstein’s equation:
\[ E = mc^2 \]

Step 3: Key Differences

\begin{tabular{|c|c|c|
\hline
Aspect & Fission & Fusion

\hline
Fuel & Uranium/Plutonium & Hydrogen isotopes

\hline
Temperature & Moderate & Very High

\hline
Energy Yield & High & Very High

\hline
Example & Nuclear reactors & Sun's energy

\hline
\end{tabular
\[ \boxed{Fission releases energy by splitting; Fusion releases energy by combining.} \] Quick Tip: Fusion releases more energy than fission but requires extreme conditions to occur.

Differences between \( n \)-type and \( p \)-type semiconductors:

\begin{table[h]
\centering
\begin{tabular{|c|c|c|
\hline
Property & \( n \)-type Semiconductor & \( p \)-type Semiconductor

\hline
Majority Carriers & Electrons & Holes

\hline
Minority Carriers & Holes & Electrons

\hline
Dopant Type & Donor (e.g., Phosphorus) & Acceptor (e.g., Boron)

\hline
Charge Carrier Mobility & Higher & Lower

\hline
Electrical Neutrality & Yes & Yes

\hline
\end{tabular
\end{table

Step 1: Electrical Neutrality

Although \( n \)-type semiconductors have excess electrons and \( p \)-type have excess holes, the total number of positive and negative charges remains equal due to the balance between ionized dopant atoms and charge carriers.
\[ \boxed{Both types of semiconductors are electrically neutral.} \] Quick Tip: Both \( n \)-type and \( p \)-type semiconductors are electrically neutral because the charge of dopants balances the free charge carriers.

(i) Forward Biasing:

Step 1: In forward bias, the \( p \)-side is connected to the positive terminal and the \( n \)-side to the negative terminal.

Step 2: The depletion region narrows, allowing current to flow.
\[ I_{forward} \approx I_s (e^{\frac{V}{kT}} - 1) \]

Step 3: Current flows from \( p \) to \( n \).
\[ \boxed{Current increases exponentially with voltage.} \]


\hrule


(ii) Reverse Biasing:

Step 1: In reverse bias, the \( p \)-side is connected to the negative terminal and the \( n \)-side to the positive terminal.

Step 2: The depletion region widens, preventing current flow.

Step 3: A small leakage current due to minority carriers flows.
\[ \boxed{Current remains nearly constant.} \] Quick Tip: In forward bias, current flows easily; in reverse bias, only leakage current flows.