UP Board Class 12 Physics Question Paper 2024 (Code 347 FW) Available- Download Solution PDF with Answer Key

The unit of magnetic field is derived from the definition of force \( F = q(\mathbf{v} \times \mathbf{B}) \), where magnetic force \( B = \frac{F}{qv} \). In terms of SI units: \[ B = \frac{N}{A \cdot m} \quad \Rightarrow \quad newton \cdot metre^{-1} \cdot ampere^{-1}. \] Quick Tip: The SI unit of the magnetic field (B) is Tesla, which is equivalent to \( newton \cdot metre^{-1} \cdot ampere^{-1} \).

The radius of a nucleus is proportional to the cube root of its mass number (\(A\)): \[ R \propto A^{1/3}. \]
For \(^{13} Al^{27}\), \[ R_1 = 3.6 \, fermi, \quad A_1 = 27, \quad A_2 = 125. \] \[ \frac{R_2}{R_1} = \left(\frac{A_2}{A_1}\right)^{1/3} \quad \Rightarrow \quad R_2 = R_1 \cdot \left(\frac{125}{27}\right)^{1/3}. \] \[ R_2 = 3.6 \cdot 2 \quad \Rightarrow \quad R_2 = 6 \, fermi. \] Quick Tip: The nuclear radius is proportional to \( A^{1/3} \), where \(A\) is the mass number.

Einstein's photoelectric equation states that: \[ h\nu = \phi + K_{max}, \]
where \(h\nu\) is the energy of the incident photon, \(\phi\) is the work function, and \(K_{max}\) is the maximum kinetic energy of the ejected electron. This conservation equation explicitly describes the conservation of energy. Quick Tip: The law of conservation of energy is fundamental in all physical processes, including photoelectric emission.

In an L-C-R circuit, the voltage \(V_O\) of the source is the vector sum of voltages across the resistor (\(V_R\)), inductor (\(V_L\)), and capacitor (\(V_C\)). Due to phase differences, the net voltage satisfies: \[ V_O^2 = (V_L - V_C)^2 + V_R^2. \] Quick Tip: Use the phasor addition technique to derive voltage relationships in L-C-R circuits.

For \(n_1 > n_2\), light rays bend away from the normal upon passing from a denser to a rarer medium. The first diagram (i) correctly demonstrates this refraction. Quick Tip: The bending of light depends on the refractive index of the materials: it bends towards the normal when entering a denser medium and away when entering a rarer medium.

The resistance \(R\) of a material is given by: \[ R \propto \frac{1}{Cross-sectional area}. \]
For a cube, the cross-sectional area scales with \(l^2\), and resistance is inversely proportional to the square of the side length: \[ R_1 : R_2 = \frac{1}{l^2} : \frac{1}{(3l)^2} \quad \Rightarrow \quad R_1 : R_2 = 9 : 1. \] Quick Tip: The resistance of a material is inversely proportional to its cross-sectional area.

Using Kirchhoff’s Current Law (KCL), the sum of currents entering a junction equals the sum of currents leaving it: \[ i = (6 \, A + 3 \, A) - (10 \, A + 3 \, A) = 4 \, A. \] Quick Tip: Kirchhoff’s Current Law states that at any junction, the sum of currents entering equals the sum of currents leaving.

When the magnetic flux linked with a circuit changes, an electromotive force (EMF) is induced, and the induced current opposes this change. Quick Tip: Lenz’s law is a consequence of the principle of conservation of energy in electromagnetic induction.

In a \(p\)-\(n\) junction, due to diffusion, electrons and holes recombine near the junction, leaving behind a region with immobile ions. This region forms the depletion layer, acting as a potential barrier for current flow. Quick Tip: The depletion layer thickness determines the behavior of the \(p\)-\(n\) junction under different biasing conditions.

Using Snell's law: \[ n_1 \sin i = n_2 \sin r, \]
where \( n_1 = 1 \) (air), \( n_2 = \sqrt{3} \), \( i = 60^\circ \): \[ \sin r = \frac{n_1 \sin i}{n_2} = \frac{1 \times \sin 60^\circ}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}. \]
Thus, \( r = 30^\circ \). Quick Tip: Snell's law helps calculate refraction at the boundary between two media of different refractive indices.

The amplitude of the electric field (\(E_0\)) and magnetic field (\(B_0\)) are related by: \[ E_0 = c B_0, \]
where \( c = 3 \times 10^8 \, m/s \). Substituting the values: \[ E_0 = 3 \times 10^8 \times 510 \times 10^{-9} = 153 \, V/m. \] Quick Tip: In electromagnetic waves, the electric and magnetic fields are related by the speed of light: \( E_0 = c B_0 \).

- Paramagnetic substances have unpaired electrons and exhibit weak magnetic moments.
- Ferromagnetic substances have domains aligned in a magnetic field and show strong magnetic behavior even when the external field is removed. Quick Tip: The retention of magnetism distinguishes ferromagnetic materials from paramagnetic materials.

1. Valence Band: The energy band filled with the highest energy electrons in a solid.
2. Conduction Band: The energy band above the valence band where electrons can move freely, contributing to conductivity.
3. Forbidden Energy Gap: The region between the valence band and conduction band where no electron can exist.

Classification of solids based on band theory:
- Conductors: Overlapping valence and conduction bands.
- Semiconductors: Small energy gap between valence and conduction bands.
- Insulators: Large energy gap, preventing electron movement. Quick Tip: The nature of the energy gap between bands determines the electrical conductivity of a material.

Key conclusions from the \( \alpha \)-particle scattering experiment:
1. Most \( \alpha \)-particles passed through the gold foil, indicating atoms are mostly empty space.
2. A few \( \alpha \)-particles were deflected at small angles, suggesting a positively charged region in the atom.
3. Very few were deflected back, confirming that the nucleus is small, massive, and positively charged. Quick Tip: Rutherford’s model of the atom replaced Thomson's "plum pudding" model by introducing the nucleus.

From de Broglie’s wavelength formula: \[ \lambda = \frac{h}{\sqrt{2mK}}, \]
where \(m\) is the mass and \(K\) is the kinetic energy. For the same wavelength: \[ K \propto \frac{1}{m}. \]
Since the proton has less mass than the neutron, its kinetic energy will be greater for the same wavelength. Quick Tip: For the same de Broglie wavelength, a lighter particle will always have higher kinetic energy.

Let the EMF of the battery be \( E \) and its internal resistance be \( r \).

The heat \( H \) is proportional to \( I^2 R \), where \( I \) is the current and \( R \) is the resistance. For both cases, the heat is equal: \[ I_1^2 \cdot 4 = I_2^2 \cdot 9. \]
Using Ohm’s law: \[ I_1 = \frac{E}{4 + r}, \quad I_2 = \frac{E}{9 + r}. \]
Substituting these: \[ \left( \frac{E}{4 + r} \right)^2 \cdot 4 = \left( \frac{E}{9 + r} \right)^2 \cdot 9. \]
Simplifying: \[ \frac{4}{(4 + r)^2} = \frac{9}{(9 + r)^2}. \]
Taking the square root: \[ \frac{2}{4 + r} = \frac{3}{9 + r}. \]
Cross-multiplying: \[ 2(9 + r) = 3(4 + r). \]
Expanding and solving: \[ 18 + 2r = 12 + 3r \quad \Rightarrow \quad r = 6 \, \Omega. \] Quick Tip: The internal resistance of a battery affects the total current and power delivered to external resistors.

• They travel in a vacuum at the speed of light (\( c = 3 \times 10^8 \, \mathrm{m/s} \)).
• They carry energy and momentum and can exert radiation pressure.

• (A) The threshold frequency (\( \nu_0 \)) is the x-intercept of the graph: \[ \nu_0 = 2.5 \times 10^{14} \, \mathrm{Hz}. \]
• (B) The work function (\( \phi \)) is given by: \[ \phi = h \nu_0 = (6.6 \times 10^{-34}) (2.5 \times 10^{14}) = 1.65 \times 10^{-19} \, \mathrm{J}. \] Converting to eV: \[ \phi = \frac{1.65 \times 10^{-19}}{1.6 \times 10^{-19}} = 1 \, \mathrm{eV}. \]

Quick Tip: Use Biot-Savart law for problems involving magnetic fields due to current elements or loops.

The maximum induced emf is related to mutual inductance \( M \) by:

emf_max = M \frac{di}{dt}.

For \( i = 10 \sin(100 \pi t) \):

\(\frac{di}{dt} = 1000 \pi \cos(100 \pi t).\)

The maximum value of \( \frac{di}{dt} \) is \( 1000 \pi \). Substituting values:

5 \pi = M (1000 \pi) \quad \Rightarrow \quad M = 0.005 \, \mathrm{H}.

For a concave mirror, the magnification \( m = -\frac{v}{u} \), where \( v \) is the image distance and \( u \) is the object distance.

Case 1 (\( m = -4 \)):

\( v = -4u. \)

The mirror formula is:

\( \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \)

Substituting \( v = -4u \):

\( \frac{1}{f} = \frac{1}{u} + \frac{1}{-4u} = \frac{3}{4u}. \)

Case 2 (\( m = -3 \), object moved by 3 cm):

\( u' = u + 3, \quad v' = -3u'. \)

Using the mirror formula:

\( \frac{1}{f} = \frac{1}{u'} + \frac{1}{v'}. \)

Substituting \( v' = -3u' \):

\( \frac{1}{f} = \frac{1}{u+3} + \frac{1}{-3(u+3)} = \frac{2}{3(u+3)}. \)

Equating \( \frac{1}{f} \) from both cases:

\( \frac{3}{4u} = \frac{2}{3(u+3)}. \)

Solving for \( u \):

\( u = 12 \, \mathrm{cm}, \quad f = 9 \, \mathrm{cm}. \)

The radius of curvature is:

\( R = 2f = 18 \, \mathrm{cm}. \)

Plane polarized light consists of waves oscillating in a single plane perpendicular to the direction of propagation.

Obtaining polarized light by refraction: When unpolarized light is incident at Brewster's angle on a transparent medium, the reflected light becomes plane polarized.
Quick Tip: Polarization can occur by reflection, refraction, or scattering. Brewster's angle is key for obtaining plane polarized light.

The total current in the circuit is:

I = \frac{E}{R + r}, where \( r \) is the internal resistance.

The terminal voltage is given by:

V = IR = \frac{ER}{R + r}.

Rearranging:

r = R \left( \frac{E}{V} - 1 \right).

The shunt resistance is:

S = \frac{IG \, G}{I - IG}, where \( G = 50 \, \Omega, I = 5 \, \mathrm{A}, IG = 0.05 \, \mathrm{A}.

Substituting the values:

S = \frac{0.05 \times 50}{5 - 0.05} = 0.5 \, \Omega.

The length of the wire is:

R = \rho \frac{l}{A} \quad \Rightarrow \quad l = \frac{R A}{\rho}.

Substituting \( R = 0.5, A = 2.7 \times 10^{-6}, \rho = 5.0 \times 10^{-7} \):

l = \frac{0.5 \times 2.7 \times 10^{-6}}{5.0 \times 10^{-7}} = 2.7 \, \mathrm{m}.

1. Forward Bias Configuration:
- The positive terminal of the power supply is connected to the \( p \)-side (anode), and the negative terminal is connected to the \( n \)-side (cathode).
- This reduces the potential barrier, allowing majority charge carriers (holes in the \( p \)-region and electrons in the \( n \)-region) to cross the junction.
- Once the applied voltage exceeds the **knee voltage** (typically \( \sim 0.7 \, V \) for silicon), a significant current flows due to the reduced barrier potential.

2. Graph:
- The relationship between forward voltage (X-axis) and forward current (Y-axis) shows an exponential increase in current after the knee voltage is reached.
\[ (Graph of forward current vs forward voltage is shown below.) \]


Quick Tip: In forward bias, a p-n junction diode behaves as a low-resistance circuit element, conducting current beyond the knee voltage.

1. Reverse Bias Configuration:
- The positive terminal of the power supply is connected to the \( n \)-side (cathode), and the negative terminal is connected to the \( p \)-side (anode).
- This increases the potential barrier, blocking the majority carriers. A small leakage current flows due to minority carriers.
- At the **breakdown voltage**, the junction experiences a sudden increase in reverse current due to mechanisms like Zener breakdown or avalanche breakdown.

2. Graph:
- The graph between reverse voltage (X-axis) and reverse current (Y-axis) shows a small leakage current initially, followed by a sharp increase at the breakdown voltage.
(Graph of reverse current vs reverse voltage is shown below.) 


Quick Tip: In reverse bias, a p-n junction diode behaves as a high-resistance circuit element, except at the breakdown voltage.

1. Bohr's Quantum Theory:
- Bohr's quantization condition states that the angular momentum \( mvr \) of an electron in the \( n \)-th orbit is given by:
\[ mvr = \frac{nh}{2\pi}, \quad where n = 1, 2, 3, \dots \]
Here, \( m \) is the mass of the electron, \( v \) is its velocity, \( r \) is the orbital radius, \( h \) is Planck's constant, and \( n \) is the quantum number.

2. Relation with De Broglie Wavelength:
- De Broglie wavelength of the electron is given by \( \lambda = \frac{h}{mv} \).
- Substituting \( \lambda \) in the above expression:
\[ 2\pi r = n\lambda. \]
Thus, the circumference of the orbit is \( n \lambda \).

3. Emission and Absorption Lines:
- The energy difference between two energy levels \( E_n = -\frac{13.6}{n^2} \, eV \) is emitted or absorbed as a photon:
\[ E = E_f - E_i = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2}. \]
- For transitions between \( n = 1 \), \( n = 2 \), and \( n = 3 \), the emission and absorption lines correspond to the Lyman and Balmer series in the spectrum.

Quick Tip: The Bohr model combines classical circular orbits with quantum conditions. Emission corresponds to the release of energy, while absorption involves gaining energy.

1. Binding Energy:
- The mass defect \( \Delta m \) is the difference between the total mass of nucleons and the actual mass of the nucleus.
\[ \Delta m = Zm_p + Nm_n - m_{nucleus}, \]
where \( Z = 8 \), \( N = 8 \), \( m_p = 1.007593 \, amu \), \( m_n = 1.008982 \, amu \), and \( m_{nucleus} = 16.00000 \, amu \).

Substituting:
\[ \Delta m = (8 \times 1.007593) + (8 \times 1.008982) - 16.00000 = 16.13176 - 16.00000 = 0.13176 \, amu. \]

2. Binding Energy:
- The binding energy is \( \Delta m \times 931 \, MeV/amu \):
\[ Binding energy = 0.13176 \times 931 = 122.65 \, MeV. \]

3. Per Nucleon:
- The binding energy per nucleon is:
\[ Binding energy per nucleon = \frac{Binding energy}{Total nucleons} = \frac{122.65}{16} = 7.67 \, MeV. \] Quick Tip: Higher binding energy per nucleon indicates a more stable nucleus. For oxygen-16, it is \( \approx 7.67 \, MeV \).

(A) Difference Between Interference and Diffraction: 

(A) Ray Diagram:

(B) Relation Between Focal Lengths:
- The magnifying power \( M \) for an astronomical telescope is:
\[ M = \frac{f_o}{f_e}, \]
where \( f_o \) is the focal length of the objective lens, and \( f_e \) is the focal length of the eyepiece lens.

- The length of the telescope \( L \) is:
\[ L = f_o + f_e. \]

3. Calculations:
- Given \( M = 5 \) and \( L = 36 \, cm \):
\[ f_o = 5f_e. \]

Substituting \( f_o = 5f_e \) into \( L = f_o + f_e \):
\[ 36 = 5f_e + f_e = 6f_e \quad \Rightarrow \quad f_e = 6 \, cm, \, f_o = 30 \, cm. \]


(C) Final Answer:
- Focal length of the objective lens: \( f_o = 30 \, cm \).
- Focal length of the eyepiece lens: \( f_e = 6 \, cm \). Quick Tip: To achieve a higher magnifying power in telescopes, increase the focal length of the objective lens or decrease that of the eyepiece.

(A) [i)] Total Capacitance of the Combination:

The initial total capacitance (without dielectric) is the series combination of \( C \) and \( 2C \):
\[ \frac{1}{C_{total, initial}} = \frac{1}{C} + \frac{1}{2C} = \frac{3}{2C} \quad \Rightarrow \quad C_{total, initial} = \frac{2C}{3}. \]

After inserting the dielectric of constant \( K \) into the capacitor of capacitance \( C \), its new capacitance becomes:
\[ C' = KC. \]

The new total capacitance is the series combination of \( C' = KC \) and \( 2C \):
\[ \frac{1}{C_{total, final}} = \frac{1}{KC} + \frac{1}{2C}. \]

Simplifying:
\[ \frac{1}{C_{total, final}} = \frac{K + 2}{2KC}. \]

Thus:
\[ C_{total, final} = \frac{2KC}{K + 2}. \]


(B) [ii)] Final Voltage Difference Across the Combination:

The charge on the initial combination is given by:
\[ Q_{initial} = C_{total, initial} \cdot V = \frac{2C}{3} \cdot V. \]

Since the battery is removed, the total charge remains the same. The final voltage \( V_{final} \) across the new combination is:
\[ V_{final} = \frac{Q_{initial}}{C_{total, final}}. \]

Substituting the values:
\[ V_{final} = \frac{\frac{2C}{3} \cdot V}{\frac{2KC}{K + 2}} = \frac{V \cdot (K + 2)}{3K}. \]


(C) [iii)] Total Energy Stored in the Combination:

The total energy stored in a capacitor is given by:
\[ U = \frac{1}{2} C_{total, final} \cdot V_{final}^2. \]

Substituting \( C_{total, final} = \frac{2KC}{K + 2} \) and \( V_{final} = \frac{V \cdot (K + 2)}{3K} \):
\[ U = \frac{1}{2} \cdot \frac{2KC}{K + 2} \cdot \left(\frac{V \cdot (K + 2)}{3K}\right)^2. \]

Simplifying:
\[ U = \frac{1}{2} \cdot \frac{2KC}{K + 2} \cdot \frac{V^2 \cdot (K + 2)^2}{9K^2}. \]

\[ U = \frac{C V^2 \cdot (K + 2)}{9K}. \]

Thus, the total energy stored is:
\[ U = \frac{C V^2 (K + 2)}{9K}. \] Quick Tip: In capacitors, the total charge remains constant when a battery is removed, and energy calculations depend on the final capacitance and voltage configuration.

(A) [i)] Definition of Electric Flux:

Electric flux is the total number of electric field lines passing through a given surface. Mathematically:
\[ \Phi_E = \int \vec{E} \cdot d\vec{A}, \]
where \( \vec{E} \) is the electric field and \( d\vec{A} \) is the area vector.

Unit of electric flux: \( volt meter (Vm) or Nm^2/C. \)


(B) [ii)] Electric Field Intensity (Using Gauss' Law):

Consider a uniformly charged thin spherical shell of total charge \( Q \) and radius \( R \).

- Outside the Shell \((r > R)\):

By Gauss' law:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}. \]

The Gaussian surface is a sphere of radius \( r \):
\[ E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0}. \]

Thus, the electric field outside the shell is:
\[ E = \frac{Q}{4\pi \epsilon_0 r^2}. \]

- Inside the Shell \((r < R)\):

Since the shell is uniformly charged, no charge is enclosed within the Gaussian surface inside the shell:
\[ E = 0 \quad for r < R. \] Quick Tip: Gauss' law simplifies electric field calculations by utilizing symmetrical Gaussian surfaces.