HINTS & SOLUTIONS

Answer: (B)
Explanation: Both assertions are factual. Conductors with current in the same direction attract due to the magnetic Lorentz force. Also, the magnetic fields produced by each wire midway between them are equal and opposite, canceling out. However, the reason (zero net field midway) is not the cause of the attraction.

Answer: (C)
Explanation: Assertion is true (doping increases charge carriers, hence conductivity). Reason is false because p-type doping (trivalent) increases the number of holes, not electrons.

Answer: (A)
Explanation: In a p-type semiconductor, holes are majority carriers, and it is formed by doping with trivalent atoms (like Boron, Aluminum).

Answer: (D)
Solution: From Faraday's Law of self-induction, $|e| = L \frac{\Delta I}{\Delta t}$.
$100 = L \left(\frac{5.0 - 0.0}{0.1}\right)$
$100 = L (50) \implies L = 2 \text{ H}$.

Answer: (C)
Explanation: The frequency of a wave is a property of the source and remains constant. In a denser medium, the speed of light decreases ($v < c$), and since $v = f\lambda$, the wavelength also decreases.

Answer: (A)
Solution: Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Given: $u = -12 \text{ cm}$, $f = +15 \text{ cm}$ (convex mirror).
$\frac{1}{v} = \frac{1}{15} - \left(\frac{1}{-12}\right) = \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60}$
$v = \frac{60}{9} = +6.67 \text{ cm}$ (formed behind the mirror).

Answer: (B)
Explanation: The question asks specifically for the electric field due to the $+q$ charge only. Since the point P is at distance $r$ from the center, the distance from $+q$ is $(r - a)$. Thus, $\vec{E}_{+q} = \frac{q}{4\pi\epsilon_{0}(r-a)^{2}}\hat{P}$.

Answer: (C)
Solution: When a wire is stretched to $n$ times its original length, the volume remains constant. New length $l' = nl$, new area $A' = A/n$.
New resistance $R' = \rho \frac{l'}{A'} = \rho \frac{nl}{A/n} = n^2 \left(\rho\frac{l}{A}\right) = n^2 R$.
Here $n=5$, so $R' = 5^2 R = 25R$.

Answer: (D)
Explanation: To convert a galvanometer to a voltmeter, its resistance must be very high so it doesn't draw current from the circuit. This is achieved by connecting a high resistance in series.

Answer: (C)
Explanation: In a purely inductive AC circuit, the current lags the applied voltage by a phase angle of $90^\circ$ or $\frac{\pi}{2}$ radians.

Answer: (D)
Solution: Using Einstein's Photoelectric Equation: $K_{max} = h\nu - \Phi$.
Energy of incident photon ($h\nu$) $= \frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}} \text{ eV} = 2.48 \text{ eV}$.
$K_{max} = 2.48 \text{ eV} - 2.14 \text{ eV} = 0.34 \text{ eV}$.
Stopping potential $V_0 = K_{max}/e = 0.34 \text{ V}$.

Answer: (A)
Solution: Using mass-energy equivalence $E = mc^2$.
Given $m = 1 \text{ mg} = 10^{-6} \text{ kg}$, $c = 3 \times 10^8 \text{ m/s}$.
$E = 10^{-6} \times (3 \times 10^8)^2 = 10^{-6} \times 9 \times 10^{16} = 9 \times 10^{10} \text{ J}$.

Solution:
Magnetic field inside a solenoid, $B = \mu_0 n I = \mu_0 \left(\frac{N}{L}\right) I$
Given: $N = 500$, $L = 0.5 \text{ m}$, $I = 5 \text{ A}$.
$B = 4\pi \times 10^{-7} \times \left(\frac{500}{0.5}\right) \times 5$
$B = 4\pi \times 10^{-7} \times 1000 \times 5 = 20\pi \times 10^{-4} \text{ T}$
$B \approx 6.28 \times 10^{-3} \text{ T}$
(1 mark for formula, 1 mark for correct calculation and unit) Solution (Alternative):
Lowest Resistance: Ammeter. (An ideal ammeter has zero resistance so it does not drop voltage and alter the main current when connected in series).
Highest Resistance: Voltmeter. (An ideal voltmeter has infinite resistance so it draws negligible current when connected in parallel across the component being measured).

Solution:
According to Kirchhoff's Voltage Law (KVL), the algebraic sum of changes in potential around any closed loop is zero ($\sum \Delta V = 0$).
When traversing the loop ABCA in a specific direction (e.g., clockwise):
$\sum \text{emf} = \sum IR$
(Student must write the equation corresponding to the specific resistors and batteries shown in the given diagram, assigning positive signs to emf if traversing from - to + terminal, and negative to IR drops in the direction of current).

Answer:
Two important properties of paramagnetic substances:

• They are feebly (weakly) attracted by an external magnetic field. They tend to move from weaker to stronger parts of the field.
• Their relative permeability ($\mu_r$) is slightly greater than 1, and their magnetic susceptibility ($\chi_m$) is a small positive value, which varies inversely with absolute temperature (Curie's Law).

Solution:
Einstein's photoelectric equation is $K_{max} = h\nu - \Phi$ (where $\Phi = h\nu_0$).

• Threshold Frequency: If $\nu < \nu_0$, $K_{max}$ is negative (impossible), proving emission cannot occur below a certain frequency.
• Kinetic Energy Dependence: Above $\nu_0$, $K_{max}$ depends linearly on frequency $\nu$, and is independent of intensity.
• Instantaneous Emission: Energy is provided in discrete packets (photons). A single collision immediately transfers enough energy to eject an electron, causing no time lag.

Solution:
Definition: One atomic mass unit (1 amu or 1 u) is defined as exactly $\frac{1}{12}$th the mass of an unexcited Carbon-12 ($^{12}C$) atom.
$1 \text{ u} = 1.6605 \times 10^{-27} \text{ kg}$.
Energy Equivalent:
Using $E = mc^2$:
In Joules: $E = (1.6605 \times 10^{-27} \text{ kg}) \times (2.9979 \times 10^8 \text{ m/s})^2 \approx 1.492 \times 10^{-10} \text{ J}$.
In MeV: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.
$E (\text{in MeV}) = \frac{1.492 \times 10^{-10}}{1.6 \times 10^{-19} \times 10^6} \approx 931.5 \text{ MeV}$.

Solution:
Let surface charge density on plate 1 be $+\sigma$ and on plate 2 be $-\sigma$.
Given $\sigma = 17.0 \times 10^{-22} \text{ C/m}^2$. Electric field due to a single infinite plate is $E = \frac{\sigma}{2\epsilon_0}$.
(a) Outer region of first plate: Fields from both plates are opposite and equal. $E_{net} = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$.
(b) Outer region of second plate: Similarly, fields are opposite and equal. $E_{net} = 0$.
(c) Between the plates: Fields from both plates point in the same direction (from + to -).
$E_{net} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$
$E_{net} = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} = 1.92 \times 10^{-10} \text{ N/C}$.

Solution:
Microwaves: These are short-wavelength radio waves, with frequencies ranging from about $1 \text{ GHz}$ to $300 \text{ GHz}$ ($\lambda \approx 1 \text{ mm}$ to $0.3 \text{ m}$). They are produced by special vacuum tubes (klystrons, magnetrons). Due to their short wavelengths, they are suitable for radar systems, satellite communication, and microwave ovens.

Infrared Waves: These waves have frequencies lower than visible light ($\lambda \approx 700 \text{ nm}$ to $1 \text{ mm}$). They are produced by hot bodies and molecules. Often called "heat waves," they cause heating effects. Applications include physical therapy, remote controls, night vision goggles, and keeping food warm.

Solution:
Electric Energy: The total work done by the source of emf in maintaining the electric current in the circuit for a given time. Energy $E = VIt = I^2Rt$.
Electric Power: The rate at which electric energy is dissipated or consumed in an electric circuit. $P = VI = I^2R = V^2/R$.
Source of Power: The power comes from the source driving the circuit (e.g., chemical energy in a battery or mechanical energy turning a generator).
Minimizing Power Loss: During transmission, the power loss in cables is $P_c = I^2 R_c$. Since transmitted power is $P = VI$, current $I = P/V$. Therefore, $P_c = (P/V)^2 R_c$. By transmitting power at a very high voltage ($V$), the current ($I$) is significantly reduced, which drastically minimizes the $I^2 R$ heat loss in the transmission cables.

Solution:
Drawbacks of Rutherford's Model:
1. Stability: According to classical electromagnetism, an electron accelerating in a circular orbit must continuously radiate energy, causing it to spiral into the nucleus. Atom should collapse.
2. Spectra: The continuous loss of energy should produce a continuous emission spectrum, but atoms emit discrete line spectra.
Bohr's Postulates:
1. Electrons revolve around the nucleus only in certain stable, non-radiating orbits (Stationary Orbits).
2. Quantization Condition: The angular momentum of the orbiting electron is an integral multiple of $\frac{h}{2\pi}$. ($L = mvr = n\frac{h}{2\pi}$).
3. Frequency Condition: Energy is emitted/absorbed only when an electron jumps from one allowed orbit to another. $h\nu = E_{initial} - E_{final}$.

Solution:
Self-induction: The phenomenon of production of induced emf in a coil when a changing current passes through it.
Self-inductance ($L$): The ratio of total magnetic flux linked with the coil to the current flowing through it ($L = \Phi/I$).
Derivation for long solenoid:
Consider a solenoid of length $l$, area $A$, and total turns $N$. Let $n = N/l$ (turns per unit length).
When current $I$ flows, the magnetic field inside is $B = \mu_0 n I$.
Total magnetic flux linked with the solenoid is $\Phi = (\text{Flux per turn}) \times N = (B \cdot A) \cdot N$
$\Phi = (\mu_0 n I) \cdot A \cdot (n l) = \mu_0 n^2 A l I$.
Since $\Phi = LI$, we get:
$L = \mu_0 n^2 A l = \frac{\mu_0 N^2 A}{l}$.

Solution:
Consider a plane wavefront AB incident on a plane interface XY separating two media of velocities $v_1$ and $v_2$ (where $v_1 > v_2$). Let $i$ be the angle of incidence.
According to Huygens principle, every point on AB acts as a secondary source. In the time $t$ the wavelet from B reaches C ($BC = v_1 t$), the wavelet from A travels a distance $AD = v_2 t$ into the second medium.
Drawing a tangent from C to this wavelet gives the refracted wavefront CD. Let $r$ be the angle of refraction.
From right $\Delta ABC$: $\sin i = \frac{BC}{AC} = \frac{v_1 t}{AC}$
From right $\Delta ADC$: $\sin r = \frac{AD}{AC} = \frac{v_2 t}{AC}$
Dividing the two equations:
$\frac{\sin i}{\sin r} = \frac{v_1 t / AC}{v_2 t / AC} = \frac{v_1}{v_2}$
Since the ratio of velocities $\frac{v_1}{v_2}$ is the relative refractive index $\mu_{21}$, we get $\frac{\sin i}{\sin r} = \mu_{21}$, which is Snell's Law of refraction.

Solution:
(i) (D) $W_{2}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r_{12}}$
(Explanation: Potential due to $q_1$ is $V = \frac{k q_1}{r_{12}}$. Work done $W_2 = q_2 V$.)

(ii) (B) $V_{1,2}=\frac{1}{4\pi\epsilon_{0}}\left(\frac{q_{1}}{r_{1P}}+\frac{q_{2}}{r_{2P}}\right)$
(Explanation: By superposition principle, potential is the scalar sum of individual potentials).

(iii) (C) $W_{3}=\frac{1}{4\pi\epsilon_{0}}\left(\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\right)$
(Explanation: Work done to bring $q_3$ against the field of $q_1$ and $q_2$.)

(iv) (A) $U=\frac{1}{4\pi\epsilon_{0}}\left(\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\right)$
(Explanation: Total energy is the sum of works done: $W_1 + W_2 + W_3$).

Solution:
Impedance of Series LCR:
Let alternating current be $I = I_0 \sin(\omega t)$. The voltage across resistor ($V_R$) is in phase with $I$. Voltage across inductor ($V_L$) leads $I$ by $\frac{\pi}{2}$. Voltage across capacitor ($V_C$) lags $I$ by $\frac{\pi}{2}$.
Using a phasor diagram, $V_L$ and $V_C$ are anti-parallel. The net reactive voltage is $(V_L - V_C)$ assuming $V_L > V_C$.
The resultant voltage $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Substituting $V_R = IR$, $V_L = I X_L$, and $V_C = I X_C$:
$V = I \sqrt{R^2 + (X_L - X_C)^2}$.
Impedance $Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
Resonance Condition:
Resonance occurs when impedance $Z$ is minimum, i.e., $X_L = X_C$, so the current amplitude becomes maximum.
$\omega_0 L = \frac{1}{\omega_0 C} \implies \omega_0^2 = \frac{1}{LC}$
Resonant angular frequency: $\omega_0 = \frac{1}{\sqrt{LC}}$.
Resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$. Solution (Alternative Numerical):
Given: $V_{rms} = 230\text{ V}$, $L = 5.0\text{ H}$, $C = 80 \mu\text{F} = 80 \times 10^{-6}\text{ F}$, $R = 40 \Omega$.
(a) Source frequency at resonance:
$\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{0.02} = 50 \text{ rad/s}$.
$f_r = \frac{\omega_r}{2\pi} = \frac{50}{2\pi} \approx 7.96 \text{ Hz}$.
(b) Impedance and amplitude of current:
At resonance, $X_L = X_C$, so Impedance $Z = R = 40 \Omega$.
$I_{rms} = \frac{V_{rms}}{Z} = \frac{230}{40} = 5.75 \text{ A}$.
Amplitude of current $I_m = I_{rms} \sqrt{2} = 5.75 \times 1.414 \approx 8.13 \text{ A}$.
(c) Potential drops:
$V_R = I_{rms} R = 5.75 \times 40 = 230 \text{ V}$.
$X_L = \omega_r L = 50 \times 5.0 = 250 \Omega$. So, $V_L = I_{rms} X_L = 5.75 \times 250 = 1437.5 \text{ V}$.
$X_C = \frac{1}{\omega_r C} = \frac{1}{50 \times 80 \times 10^{-6}} = 250 \Omega$. So, $V_C = I_{rms} X_C = 5.75 \times 250 = 1437.5 \text{ V}$.
Potential drop across LC combination: $V_{LC} = V_L - V_C = 1437.5 - 1437.5 = 0 \text{ V}$. (Proved).

Solution:
(i) Power of lens $P = \frac{1}{f (\text{in m})} = \frac{1}{0.5} = +2 \text{ D}$.
(ii) Given: $R_1 = +10 \text{ cm}$, $R_2 = -15 \text{ cm}$ (double convex lens). $f = +12 \text{ cm}$.
Using Lens Maker's formula: $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
$\frac{1}{12} = (\mu - 1)\left(\frac{1}{10} - \frac{1}{-15}\right) = (\mu - 1)\left(\frac{1}{10} + \frac{1}{15}\right)$
$\frac{1}{12} = (\mu - 1)\left(\frac{3 + 2}{30}\right) = (\mu - 1)\left(\frac{5}{30}\right) = (\mu - 1)\left(\frac{1}{6}\right)$
$\mu - 1 = \frac{6}{12} = 0.5 \implies \mu = 1.5$.
(iii) $f_a = 20 \text{ cm}$. Refractive index of glass wrt air $^a\mu_g = 1.5$. Water wrt air $^a\mu_w = 1.33 \approx \frac{4}{3}$.
Refractive index of glass wrt water $^w\mu_g = \frac{^a\mu_g}{^a\mu_w} = \frac{1.5}{4/3} = \frac{1.5 \times 3}{4} = 1.125$.
In air: $\frac{1}{f_a} = (^a\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \implies \frac{1}{20} = (0.5)K$
In water: $\frac{1}{f_w} = (^w\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \implies \frac{1}{f_w} = (1.125 - 1)K = (0.125)K$
Dividing the two: $\frac{f_w}{20} = \frac{0.5}{0.125} = 4 \implies f_w = 4 \times 20 = 80 \text{ cm}$. Solution (Alternative):
Lens Maker's Formula Derivation:
Assumptions: Thin lens, small aperture, point object on principal axis, paraxial rays.
Sign Convention: Distances measured from optical center. Direction of incident light is positive.
Refraction at 1st spherical surface (radius $R_1$): Object at $O$, image forms at $I_1$ (virtual object for 2nd surface).
$\frac{\mu_2}{v_1} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1}$ ... (Eq 1)
Refraction at 2nd surface (radius $R_2$): $I_1$ acts as object, final image at $I$.
$\frac{\mu_1}{v} - \frac{\mu_2}{v_1} = \frac{\mu_1 - \mu_2}{R_2}$ ... (Eq 2)
Adding (1) and (2):
$\mu_1 \left(\frac{1}{v} - \frac{1}{u}\right) = (\mu_2 - \mu_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
If object is at infinity ($u = -\infty$), image forms at focus ($v = f$):
$\frac{\mu_1}{f} = (\mu_2 - \mu_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \implies \frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
Numerical: Given $\mu = 1.55$, $f = +20 \text{ cm}$. Double convex: $R_1 = +R$, $R_2 = -R$.
$\frac{1}{20} = (1.55 - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (0.55)\left(\frac{2}{R}\right) = \frac{1.1}{R}$.
$R = 20 \times 1.1 = 22 \text{ cm}$.

Solution:
(i) Intrinsic Semiconductors: Pure semiconductors free from any significant impurity species. The electrical conductivity is solely determined by the thermal generation of charge carriers. Example: Pure Si or Ge.

(ii) Difference between C, Si, and Ge: Despite having the same diamond lattice structure (each atom bonded to 4 neighbors), the energy band gap ($E_g$) between valence and conduction bands differs significantly. For Carbon (Diamond), $E_g \approx 5.4 \text{ eV}$, which is too high for thermal energy at room temperature to break bonds and excite electrons. Hence it acts as an insulator. For Si ($E_g \approx 1.1 \text{ eV}$) and Ge ($E_g \approx 0.7 \text{ eV}$), the gap is small enough that sufficient thermal energy is available at room temperature to push electrons into the conduction band, making them intrinsic semiconductors.

(iii) Generation of Charge Carriers:
At $T = 0 \text{ K}$, all electrons are tightly bound in covalent bonds (valence band is full, conduction band is empty). The crystal acts as a perfect insulator.
As temperature increases ($T > 0 \text{ K}$), thermal energy causes atomic lattice vibrations. Some covalent bonds break, freeing electrons. These freed electrons transition into the conduction band and can move under an electric field.
Simultaneously, the vacancy left behind by the electron in the covalent bond acts as an effective positive charge, called a "hole", in the valence band. Thus, thermal excitation generates electron-hole pairs. In intrinsic semiconductors, $n_e = n_h$ (number density of electrons equals number density of holes).