HINTS & SOLUTIONS

Answer: (c) $\vec{P}\times\vec{E}$
Explanation: The torque on a dipole in a uniform electric field is given by the cross product of the dipole moment vector and the electric field vector. It tends to align the dipole with the field.

Answer: (a) Drift velocity alone.
Explanation: The random thermal velocity of electrons averages to zero and does not constitute a net current. Current is determined solely by the slow, net directional motion of electrons under an applied electric field, known as drift velocity ($I = neAv_d$).

Answer: (b) Fleming left hand rule.
Explanation: Fleming's Left-Hand Rule is used to find the direction of the magnetic force on a current-carrying conductor (or moving charge) in a magnetic field. The thumb represents Force (F), the forefinger Magnetic Field (B), and the middle finger Current (I).

Answer: (b) Transmission lines can be made thinner.
Explanation: While higher voltage lowers the current (reducing $I^2R$ loss), at extremely high voltages, the "skin effect" and "corona discharge" become significant. To mitigate corona losses and electrical breakdown of air, transmission lines must actually be made thicker (or use bundled conductors). Therefore, stating they can be made thinner is practically incorrect for high-voltage transmission.

Answer: (a) increases when they are brought nearer and is same as $M_{21}$ of coil 2 with respect to coil 1.
Explanation: Mutual inductance depends on the geometric proximity and orientation of the coils. Bringing them closer increases the flux linkage, thereby increasing M. By the reciprocity theorem, $M_{12} = M_{21}$.

Answer: (a) partially plane polarised
Explanation: If light is unpolarised, intensity remains constant upon rotation. If it is completely plane polarised, intensity drops to zero at $90^\circ$ (crossed position). If intensity varies but never reaches zero, it indicates a mixture of unpolarised and polarised light (partially polarised).

Answer: (b) -25 cm
Solution: Focal length $f = \frac{1}{P(\text{in D})} \text{ meters}$.
$f = \frac{1}{-4.0} \text{ m} = -0.25 \text{ m} = -25 \text{ cm}$. The negative sign confirms it is a concave lens.

Answer: (a) $2.76\times10^{-34}m$
Solution: $\lambda = \frac{h}{mv}$.
$\lambda = \frac{6.63 \times 10^{-34}}{0.12 \times 20} = \frac{6.63 \times 10^{-34}}{2.4} \approx 2.762 \times 10^{-34} \text{ m}$.

Answer: (c) nuclear fission
Explanation: An atom bomb relies on the uncontrolled chain reaction of nuclear fission, where heavy unstable nuclei (like Uranium-235 or Plutonium-239) split into lighter nuclei, releasing massive amounts of energy.

Answer: (c) micrometer
Explanation: The thickness of the depletion region in a typical p-n junction diode is very small, usually in the order of $10^{-6} \text{ m}$ to $10^{-7} \text{ m}$, which corresponds to about one-tenth of a micrometer ($0.1 \text{ }\mu\text{m}$).

Answer: (b) If both assertion and reason are true but reason is not a correct explanation.
Explanation: The assertion is true because $\vec{F} = q(\vec{v} \times \vec{B})$, meaning force is perpendicular to $\vec{B}$. The reason is also true because $\vec{F} = q\vec{E}$, meaning electric force is parallel/anti-parallel to $\vec{E}$. However, the behavior of electric force is not the reason for the perpendicular nature of the magnetic force.

Answer: (b) If both assertion and reason are true but reason is not a correct explanation.
Explanation: The band gap in Si ($1.1 \text{ eV}$) is indeed greater than in Ge ($0.7 \text{ eV}$). It is also true that in an intrinsic semiconductor, $n_e = n_h$. However, the equality of carriers has nothing to do with why the structural band gap energies of the two elements differ.

Solution:
Resistivity ($\rho$): It is defined as the resistance offered by a conductor of unit length and unit cross-sectional area. $\rho = R\frac{A}{l}$.
Factors: It depends strictly on (i) the nature of the material (electron density) and (ii) the temperature. It is independent of the physical dimensions of the conductor.
SI Unit: Ohm-meter ($\Omega \cdot m$).
Least Resistivity: Silver (Ag) has the lowest resistivity among all metals.

Solution:
Magnetic Field Lines: These are imaginary continuous curves in a magnetic field such that the tangent at any point gives the direction of the net magnetic field at that point.
Why they don't cross: If two magnetic field lines were to intersect at a point, there would be two tangents drawn at that single point. This would mean the magnetic field has two different directions at the same point simultaneously, which is physically impossible.

Solution:
Effect: The photoelectric current is directly proportional to the intensity of incident light (provided the frequency is above the threshold frequency). This is because higher intensity means more photons strike the surface per second, ejecting proportionally more photoelectrons.
Graph description: The graph of Photocurrent (y-axis) versus Intensity of light (x-axis) is a straight line passing through the origin.

Solution:
Magnetic force on a current-carrying wire is $F = I L B \sin\theta$.
Given: $I = 10 \text{ A}$, $L = 3.0 \text{ cm} = 0.03 \text{ m}$, $B = 0.27 \text{ T}$.
The wire is perpendicular to the axis of the solenoid, so the angle between the length of the wire and the magnetic field is $\theta = 90^\circ$ ($\sin 90^\circ = 1$).
$F = 10 \times 0.03 \times 0.27 \times 1$
$F = 0.3 \times 0.27 = 0.081 \text{ N}$. The direction is given by Fleming's Left-Hand Rule. Solution (Alternative):
Differences:
1. Voltage & Current: Step-up increases AC voltage and decreases current. Step-down decreases voltage and increases current.
2. Turns Ratio: In step-up, secondary turns > primary turns ($N_s > N_p$). In step-down, $N_s < N_p$.
Conservation of Energy: No, a step-up transformer does not contradict the principle of conservation of energy. While it steps up the voltage, it simultaneously steps down the current such that the output power ($V_s I_s$) is almost equal to the input power ($V_p I_p$), minus minor heat losses.

Solution:
Self Inductance of Solenoid: It is the property of the solenoid by virtue of which it opposes any change in the strength of the current flowing through it by inducing an emf in itself.
Energy Stored Derivation:
When current $I$ grows in an inductor, an opposing induced emf $e = -L \frac{dI}{dt}$ is set up.
To maintain the flow of current, the external source must do work against this emf.
Rate of doing work (Power) $P = \frac{dW}{dt} = |e| I = \left(L \frac{dI}{dt}\right) I$.
Small work done in time $dt$ is $dW = L I \, dI$.
Total work done in establishing current from $0$ to $I$ is:
$W = \int_{0}^{I} L I \, dI = L \left[\frac{I^2}{2}\right]_{0}^{I} = \frac{1}{2} L I^{2}$.
This work done is stored in the inductor in the form of magnetic potential energy. $U = \frac{1}{2} L I^{2}$.

Solution:
Huygens' Principle: 1. Every point on a given wavefront acts as a fresh source of new disturbances called secondary wavelets. 2. These wavelets spread out in all directions with the speed of light. 3. The new wavefront is the forward envelope (tangent) of these secondary wavelets at any instant.
Proof of Reflection: Let a plane wavefront AB strike a reflecting surface XY at angle $i$. While wavelet from B travels to C in time $t$ ($BC=vt$), the wavelet from A reflects and spreads a distance $AD=vt$. Drawing a tangent from C to this spherical arc at D gives the reflected wavefront CD at angle $r$.
In $\Delta ABC$ and $\Delta ADC$:
$AC$ is common.
$\angle B = \angle D = 90^\circ$.
$BC = AD = vt$.
By RHS congruence, $\Delta ABC \cong \Delta ADC$. Therefore, $\angle BAC = \angle DCA \implies \angle i = \angle r$. (Law of Reflection).
Validity: Yes, Huygens' principle is purely a kinematic theory and is valid for all types of waves, including longitudinal sound waves.

Solution:
Nuclear Fusion: It is the process in which two lighter nuclei combine at extremely high temperatures and pressures to form a single, heavier, and more stable nucleus, releasing a massive amount of energy (due to mass defect).
Differences:
1. Process: Fusion is the combining of light nuclei; Fission is the splitting of a heavy nucleus.
2. Conditions: Fusion requires extreme temperature/pressure (millions of Kelvin); Fission can occur at room temperature by bombarding with thermal neutrons.
Examples:
Fusion: Combination of Deuterium and Tritium to form Helium (energy source of the Sun and Hydrogen bombs).
Fission: Splitting of Uranium-235 when bombarded by a slow neutron into Barium and Krypton (used in nuclear reactors).

Solution:
Given charges: $q_A = 2.5 \times 10^{-7} \text{ C}$ at $(0, 0, -15\text{ cm})$ and $q_B = -2.5 \times 10^{-7} \text{ C}$ at $(0, 0, +15\text{ cm})$.
Total Charge:
$Q_{total} = q_A + q_B = 2.5 \times 10^{-7} - 2.5 \times 10^{-7} = 0$.
Electric Dipole Moment ($\vec{p}$):
Magnitude $p = q \times 2a$, where $2a$ is the distance between the charges.
Distance $2a = |+15 - (-15)| = 30 \text{ cm} = 0.30 \text{ m}$.
$p = (2.5 \times 10^{-7} \text{ C}) \times (0.30 \text{ m}) = 0.75 \times 10^{-7} \text{ C m} = 7.5 \times 10^{-8} \text{ C m}$.
The direction of dipole moment is always from the negative charge to the positive charge. Here, from $B (+z)$ to $A (-z)$. Thus, the direction is along the negative z-axis.

Solution:
Electromagnetic Waves: Waves consisting of oscillating electric and magnetic fields that are perpendicular to each other and perpendicular to the direction of wave propagation.
Production: EM waves are produced by accelerated or oscillating electric charges. An oscillating charge produces an oscillating electric field, which in turn produces an oscillating magnetic field, propagating through space.
Properties:

• They do not require any material medium to travel (can travel through a vacuum).
• They travel with the speed of light in a vacuum ($c = 3 \times 10^8 \text{ m/s}$).
• They are transverse in nature.
• They carry energy and momentum, which are shared equally between the electric and magnetic fields.

Solution:
Wheatstone Bridge: An arrangement of four resistances (P, Q, R, S) in the form of a quadrilateral used to accurately determine an unknown resistance.
Balancing Condition: A bridge is balanced when the galvanometer shows zero deflection ($I_g = 0$), meaning points B and D are at the same potential.
Derivation: Apply Kirchhoff's Loop Rule to the closed loops in a balanced state.
Loop ABDA: $I_1 P + 0 - I_2 R = 0 \implies I_1 P = I_2 R$ ... (Eq 1)
Loop BCDB: $I_1 Q - 0 - I_2 S = 0 \implies I_1 Q = I_2 S$ ... (Eq 2)
(Since $I_g = 0$, current through P equals Q, and current through R equals S).
Dividing Eq 1 by Eq 2:
$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \implies \frac{P}{Q} = \frac{R}{S}$.
This is the required condition for a balanced Wheatstone bridge.

Solution:
Inferences from Rutherford's Experiment:

1. Most $\alpha$-particles passed straight through the gold foil undeflected, indicating that most of the space inside an atom is empty.
2. A very small fraction of $\alpha$-particles were deflected by large angles (some even $>90^\circ$). This indicates that the entire positive charge and almost all the mass of the atom are concentrated in an extremely small, dense central core called the nucleus.

Graph description: A graph of Number of scattered particles $N(\theta)$ on the y-axis versus Scattering angle $\theta$ on the x-axis shows a rapid exponential-like decay. $N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$. It starts very high near $0^\circ$ and drops drastically as $\theta$ approaches $180^\circ$.

Solution:
(A) (a) increases
(Explanation: In parallel, equivalent capacitance is the sum of individual capacitances $C_p = C_1 + C_2 + ...$, so the overall capacity increases).

(B) (c) the potential increases
(Explanation: If an isolated charged capacitor is pulled apart, charge Q is conserved. Capacitance $C = \epsilon_0 A / d$ decreases as $d$ increases. Since $V = Q/C$, if C decreases, V must increase).

(C) (b) $\frac{1}{2}n~CV^{2}$
(Explanation: Equivalent capacitance in parallel is $nC$. The energy stored is $U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} (nC) V^2$).

(D) (d) $7.5\times10^{-8}J$
(Explanation: Energy $U = \frac{1}{2} C V^2 = \frac{1}{2} \times (15 \times 10^{-12} \text{ F}) \times (100 \text{ V})^2$
$U = 7.5 \times 10^{-12} \times 10000 = 7.5 \times 10^{-8} \text{ Joules}$).

Solution:
Refraction at Convex Spherical Surface:
Consider a point object O placed on the principal axis in a rarer medium ($n_1$). The ray strikes the convex surface of a denser medium ($n_2$) at point A, bending towards the normal to form a real image I.
Let $\alpha, \beta, \gamma$ be the angles made by the object ray, image ray, and normal with the principal axis respectively.
For small aperture, angles are small: $\alpha \approx \frac{AP}{PO}$, $\beta \approx \frac{AP}{PI}$, $\gamma \approx \frac{AP}{PC}$.
From the geometry of the triangle, exterior angle $i = \alpha + \gamma$ and exterior angle $\gamma = r + \beta \implies r = \gamma - \beta$.
By Snell's Law for small angles: $n_1 i = n_2 r$.
Substituting $i$ and $r$: $n_1(\alpha + \gamma) = n_2(\gamma - \beta)$.
$n_1\left(\frac{AP}{PO} + \frac{AP}{PC}\right) = n_2\left(\frac{AP}{PC} - \frac{AP}{PI}\right)$.
Dividing by AP: $\frac{n_1}{PO} + \frac{n_2}{PI} = \frac{n_2 - n_1}{PC}$.
Applying sign convention: $PO = -u$, $PI = +v$, $PC = +R$.
$\frac{n_1}{-u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R} \implies \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. Solution (Alternative Numerical):
(a) Convex Mirror:
Given: $h_o = 4.5 \text{ cm}$, $u = -12 \text{ cm}$, $f = +15 \text{ cm}$.
Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \left(\frac{1}{-12}\right) = \frac{1}{15} + \frac{1}{12}$
$\frac{1}{v} = \frac{4 + 5}{60} = \frac{9}{60} \implies v = \frac{60}{9} = +6.67 \text{ cm}$. (Image is virtual, behind the mirror).
Magnification $m = \frac{-v}{u} = \frac{-(60/9)}{-12} = \frac{60}{108} = \frac{5}{9} \approx +0.55$.
As the needle is moved farther away ($u \rightarrow -\infty$), the virtual image moves towards the principal focus ($v \rightarrow 15\text{ cm}$) and its size goes on shrinking until it becomes a point image.

(b) Lenses in Contact:
Convex lens $f_1 = +30 \text{ cm}$. Concave lens $f_2 = -20 \text{ cm}$.
Equivalent focal length: $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{30} + \frac{1}{-20} = \frac{2 - 3}{60} = \frac{-1}{60}$.
$F = -60 \text{ cm}$. Since the equivalent focal length is negative, the system acts as a diverging lens.

Solution:
(a) Doping & Band Theory:
Doping: The deliberate addition of desirable impurities to an intrinsic semiconductor to vastly increase its conductivity is called doping.
Band Theory Distinctions:
1. Conductor: The valence band and conduction band overlap ($E_g = 0$). Electrons are free to move at all temperatures.
2. Insulator: Large forbidden energy gap ($E_g > 3 \text{ eV}$) between a full valence band and an empty conduction band. Thermal energy cannot bridge this gap.
3. Semiconductor: Small forbidden energy gap ($E_g \approx 1 \text{ eV}$). At 0K it is an insulator, but at room temperature, some electrons gain enough energy to jump to the conduction band, allowing partial conduction.

(b) Formation of n-type Semiconductor:
An intrinsic semiconductor (like Si or Ge, which are tetravalent) is doped with a pentavalent impurity (like Phosphorus, Arsenic, or Antimony). Four of the five valence electrons of the impurity atom form covalent bonds with four neighboring host atoms. The fifth electron is loosely bound and easily excited into the conduction band at room temperature. Thus, electrons become the majority charge carriers, creating an n-type semiconductor.

Solution:
Transformer:
Principle: It works on the principle of mutual induction. An alternating current in the primary coil generates a varying magnetic flux, which links with the secondary coil and induces an alternating emf in it.
Construction: It consists of a primary coil and a secondary coil wound on a laminated soft iron core to provide a low reluctance path for the magnetic flux.
Working: $V_p = -N_p \frac{d\phi}{dt}$ and $V_s = -N_s \frac{d\phi}{dt}$. Therefore, $\frac{V_s}{V_p} = \frac{N_s}{N_p}$. If $N_s > N_p$, it's a step-up transformer; if $N_s < N_p$, it's step-down.
Four Sources of Energy Loss:
1. Copper Loss: Heat generated ($I^2R$) in the resistance of the windings.
2. Eddy Current Loss: Induced circulating currents in the iron core dissipate heat.
3. Hysteresis Loss: Energy spent in reversing the magnetization of the core during AC cycles.
4. Flux Leakage: Not all magnetic flux generated in the primary successfully links to the secondary.
Efficiency: Due to the unavoidable presence of these heat and magnetic losses, the output power is always less than the input power, making efficiency always less than 1 (100%). Solution (Alternative Numerical):
Given: $V_0 = 283 \text{ V}$, $f = 50 \text{ Hz}$, $R = 3 \text{ }\Omega$.
$L = 25.48 \text{ mH} = 25.48 \times 10^{-3} \text{ H}$, $C = 796 \text{ }\mu\text{F} = 796 \times 10^{-6} \text{ F}$.

1. Impedance ($Z$):
Angular frequency $\omega = 2\pi f = 2 \times 3.14 \times 50 = 314 \text{ rad/s}$.
Inductive Reactance $X_L = \omega L = 314 \times 25.48 \times 10^{-3} \approx 8.0 \text{ }\Omega$.
Capacitive Reactance $X_C = \frac{1}{\omega C} = \frac{1}{314 \times 796 \times 10^{-6}} = \frac{10^6}{249944} \approx 4.0 \text{ }\Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{3^2 + (8 - 4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ }\Omega$.

2. Power Dissipated ($P$):
RMS Voltage $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{283}{1.414} \approx 200 \text{ V}$.
RMS Current $I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{5} = 40 \text{ A}$.
Power $P = I_{rms}^2 R = (40)^2 \times 3 = 1600 \times 3 = 4800 \text{ W}$.