HINTS & SOLUTIONS

Answer: (a) $M^{-1}L^{-3}T^{4}A^{2}$
Explanation: From Coulomb's Law, $\epsilon_{0} = \frac{1}{4\pi F}\frac{q_1q_2}{r^2}$. Units are $C^{2}N^{-1}m^{-2}$. Dimensions: $\frac{[AT]^{2}}{[MLT^{-2}][L^{2}]} = \frac{A^{2}T^{2}}{ML^{3}T^{-2}} = M^{-1}L^{-3}T^{4}A^{2}$.

Answer: (c) wire of cross-sectional area 2A
Explanation: Resistance $R = \rho \frac{l}{A}$. Case 1: $R_1 = \rho \frac{l}{A}$. Case 2: $R_2 = \rho \frac{2l}{A/2} = 4\rho \frac{l}{A}$. Case 3: $R_3 = \rho \frac{l/2}{2A} = \frac{1}{4}\rho \frac{l}{A}$. Minimum resistance is in the third case.

Answer: (b) repel each other
Explanation: According to the magnetic Lorentz force and right-hand grip rule, parallel currents attract, while anti-parallel (opposite direction) currents repel each other.

Answer: (c) 212 $\Omega$
Solution: Capacitive reactance $X_C = \frac{1}{2\pi f C}$.
$X_C = \frac{1}{2 \times 3.14 \times 50 \times 15 \times 10^{-6}} = \frac{1}{314 \times 15 \times 10^{-6}} = \frac{10^6}{4710} \approx 212.3 \text{ }\Omega$.

Answer: (d) energy
Explanation: The mechanical work done in moving a magnet against the opposing force (due to the induced current's magnetic field as per Lenz's law) is converted into electrical energy, satisfying the law of conservation of energy.

Answer: (d) $\frac{I_{0}}{2}$
Explanation: According to Malus' Law integration over all angles, when unpolarised light of intensity $I_0$ passes through a single polaroid, exactly half of the intensity is absorbed, leaving the transmitted intensity as $I_0/2$.

Answer: (a) +40 cm
Solution: Focal length $f = \frac{1}{P(\text{in D})} \text{ meters}$.
$f = \frac{1}{+2.5} \text{ m} = +0.4 \text{ m} = +40 \text{ cm}$.

Answer: (d) $6.59\times10^{-34}Js$
Solution: From Einstein's equation, $eV_s = h\nu - \Phi$. Rearranging gives $V_s = \left(\frac{h}{e}\right)\nu - \frac{\Phi}{e}$.
The slope of the graph is $\frac{h}{e} = 4.12 \times 10^{-15} \text{ V s}$.
$h = \text{slope} \times e = 4.12 \times 10^{-15} \times 1.6 \times 10^{-19} = 6.592 \times 10^{-34} \text{ J s}$.

Answer: (c) neutron is unstable
Explanation: Inside the nucleus, strong forces keep neutrons stable. However, a free neutron outside a nucleus is unstable and undergoes beta decay into a proton, an electron, and an antineutrino, with a mean lifetime of about 15 minutes.

Answer: (a) p-type semiconductor
Explanation: When a trivalent atom (e.g., Boron, Aluminum) is doped into a tetravalent lattice (e.g., Silicon), it creates a deficiency of one electron, creating a "hole" which acts as a positive charge carrier, forming a p-type semiconductor.

Answer: (d) Both assertion and reason are false.
Explanation: Assertion is false because a *stationary* charge produces only an electric field, not a magnetic field.
Reason is false because *moving* charges produce both an electric field AND a magnetic field in the surrounding space.

Answer: (b) If both assertion and reason are true but reason is not a correct explanation.
Explanation: Both statements are technically correct definitions. However, the reason a diode can act as a rectifier is specifically because of its unidirectional current property (allowing current in forward bias, blocking in reverse bias), not merely because a rectifier is defined as an AC to DC converter.

Solution:
Mobility ($\mu$): The mobility of charge carriers is defined as the magnitude of drift velocity acquired by them per unit applied electric field. $\mu = \frac{v_{d}}{E}$. Its SI unit is $m^{2}V^{-1}s^{-1}$.
Relation with Relaxation Time: Since drift velocity $v_{d} = \frac{eE\tau}{m}$, substituting this into the mobility formula gives $\mu = \frac{e\tau}{m}$, where $e$ is the charge, $m$ is the mass of the carrier, and $\tau$ is the average relaxation time.

Solution:
Magnetic Intensity ($H$): It is the degree to which a magnetic field can magnetize a material. It is defined as the external magnetizing field independent of the material medium. $H = \frac{B_{0}}{\mu_{0}}$.
Intensity of Magnetisation ($M$ or $I$): It is defined as the magnetic dipole moment developed per unit volume of the material when subjected to a magnetizing field. $M = \frac{m_{net}}{V}$.

Solution:
Photoelectric Effect: The phenomenon of emission of electrons from a metal surface when light of suitable frequency (greater than the threshold frequency) falls on it.
Einstein's Equation: According to Planck's quantum theory, light consists of packets of energy called photons ($E = h\nu$). When a photon strikes an electron, a part of its energy is used to overcome the surface barrier (Work Function, $\Phi_{0} = h\nu_{0}$), and the remaining energy imparts kinetic energy ($K_{max}$) to the emitted electron. By conservation of energy: $h\nu = \Phi_{0} + K_{max} \implies K_{max} = h\nu - h\nu_{0} = \frac{1}{2}mv_{max}^{2}$.

Solution:
Magnetic field due to a long straight wire: $B = \frac{\mu_{0} I}{2\pi r}$.
Given: $I = 35 \text{ A}$, $r = 20 \text{ cm} = 0.2 \text{ m}$.
$B = \frac{4\pi \times 10^{-7} \times 35}{2\pi \times 0.2} = \frac{2 \times 10^{-7} \times 35}{0.2}$
$B = \frac{70 \times 10^{-7}}{0.2} = 350 \times 10^{-7} \text{ T} = 3.5 \times 10^{-5} \text{ T}$. Solution (Alternative):
The capacitive reactance offered by a capacitor is $X_{C} = \frac{1}{2\pi f C}$.
For Direct Current (DC), the frequency $f = 0$. Therefore, $X_{C} = \frac{1}{0} = \infty$. It offers infinite resistance and blocks DC.
For Alternating Current (AC), $f$ has a finite non-zero value, so $X_{C}$ is finite, allowing AC to easily pass through it.

Solution:
Principle: AC Generator works on the principle of Electromagnetic Induction. When a closed coil is rotated in a uniform magnetic field, the magnetic flux linked with it changes continuously, inducing an alternating emf.
Construction: It consists of four main parts:

• Armature: A rectangular coil of many turns of insulated copper wire wound on a soft iron core.
• Field Magnet: A strong permanent magnet or electromagnet with concave poles (N and S) providing a uniform radial magnetic field.
• Slip Rings ($R_{1}, R_{2}$): Two hollow metallic rings to which the ends of the armature coil are connected. They rotate with the coil.
• Brushes ($B_{1}, B_{2}$): Stationary carbon rods pressing against the slip rings to pass the induced current to the external circuit.

Solution:
Interference: The phenomenon of redistribution of light energy due to the superposition of light waves from two coherent sources.
Conditions for Constructive Interference (Maxima):
The phase difference ($\phi$) between the two waves must be an even multiple of $\pi$: $\phi = 2n\pi$ where $n = 0, 1, 2...$
The path difference ($\Delta x$) must be an integral multiple of wavelength: $\Delta x = n\lambda$.
Conditions for Destructive Interference (Minima):
The phase difference ($\phi$) must be an odd multiple of $\pi$: $\phi = (2n+1)\pi$ where $n = 0, 1, 2...$
The path difference ($\Delta x$) must be an odd integral multiple of half wavelength: $\Delta x = (n + \frac{1}{2})\lambda$.

Solution:
(a) Nuclear Forces: The strong attractive forces between nucleons (protons and neutrons) inside a nucleus that hold them together against the massive electrostatic repulsion of protons.
Properties: 1. Strongest force in nature. 2. Very short range (operative up to 2-3 fm). 3. Charge independent (same for p-p, n-n, p-n). 4. Spin dependent and non-central.
(b) Conversion:
Using $E = mc^2$, where $m = 1 \text{ a.m.u.} = 1.6605 \times 10^{-27} \text{ kg}$.
$E = (1.6605 \times 10^{-27}) \times (3 \times 10^8)^2 \text{ Joules} \approx 1.49 \times 10^{-10} \text{ J}$.
To convert to MeV, divide by $1.6 \times 10^{-13} \text{ J/MeV}$.
$E \approx 931.5 \text{ MeV}$.

Solution:
Given: Dipole moment magnitude $p = 4 \times 10^{-9} \text{ C m}$ (assuming standard units), angle $\theta = 30^{\circ}$, Electric field $E = 5 \times 10^{4} \text{ N C}^{-1}$.
The magnitude of torque acting on the dipole is given by $\tau = pE \sin\theta$.
$\tau = (4 \times 10^{-9}) \times (5 \times 10^{4}) \times \sin(30^{\circ})$
$\tau = (20 \times 10^{-5}) \times 0.5$
$\tau = 10 \times 10^{-5} \text{ N m} = 10^{-4} \text{ N m}$.

Solution:
Displacement Current ($I_d$): It is that current which comes into existence in a region wherever the electric field and hence the electric flux is changing with time. $I_d = \epsilon_{0} \frac{d\phi_{E}}{dt}$.
Four Maxwell's Equations:

1. Gauss's Law for Electricity: $\oint \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_{0}}$
2. Gauss's Law for Magnetism: $\oint \vec{B} \cdot d\vec{S} = 0$ (Monopoles do not exist)
3. Faraday's Law of EMI: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_{B}}{dt}$
4. Ampere-Maxwell Law: $\oint \vec{B} \cdot d\vec{l} = \mu_{0} \left(I_c + \epsilon_{0} \frac{d\phi_{E}}{dt}\right)$

Solution:
Internal Resistance ($r$): The resistance offered by the electrolyte and electrodes of a cell to the flow of current within the cell itself.
Derivation:
Consider a cell of emf $E$ and internal resistance $r$ connected across an external resistance $R$. Current $I$ flows.
The total resistance of circuit $= R + r$. By Ohm's law, $I = \frac{E}{R+r} \implies E = I(R+r)$.
The terminal potential difference across $R$ is $V = IR$.
Substituting $I = \frac{V}{R}$ into the first equation:
$E = \frac{V}{R}(R+r) \implies \frac{E}{V} = \frac{R+r}{R} = 1 + \frac{r}{R}$
$\frac{r}{R} = \frac{E}{V} - 1 \implies r = \left(\frac{E}{V} - 1\right)R$.

Solution:
(a) Bohr's Postulates:

• Electrons revolve around the nucleus in specific, non-radiating circular orbits called stationary states.
• The angular momentum of the orbiting electron is quantized: $L = mvr = n\frac{h}{2\pi}$.
• Energy is absorbed/emitted as a photon when an electron jumps between orbits: $h\nu = E_{f} - E_{i}$.

(b) Energy Breakdown:
Given Total Energy $E = -13.6 \text{ eV}$.
In a hydrogen atom, Kinetic Energy is equal to the negative of Total Energy: $K = -E = -(-13.6 \text{ eV}) = +13.6 \text{ eV}$.
Potential Energy is twice the Total Energy: $U = 2E = 2(-13.6 \text{ eV}) = -27.2 \text{ eV}$.

Solution:
(A) (c) $90^{\circ}$
(Explanation: The paragraph states "electric field E at a point is perpendicular to equipotential surface". Angle is 90 degrees).

(B) (d) Both scalars.
(Explanation: Electric potential is work done per unit charge (scalar). Electric flux is the dot product of two vectors, $\vec{E} \cdot \vec{A}$, resulting in a scalar).

(C) (a) Concentric spheres
(Explanation: At a very great distance, any collection of charges with a non-zero net charge appears essentially as a single point charge. The equipotential surfaces for a point charge are concentric spheres).

(D) (b) 2.5 J
(Explanation: Energy acquired $U = qV = (5 \text{ C}) \times (0.5 \text{ V}) = 2.5 \text{ Joules}$).

Solution:
Compound Microscope & Magnification:
A compound microscope consists of two convex lenses: an objective lens (short focal length and small aperture) and an eyepiece (moderate focal length and aperture).
Working: The object is placed just outside the focus of the objective, forming a real, inverted, and magnified image ($I_1$). The eyepiece is adjusted so $I_1$ lies within its focal length, acting as a simple magnifier to produce a highly magnified, virtual final image ($I_2$) at the least distance of distinct vision (D) or at infinity.
Magnifying Power ($M$): $M = m_o \times m_e$. For final image at D: $M = \left(\frac{L}{f_o}\right) \left(1 + \frac{D}{f_e}\right)$.
To increase M: Both the focal length of the objective ($f_o$) and the eyepiece ($f_e$) should be small. Solution (Alternative Numerical):
(a) Lens Maker's Formula:
Given: Refractive index $\mu = 1.55$, Focal length $f = +20 \text{ cm}$. For a double convex lens with same radii, $R_{1} = +R$ and $R_{2} = -R$.
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right) \implies \frac{1}{20} = (1.55 - 1)\left(\frac{1}{R} - \frac{1}{-R}\right)$
$\frac{1}{20} = (0.55)\left(\frac{2}{R}\right) = \frac{1.1}{R} \implies R = 20 \times 1.1 = 22 \text{ cm}$.

(b) Lenses in Contact:
Converging lens $f_{1} = +f$. Diverging lens $f_{2} = -f$.
Power of combination $P = P_{1} + P_{2} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{f} - \frac{1}{f} = 0$.
Equivalent focal length $\frac{1}{F} = 0 \implies F = \infty$.
The system behaves neither as converging nor diverging, but as a plain glass plate.

Solution:
(a) Rectifier & Full Wave Rectification:
Rectifier: A device which converts alternating current (AC) into direct current (DC) utilizing the unidirectional conduction property of a p-n junction diode.
Full Wave Rectifier: It rectifies both halves of the AC input cycle. It uses a center-tapped step-down transformer and two diodes ($D_{1}, D_{2}$).
Working: During the positive half cycle of AC, terminal A is positive and B is negative. $D_{1}$ is forward biased (conducts) and $D_{2}$ is reverse biased (does not conduct). Current flows through the load resistor ($R_L$). During the negative half cycle, A is negative and B is positive. $D_{1}$ is reverse biased, $D_{2}$ is forward biased. Current again flows through $R_L$ in the same direction. Thus, unidirectional pulsing DC is achieved for both cycles.

(b) Output Frequencies:
Input Frequency $= 50 \text{ Hz}$.
Half Wave Rectifier: Yields one pulse per cycle. Output frequency $= \text{Input frequency} = 50 \text{ Hz}$.
Full Wave Rectifier: Yields two pulses per cycle. Output frequency $= 2 \times \text{Input frequency} = 2 \times 50 = 100 \text{ Hz}$.

Solution:
Transformer:
Principle: Based on mutual induction. When AC passes through the primary coil, the changing magnetic flux induces an alternating emf in the secondary coil.
Construction: Consists of two coils (primary and secondary) insulated from each other and wound on a laminated soft iron core.
Energy Losses: 1. Copper loss (Joule heating in wires). 2. Iron/Eddy current loss (heating in core). 3. Hysteresis loss (energy lost in repeated magnetization cycles). 4. Flux leakage (all primary flux doesn't reach secondary).
Efficiency < 1: Due to these inevitable physical energy losses, output power is always slightly less than input power.
Solution (Alternative Numerical):
Given: $V_{0} = 283 \text{ V}$, $f = 50 \text{ Hz}$, $R = 3 \text{ }\Omega$, $L = 25.48 \text{ mH} = 25.48 \times 10^{-3} \text{ H}$, $C = 796 \text{ }\mu\text{F} = 796 \times 10^{-6} \text{ F}$.
Impedance ($Z$):
$X_{L} = 2\pi f L = 2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} = 314 \times 0.02548 \approx 8 \text{ }\Omega$.
$X_{C} = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}} = \frac{1}{314 \times 796 \times 10^{-6}} = \frac{10^6}{249944} \approx 4 \text{ }\Omega$.
$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}} = \sqrt{3^{2} + (8 - 4)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ }\Omega$.

Power Dissipated ($P$):
$V_{rms} = \frac{V_{0}}{\sqrt{2}} = \frac{283}{1.414} \approx 200 \text{ V}$.
Current $I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{5} = 40 \text{ A}$.
Power $P = I_{rms}^{2} R = (40)^{2} \times 3 = 1600 \times 3 = 4800 \text{ W}$.