HINTS & SOLUTIONS

Answer: (b) $NC^{-1}m^{2}$
Explanation: Electric flux is defined as $\phi = \vec{E} \cdot \vec{A}$. The SI unit of Electric field ($E$) is N/C and Area ($A$) is m². Thus, the unit of flux is $Nm^{2}C^{-1}$.

Answer: (c) conservation of charge and the fact that there is no accumulation of charges at a junction.
Explanation: Kirchhoff's Current Law (junction rule) states that total current entering a junction equals total current leaving, which directly implies the conservation of electric charge.

Answer: (d) $\vec{B}\perp\vec{v}$
Explanation: According to the vector form of Biot-Savart law for a moving charge, $d\vec{B} = \frac{\mu_{0}}{4\pi} \frac{q(\vec{v} \times \vec{r})}{r^{3}}$. The cross product indicates that the magnetic field is perpendicular to both the velocity vector and the position vector.

Answer: (a) 500V
Solution: Transformer ratio: $\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}$.
$\frac{400}{V_{p}} = \frac{400}{500} \implies V_{p} = 500 \text{ V}$.

Answer: (c) 53 volt
Solution: Induced emf $|e| = \frac{d\phi}{dt} = \frac{d}{dt}(5t^{2} + 3t + 16) = 10t + 3$.
At $t = 5\text{ s}$, $|e| = 10(5) + 3 = 53 \text{ V}$.

Answer: (b) intensity of light
Explanation: Polaroids cut off vibrations of the electric field in a specific plane, thereby reducing the overall intensity (brightness) of the transmitted light to prevent glare.

Answer: (c) Power of a lens
Explanation: Power of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distance from the optical centre. $P = \frac{1}{f}$.

Answer: (b) work function
Explanation: Work function ($\Phi_{0}$) is the minimum threshold energy required by an electron to just overcome the surface barrier of a metal and escape.

Answer: (b) electrostatic force between protons are repulsive
Explanation: As the nucleus gets larger, the long-range Coulomb repulsion between protons increases. To keep the nucleus stable, extra neutrons are needed to provide additional short-range attractive strong nuclear forces without adding any repulsive forces.

Answer: (c) drift of charges
Explanation: In reverse bias, the majority carrier diffusion is opposed. The tiny reverse saturation current that flows is due to the drift of minority charge carriers swept across the junction by the strong electric field.

Answer: (a)
Explanation: When current passes through a spring, adjacent turns act like parallel wires carrying current in the same direction. According to the magnetic Lorentz force, parallel currents attract, compressing the spring. Assertion and Reason are both true, and Reason is the correct explanation.

Answer: (d) Both assertion and reason are false.
Explanation: The thickness of the depletion layer is not fixed; it decreases in forward bias and increases in reverse bias. Furthermore, the depletion layer *is* composed entirely of immobile charged carrier ions (positive donors and negative acceptors).

Solution:
Conductivity ($\sigma$): It is defined as the reciprocal of the electrical resistivity of the material of a conductor. $\sigma = \frac{1}{\rho}$. Its SI unit is $\Omega^{-1}m^{-1}$ or $Sm^{-1}$.
Variation with Temperature: For metallic conductors, as temperature increases, the thermal agitation of positive ions in the lattice increases. This increases the collision frequency of free electrons, leading to a decrease in average relaxation time ($\tau$). Since $\sigma = \frac{ne^{2}\tau}{m}$, a decrease in $\tau$ results in a decrease in conductivity.

Solution:
Relative Permeability ($\mu_{r}$): It is the ratio of the magnetic permeability of the material ($\mu$) to the magnetic permeability of free space ($\mu_{0}$). Mathematically, $\mu_{r} = \frac{\mu}{\mu_{0}}$. It is a dimensionless quantity.
Relation with Susceptibility: Relative permeability is related to magnetic susceptibility ($\chi_{m}$) by the relation: $\mu_{r} = 1 + \chi_{m}$.

Solution:
Matter Waves: The waves associated with moving material particles (like electrons, protons, etc.) are called matter waves or de-Broglie waves.
de-Broglie Wavelength of a Photon: A photon behaves as a wave with momentum $p = \frac{E}{c}$. From Planck's quantum theory, $E = h\nu = \frac{hc}{\lambda}$. Substituting this gives the de-Broglie wavelength expression: $\lambda = \frac{h}{p}$.

Solution:
Magnetic field inside a solenoid is given by $B = \mu_{0} n I = \mu_{0} \left(\frac{N}{l}\right) I$.
Given: $N = 500$, $l = 0.5 \text{ m}$, $I = 5 \text{ A}$.
$B = 4\pi \times 10^{-7} \times \left(\frac{500}{0.5}\right) \times 5$
$B = 4\pi \times 10^{-7} \times 1000 \times 5 = 20\pi \times 10^{-4} \text{ T}$
$B \approx 6.28 \times 10^{-3} \text{ T}$. Solution (Alternative):
Inductive Reactance ($X_{L}$): The opposition offered by an inductor to the flow of alternating current is called inductive reactance. It is given by $X_{L} = \omega L = 2\pi f L$.
For Direct Current (DC): The frequency ($f$) of steady direct current is zero. Putting $f = 0$ in the formula gives $X_{L} = 2\pi(0)L = 0$. Thus, an ideal inductor offers zero resistance to DC.

Solution:
Mutual Induction: It is the phenomenon of production of induced emf in one coil due to the change in current (and hence magnetic flux) in a neighboring coil.
Coefficient of Mutual Induction ($M$): It is numerically equal to the magnetic flux linked with the secondary coil when a unit current flows through the primary coil. $\phi_{2} = M I_{1}$.
SI Unit: Henry (H).
Expression (for two long coaxial solenoids): $M = \mu_{0} n_{1} n_{2} \pi r_{1}^{2} l$, where $n_{1}, n_{2}$ are turns per unit length, $r_{1}$ is the radius of inner solenoid, and $l$ is their common length.

Solution:
Wavefront: It is the continuous locus of all such particles of the medium which are vibrating in the same phase at any instant.
Proof of Snell's Law:
Let a plane wavefront AB be incident on a plane interface separating medium 1 (velocity $v_{1}$) and medium 2 (velocity $v_{2}$). Let $i$ be the angle of incidence.
By Huygens' principle, the time $t$ taken by light to travel from B to C is $t = \frac{BC}{v_{1}}$. In the same time, the secondary wavelet from A travels a distance $AD = v_{2}t$ in medium 2. Drawing a tangent from C to this arc 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: $\frac{\sin i}{\sin r} = \frac{v_{1}t/AC}{v_{2}t/AC} = \frac{v_{1}}{v_{2}} = \mu_{21}$ (relative refractive index). This is Snell's Law.

Solution:
Mass Defect ($\Delta M$): It is the difference between the sum of the masses of the individual nucleons (protons and neutrons) making up a nucleus and the actual rest mass of the nucleus. $\Delta M = [Z m_{p} + (A-Z) m_{n}] - M_{N}$.
Binding Energy per Nucleon ($\Delta E_{bn}$): It is the average energy required to extract one nucleon from the nucleus. $\Delta E_{bn} = \frac{\Delta M c^{2}}{A}$.
Graph Features:

• The curve has a broad maximum in the mass number range $30 < A < 170$, indicating these intermediate nuclei are highly stable. The peak is at Iron ($^{56}Fe$) with $\approx 8.8 \text{ MeV}$.
• For light nuclei ($A < 30$) and heavy nuclei ($A > 170$), the BE/nucleon is lower, meaning they are relatively less stable, setting the stage for nuclear fusion and fission respectively to attain stability.

Solution:
(a) Linear Charge Density:
Electric field of infinite line charge: $E = \frac{\lambda}{2\pi\epsilon_{0}r} = \frac{2k\lambda}{r}$.
Given $E = 9\times10^{4} \text{ NC}^{-1}$, $r = 2 \text{ cm} = 0.02 \text{ m}$, $k = 9\times10^{9}$.
$9\times10^{4} = \frac{2 \times 9\times10^{9} \times \lambda}{0.02} \implies 9\times10^{4} = 9\times10^{11} \times \lambda$
$\lambda = 10^{-7} \text{ C/m} = 0.1 \mu\text{C/m}$.
(b) Electric Flux through Gaussian Surface:
By Gauss's Law, the net electric flux passing through any closed surface is independent of its shape or size, depending only on the enclosed charge.
$\phi_{E} = \frac{q_{enclosed}}{\epsilon_{0}} = \frac{2.0 \times 10^{-6} \text{ C}}{8.854 \times 10^{-12} \text{ C}^{2}\text{N}^{-1}\text{m}^{-2}}$.
$\phi_{E} \approx 2.26 \times 10^{5} \text{ Nm}^{2}\text{C}^{-1}$.

Solution:
Electromagnetic Spectrum: The orderly distribution of electromagnetic waves according to their wavelength or frequency into distinct groups having widely differing properties is called the EM spectrum.
Main Parts: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, and Gamma rays.
Uses of Microwaves:

• Used in radar systems for aircraft navigation and speed detection due to their short wavelengths.
• Used in microwave ovens for cooking food, as their frequency matches the resonant frequency of water molecules.

Solution:
Cell: A device that provides the necessary potential difference to an electric circuit to maintain a continuous flow of current in it (converts chemical energy to electrical energy).
Condition for Maximum Current in Series Grouping:
Let $n$ identical cells, each of emf $E$ and internal resistance $r$, be connected in series with an external resistance $R$.
Total emf = $nE$. Total internal resistance = $nr$.
Current in the circuit, $I = \frac{nE}{R + nr}$.
Case 1: If $R \gg nr$, then $R + nr \approx R$, so $I \approx n\left(\frac{E}{R}\right)$. (Current is $n$ times the current from a single cell).
Case 2: If $nr \gg R$, then $R + nr \approx nr$, so $I \approx \frac{nE}{nr} = \frac{E}{r}$. (Current is equal to a single cell).
Conclusion: To draw maximum current from a series combination of cells, the external resistance must be very large compared to the total internal resistance of the cells.

Solution:
de-Broglie's Explanation: According to de-Broglie, an electron orbiting the nucleus behaves as a standing matter wave. For an electron wave to survive without destructive interference, the circumference of its circular orbit must contain an integral number of wavelengths.
$2\pi r = n\lambda$ (where $n = 1, 2, 3...$)
From de-Broglie equation, wavelength $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Substituting this: $2\pi r = n\left(\frac{h}{mv}\right)$.
Rearranging the terms: $mvr = n\left(\frac{h}{2\pi}\right)$.
This is precisely Bohr's second postulate of quantization of angular momentum ($L = n\frac{h}{2\pi}$).

Solution:
(A) (b) potential
(Explanation: In parallel connection, the potential difference across all capacitors remains the same, equal to the source voltage).

(B) (b) $2.5\times10^{-9}C$
(Explanation: $Q = CV = (10 \times 10^{-12} \text{ F}) \times 250 \text{ V} = 2500 \times 10^{-12} \text{ C} = 2.5 \times 10^{-9} \text{ C}$).

(C) (a) charge on the positive plate
(Explanation: By convention, the "charge on a capacitor" refers to the magnitude of charge on either plate, which is the charge on the positive plate. The net charge is zero).

(D) (a) $5~\mu F$
(Explanation: Capacitance with dielectric is $C_{m} = K \cdot C_{air}$. So, $10 = 2 \times C_{air} \implies C_{air} = 5 \mu\text{F}$).

Solution:
Lens Maker's Formula Derivation:
Assumptions: Thin lens, paraxial rays, small aperture, point object placed on the principal axis.
Sign Convention: Distances in the direction of incident light are positive. Distances against it are negative.
Let a convex lens be made of material of refractive index $n_{2}$, placed in a medium of refractive index $n_{1}$.
For refraction at the first surface (Radius $R_{1}$): An object at O forms an image at $I_{1}$. This refraction is from rarer to denser.
$\frac{n_{2}}{v_{1}} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R_{1}}$ ... (Eq 1)
For refraction at the second surface (Radius $R_{2}$): The image $I_{1}$ acts as a virtual object for the second surface, forming the final image at $I$. This refraction is from denser to rarer.
$\frac{n_{1}}{v} - \frac{n_{2}}{v_{1}} = \frac{n_{1} - n_{2}}{R_{2}}$ ... (Eq 2)
Adding (1) and (2):
$n_{1}\left(\frac{1}{v} - \frac{1}{u}\right) = (n_{2} - n_{1})\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$
Dividing by $n_{1}$, and if object is at infinity ($u = -\infty$), the image forms at the focus ($v = f$):
$\frac{1}{f} = \left(\frac{n_{2}}{n_{1}} - 1\right)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right) = (n - 1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$. Solution (Alternative Numerical):
(a) Concave Lens:
Given: $h_{o} = +3.0 \text{ cm}$, $u = -14 \text{ cm}$, $f = -21 \text{ cm}$.
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-21} + \frac{1}{-14}$
$\frac{1}{v} = \frac{-2 - 3}{42} = \frac{-5}{42} \implies v = -8.4 \text{ cm}$.
Image is formed 8.4 cm in front of the lens. It is virtual and erect. If the object is moved further away, the image moves towards the principal focus ($21 \text{ cm}$ on the same side) and its size goes on decreasing.
(b) Double Convex Lens:
Given: $R_{1} = +10 \text{ cm}$, $R_{2} = -15 \text{ cm}$, $f = +12 \text{ cm}$.
Lens Maker's formula: $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$
$\frac{1}{12} = (\mu - 1)\left(\frac{1}{10} - \frac{1}{-15}\right) = (\mu - 1)\left(\frac{3 + 2}{30}\right)$
$\frac{1}{12} = (\mu - 1)\left(\frac{5}{30}\right) = (\mu - 1)\left(\frac{1}{6}\right)$
$\mu - 1 = \frac{6}{12} = 0.5 \implies \mu = 1.5$.

Solution:
(a) Depletion Layer: During the formation of a p-n junction, electrons diffuse from the n-side to the p-side and holes from the p-side to the n-side due to concentration gradients. This recombination creates a small region near the junction devoid of mobile charge carriers, containing only uncompensated immobile positive and negative ions. This region is called the depletion layer.
(b) V-I Characteristics:
Forward Bias: The p-side is connected to the positive terminal and n-side to the negative terminal of a battery via a rheostat. A voltmeter is connected in parallel across the diode, and a milliammeter in series. As voltage increases from zero, current remains nearly zero until the barrier potential is overcome (knee voltage). Thereafter, current increases exponentially.
Reverse Bias: The p-side is connected to the negative terminal and n-side to the positive terminal. A microammeter is used due to the very small drift current. The current remains constant (reverse saturation current) as voltage increases, until breakdown voltage is reached, where current sharply spikes.

Solution:
AC Applied to an Inductor:
Let applied voltage be $V = V_{0} \sin\omega t$.
By Kirchhoff's loop rule, the applied voltage and the induced back emf must sum to zero (assuming zero ideal resistance): $V - L\frac{dI}{dt} = 0$
$\frac{dI}{dt} = \frac{V_{0}}{L} \sin\omega t \implies dI = \frac{V_{0}}{L} \sin\omega t dt$
Integrating both sides: $I = \int \frac{V_{0}}{L} \sin\omega t dt = -\frac{V_{0}}{\omega L} \cos\omega t$
Using trigonometric identity $-\cos\theta = \sin(\theta - \frac{\pi}{2})$:
$I = \frac{V_{0}}{\omega L} \sin(\omega t - \frac{\pi}{2}) = I_{0} \sin(\omega t - \frac{\pi}{2})$.
Here, current lags the voltage by a phase angle of $\frac{\pi}{2}$ ($90^{\circ}$).
Inductive Reactance: The peak current is $I_{0} = \frac{V_{0}}{\omega L}$. Comparing with Ohm's law ($I = \frac{V}{R}$), the quantity $\omega L$ acts as the effective resistance. This is the Inductive Reactance $X_{L} = \omega L$. Solution (Alternative Numerical):
(a) Resistor connected to AC:
Given: $R = 100 \text{ }\Omega$, $V_{rms} = 220 \text{ V}$, $f = 50 \text{ Hz}$.
R.M.S value of current: $I_{rms} = \frac{V_{rms}}{R} = \frac{220}{100} = 2.2 \text{ A}$.
Net power consumed: $P_{avg} = V_{rms} I_{rms} \cos\phi$. For a pure resistor, $\cos\phi = 1$.
$P_{avg} = 220 \times 2.2 = 484 \text{ W}$.

(b) Light Bulb Rating:
Given: Power $P = 100 \text{ W}$, Voltage $V = 220 \text{ V}$.
Resistance of the bulb: $P = \frac{V^{2}}{R} \implies R = \frac{V^{2}}{P} = \frac{(220)^{2}}{100} = \frac{48400}{100} = 484 \text{ }\Omega$.
R.M.S current: $I_{rms} = \frac{V_{rms}}{R} = \frac{220}{484} \approx 0.454 \text{ A}$ (or use $P = VI \implies I = \frac{100}{220}$).