HINTS & SOLUTIONS

Answer: (c) $N^{-1}m^{-2}C^{2}$
Explanation: From Coulomb's Law, force $F = \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r^{2}}$. Rearranging for permittivity gives $\epsilon_{0} = \frac{q_{1}q_{2}}{4\pi F r^{2}}$. Substituting the units yields $\frac{C \cdot C}{N \cdot m^{2}} = C^{2}N^{-1}m^{-2}$.

Answer: (d) 0.273 A
Solution: Power $P = VI$.
Current $I = \frac{P}{V} = \frac{60}{220} = \frac{3}{11} \approx 0.2727 \text{ A} \approx 0.273 \text{ A}$.

Answer: (a) South to North
Explanation: By convention, the magnetic dipole moment vector points from the magnetic South pole to the magnetic North pole inside the magnet (analogous to electric dipole moment pointing from negative to positive charge).

Answer: (c) D.C.
Explanation: A transformer works on the principle of mutual induction, which requires a constantly changing magnetic flux. Direct Current (D.C.) provides a steady magnetic field, meaning no flux change and therefore no induced EMF.

Answer: (b) charge
Explanation: Electromagnetic waves are made of oscillating electric and magnetic fields. They carry energy, momentum, and information through space, but they are chargeless and do not transport electrical charge.

Answer: (c) $\frac{f_{1}f_{2}}{f_{1}+f_{2}}$
Solution: For two thin lenses in contact, the equivalent focal length is given by $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$.
$\frac{1}{F} = \frac{f_{2} + f_{1}}{f_{1}f_{2}} \implies F = \frac{f_{1}f_{2}}{f_{1}+f_{2}}$.

Answer: (a) dual nature
Explanation: The de-Broglie equation ($\lambda = \frac{h}{p}$) links a wave property (wavelength, $\lambda$) to a particle property (momentum, $p$), formally stating the dual nature of matter.

Answer: (c) Nuclear fusion
Explanation: A hydrogen bomb relies on the uncontrolled nuclear fusion of hydrogen isotopes (deuterium and tritium) to release massive amounts of energy.

Answer: (b) Immobile ions
Explanation: The depletion layer is a region near the p-n junction stripped of mobile charge carriers (electrons and holes). It contains only the uncovered, fixed (immobile) positive donor ions and negative acceptor ions.

Answer: (c) 55 km
Solution: The Line-of-Sight (LOS) distance is $d = \sqrt{2hR}$.
$d = \sqrt{2 \times 240 \times 6.4 \times 10^{6}} = \sqrt{480 \times 6.4 \times 10^{6}} = \sqrt{3072 \times 10^{6}} \text{ m}$.
$d \approx 55.42 \times 10^{3} \text{ m} = 55.42 \text{ km}$. Rounding to nearest option yields 55 km.

Answer: (c) difference between apparent and real depth of a pond
Explanation: Brilliance of a diamond, optical fibres, and mirages are all applications/effects of Total Internal Reflection. The apparent shallowing of a pond is strictly due to simple refraction of light.

Answer: (d) Photoelectric effect
Explanation: Interference, diffraction, and polarisation can only be explained by treating light as a wave. The photoelectric effect can only be explained by treating light as a particle (photon), therefore it does not support wave nature.

Solution:
Electric field lines indicate the direction of the net electric field at any point. If two lines were to cross, it would mean that at the point of intersection, there are two different directions for the electric field (represented by drawing two tangents). Since a net electric field at a single point can only have one unique direction, two lines can never intersect.

Solution:
The opposition offered by a capacitor is called capacitive reactance, given by $X_{C} = \frac{1}{2\pi f C}$, where $f$ is frequency.
For direct current (D.C.), the frequency $f = 0$. Substituting this into the formula gives $X_{C} = \frac{1}{0} = \infty$. Thus, a capacitor offers infinite resistance to D.C., effectively blocking it.

Solution:
Average electric energy density $u_{E} = \frac{1}{2}\epsilon_{0}E^{2}_{rms}$.
In an EM wave, $E_{rms} = cB_{rms}$ and the speed of light $c = \frac{1}{\sqrt{\mu_{0}\epsilon_{0}}}$.
Substituting $E_{rms}$: $u_{E} = \frac{1}{2}\epsilon_{0}(cB_{rms})^{2} = \frac{1}{2}\epsilon_{0}c^{2}B^{2}_{rms}$.
Substituting $c^{2}$: $u_{E} = \frac{1}{2}\epsilon_{0} \left(\frac{1}{\mu_{0}\epsilon_{0}}\right) B^{2}_{rms} = \frac{B^{2}_{rms}}{2\mu_{0}}$.
Since $\frac{B^{2}_{rms}}{2\mu_{0}}$ is the average magnetic energy density ($u_{B}$), we have shown $u_{E} = u_{B}$. Solution (Alternative Numerical):
Using the relation $\lambda = \frac{c}{\nu}$ where $c = 3\times10^{8} \text{ m/s}$.
For $\nu_{1} = 7.5 \text{ MHz} = 7.5 \times 10^{6} \text{ Hz}$: $\lambda_{1} = \frac{3\times10^{8}}{7.5\times10^{6}} = \frac{300}{7.5} = 40 \text{ m}$.
For $\nu_{2} = 12 \text{ MHz} = 12 \times 10^{6} \text{ Hz}$: $\lambda_{2} = \frac{3\times10^{8}}{12\times10^{6}} = \frac{300}{12} = 25 \text{ m}$.
The corresponding wavelength band is 25 m to 40 m.

Solution:
Einstein's Photoelectric Equation:
According to Planck's quantum theory, light consists of energy packets called photons, each with energy $E = h\nu$.
When a photon hits a metal surface, its energy is used in two ways:
1. A part of the energy is used to free the electron from the surface barrier. This minimum energy is the work function ($\Phi_{0} = h\nu_{0}$).
2. The remaining energy provides kinetic energy ($K_{max}$) to the emitted photoelectron.
By conservation of energy: $E = \Phi_{0} + K_{max}$
$h\nu = h\nu_{0} + \frac{1}{2}mv^{2}_{max} \implies K_{max} = h(\nu - \nu_{0})$.

Solution:
NOR Gate:
Boolean Expression: $Y = \overline{A + B}$
Logic Symbol: An OR gate symbol with a small bubble (inversion) at the output.
Truth Table:

A	B	Y
0	0	1
0	1	0
1	0	0
1	1	0

Solution (Alternative):
p-n junction diode: A semiconductor device formed by joining a p-type semiconductor and an n-type semiconductor, which allows current to flow easily in one direction but severely restricts it in the opposite direction.
Complete Characteristic:
It consists of a graph plotting Forward Current (mA) vs Forward Voltage (+V) in the first quadrant, showing an exponential rise after the "knee voltage". In the third quadrant, it plots Reverse Current ($\mu$A) vs Reverse Voltage (-V), showing a tiny constant saturation current until the "breakdown voltage" is reached, where current sharply drops.

Solution:
Total Internal Reflection (TIR): The phenomenon where a light ray traveling from a denser medium to a rarer medium is completely reflected back into the denser medium if it strikes the interface at an angle greater than the critical angle.
Conditions for TIR: 1. Light must travel from an optically denser medium to an optically rarer medium. 2. The angle of incidence in the denser medium must be strictly greater than the critical angle ($i > i_{c}$) for that pair of media. Solution (Alternative):
Blue Colour of Sky: As sunlight passes through the atmosphere, it undergoes Rayleigh scattering by air molecules. According to Rayleigh's law, the intensity of scattered light is inversely proportional to the fourth power of wavelength ($I \propto \frac{1}{\lambda^{4}}$). Since blue light has a much shorter wavelength than red light, it gets scattered much more strongly in all directions. This abundant scattered blue light is what our eyes see when looking at the sky.

Solution:
Drift Velocity ($v_{d}$): The average velocity with which free electrons in a conductor get drifted towards the positive end of the conductor under the influence of an applied external electric field.
Derivation:
Let $E$ be the applied electric field. The force on each electron is $\vec{F} = -e\vec{E}$.
Acceleration of each electron is $\vec{a} = \frac{\vec{F}}{m} = -\frac{e\vec{E}}{m}$.
The drift velocity is the average of the final velocities of all electrons after their respective collisions: $\vec{v}_{d} = \frac{\vec{v}_{1} + \vec{v}_{2} + ... + \vec{v}_{n}}{n}$.
Using $\vec{v} = \vec{u} + \vec{a}t$, we get: $\vec{v}_{d} = \frac{(\vec{u}_{1} + \vec{a}\tau_{1}) + ... + (\vec{u}_{n} + \vec{a}\tau_{n})}{n}$.
$\vec{v}_{d} = \left(\frac{\vec{u}_{1} + ... + \vec{u}_{n}}{n}\right) + \vec{a}\left(\frac{\tau_{1} + ... + \tau_{n}}{n}\right)$.
Since average thermal velocity is zero, $\vec{v}_{d} = 0 + \vec{a}\tau$ (where $\tau$ is average relaxation time).
Substituting $\vec{a}$, we get $\vec{v}_{d} = -\frac{e\vec{E}}{m}\tau$. (Magnitude $v_{d} = \frac{eE\tau}{m}$).

Solution:
(a) Internal Resistance: The resistance offered by the electrolyte and electrodes of a cell to the flow of electric current through it.
(b) Net Resistance calculation:
Note: The attached image "2024b_q20.JPG" is a standard Wheatstone bridge. Assuming standard labeling where the central galvanometer arm is balanced ($P/Q = R/S$), the central resistance is ignored. The top two are in series, the bottom two are in series, and these two combinations are in parallel. Since specific numerical values from the image aren't visible in this text format, you must apply the formula $R_{eq} = \frac{(R_1+R_2)(R_3+R_4)}{(R_1+R_2)+(R_3+R_4)}$ to find the final resistance between A and B.

Solution:
Magnetic Elements of Earth: The three quantities required to completely specify the Earth's magnetic field at a given location.
1. Magnetic Declination ($\theta$): The angle between the magnetic meridian and the geographic meridian at a place.
2. Angle of Dip or Inclination ($\delta$): The angle made by the Earth's total magnetic field with the horizontal direction in the magnetic meridian.
3. Horizontal Component ($B_{H}$): The component of the Earth's total magnetic field ($B$) along the horizontal direction. $B_{H} = B\cos\delta$. Solution (Alternative):
Voltmeter: An instrument used to measure the potential difference across two points in an electrical circuit. It must have high resistance and is always connected in parallel.
Conversion: A galvanometer is converted into a voltmeter by connecting a very high resistance ($R$) in series with it.
Let $G$ be galvanometer resistance, $I_{g}$ be full-scale deflection current, and $V$ be the desired voltage range.
By Ohm's law, total voltage $V = I_{g}(G + R)$.
Rearranging to find required series resistance: $R = \frac{V}{I_{g}} - G$.

Solution:
Solenoid: A long, tightly wound helical coil of insulated wire. When current flows through it, it produces a strong, uniform magnetic field inside.
Expression via Ampere's Law:
Consider a long solenoid with $n$ turns per unit length carrying current $I$. To find the magnetic field $B$ well inside, consider a rectangular Amperian loop $abcd$ where side $ab$ of length $l$ lies along the axis, and $cd$ lies far outside.
By Ampere's Circuital Law: $\oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{enclosed}$.
The line integral $\oint \vec{B} \cdot d\vec{l} = \int_{a}^{b} \vec{B}\cdot d\vec{l} + \int_{b}^{c} \vec{B}\cdot d\vec{l} + \int_{c}^{d} \vec{B}\cdot d\vec{l} + \int_{d}^{a} \vec{B}\cdot d\vec{l}$.
For $bc$ and $da$, $\vec{B} \perp d\vec{l}$, so integral is 0. For $cd$, $B \approx 0$ outside, so integral is 0.
For $ab$, $\vec{B} \parallel d\vec{l}$, so $\int_{a}^{b} B dl \cos(0) = Bl$.
Thus, $\oint \vec{B} \cdot d\vec{l} = Bl$.
Number of turns in length $l$ is $nl$. Current enclosed $I_{enclosed} = (nl)I$.
Substituting back: $Bl = \mu_{0} (nlI) \implies B = \mu_{0} n I$.

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 right $\Delta ABC$ and $\Delta ADC$:
$AC$ is the common hypotenuse.
$\angle B = \angle D = 90^\circ$.
$BC = AD = vt$.
By RHS congruence rule, $\Delta ABC \cong \Delta ADC$. Therefore, corresponding angles are equal: $\angle BAC = \angle DCA \implies \angle i = \angle r$. This proves the law of reflection.

Solution:
Binding Energy Curve: A graph plotting Binding Energy per Nucleon ($\Delta E_{bn}$) on the y-axis against Mass Number ($A$) on the x-axis.
Explanation: 1. The curve is relatively low for light nuclei ($A < 30$), indicating lesser stability. They undergo fusion to become stable. 2. It rises and forms a broad, flat maximum for intermediate nuclei ($30 < A < 170$) at about $8.0$ to $8.8$ MeV, showing they are highly stable (Peak at Fe-56). 3. It drops slowly for heavy nuclei ($A > 170$), indicating lower stability, explaining why they undergo nuclear fission. Solution (Alternative Numerical):
Nuclear Reaction: $_{3}Li^{6} + _{0}n^{1} \rightarrow _{2}He^{4} + _{1}H^{3}$.
Energy Released (Q-value):
Mass of reactants $= 6.015126 + 1.0086654 = 7.0237914 \text{ amu}$.
Mass of products $= 4.0026044 + 3.016049 = 7.0186534 \text{ amu}$.
Mass defect $\Delta m = \text{Mass of reactants} - \text{Mass of products}$
$\Delta m = 7.0237914 - 7.0186534 = 0.005138 \text{ amu}$.
Energy Released $E = \Delta m \times 931.5 \text{ MeV} = 0.005138 \times 931.5 \approx 4.786 \text{ MeV}$.

Solution:
Transistor as an Oscillator: An oscillator converts DC energy into AC energy without any external input signal. It uses a tuned tank circuit (LC circuit) and an amplifier with positive feedback.
Working: When switch is closed, a surge of collector current charges the capacitor in the LC tank circuit. The LC circuit begins to produce damped electrical oscillations of frequency $f = \frac{1}{2\pi\sqrt{LC}}$. The inductive coupling between coils feeds a fraction of this oscillating output voltage back into the base-emitter (input) circuit. Because of the transistor's properties and correct coil wiring, this feedback is *positive* (in-phase). The transistor amplifies this feedback, supplying enough energy back to the tank circuit to overcome resistive losses, sustaining continuous, undamped oscillations. Solution (Alternative):
Distinction between n-type and p-type:

Feature	n-type Semiconductor	p-type Semiconductor
Dopant	Pentavalent (e.g., Phosphorus, Arsenic)	Trivalent (e.g., Boron, Aluminum)
Majority Carriers	Electrons ($n_e \gg n_h$)	Holes ($n_h \gg n_e$)
Energy Levels	Donor level lies just below the conduction band.	Acceptor level lies just above the valence band.

Solution:
Proof for LOS distance $d$:
Consider a transmission tower of height $h$ on the surface of the Earth of radius $R$. The signal travels in a straight line, touching the Earth's surface at a distance $d$.
In the right-angled triangle formed by the Earth's center, the base of the tower, and the point where the signal touches the horizon:
Hypotenuse $= R + h$, Base $= R$, Perpendicular $= d$.
By Pythagoras Theorem: $(R+h)^{2} = R^{2} + d^{2}$
$R^{2} + h^{2} + 2Rh = R^{2} + d^{2}$
$d^{2} = 2Rh + h^{2}$
Since the height of the tower $h$ is extremely small compared to the radius of the Earth $R$ ($h \ll R$), the term $h^{2}$ is negligibly small and can be dropped.
Therefore, $d^{2} \approx 2Rh \implies d = \sqrt{2hR}$.

Solution:
Parallel Plate Capacitor: It consists of two large plane parallel conducting plates separated by a small distance, used to store large amounts of electric charge and energy.
Derivation with Dielectric Slab:
Let plates have area $A$ and separation $d$. Charge on plates is $+Q$ and $-Q$.
Without dielectric, initial field is $E_{0} = \frac{\sigma}{\epsilon_{0}} = \frac{Q}{A\epsilon_{0}}$.
A dielectric slab of thickness $t$ ($t < d$) and constant $K$ is introduced. Inside the slab, the field is reduced due to polarization: $E = \frac{E_{0}}{K}$.
The potential difference $V$ across the plates is the work done in moving a unit charge. It has two parts: air region ($d-t$) and dielectric region ($t$).
$V = E_{0}(d-t) + Et = E_{0}(d-t) + \frac{E_{0}}{K}t = E_{0}\left[(d-t) + \frac{t}{K}\right]$.
Substituting $E_{0}$: $V = \frac{Q}{A\epsilon_{0}}\left[d - t(1 - \frac{1}{K})\right]$.
Capacitance $C = \frac{Q}{V} = \frac{A\epsilon_{0}}{d - t\left(1 - \frac{1}{K}\right)}$. Solution (Alternative):
Electric Dipole: A pair of equal and opposite point charges ($+q, -q$) separated by a small vector distance ($2a$).
Torque on Dipole:
Let a dipole be placed at an angle $\theta$ in a uniform electric field $\vec{E}$.
Force on $+q$ is $F_{+} = qE$ (along $\vec{E}$). Force on $-q$ is $F_{-} = -qE$ (opposite to $\vec{E}$).
These equal, opposite, non-collinear forces form a couple that tries to align the dipole. Torque $\tau = \text{Force} \times \text{perpendicular distance}$.
From geometry, perpendicular distance $= 2a \sin\theta$.
$\tau = (qE) \times (2a \sin\theta) = (q \times 2a) E \sin\theta$.
Since dipole moment $p = q(2a)$, $\tau = pE \sin\theta$. In vector form, $\vec{\tau} = \vec{p} \times \vec{E}$.
Net Force: Since $F_{+}$ and $F_{-}$ are perfectly equal in magnitude and opposite in direction in a uniform field, the net translatory force is strictly zero ($F_{net} = qE - qE = 0$).

Solution:
Power in a.c. LCR Circuit:
Let the alternating voltage be $V = V_{0}\sin\omega t$.
In an LCR circuit, the current leads or lags the voltage by a phase angle $\phi$. Let $I = I_{0}\sin(\omega t - \phi)$.
Instantaneous power $P_{inst} = VI = V_{0}I_{0}\sin\omega t \sin(\omega t - \phi)$.
Using $2\sin A\sin B = \cos(A-B) - \cos(A+B)$:
$P_{inst} = \frac{V_{0}I_{0}}{2} [\cos\phi - \cos(2\omega t - \phi)]$.
The average power over a complete cycle involves averaging this expression. The average value of $\cos(2\omega t - \phi)$ over a full cycle is zero.
Average Power $P_{av} = \frac{V_{0}I_{0}}{2}\cos\phi = \left(\frac{V_{0}}{\sqrt{2}}\right)\left(\frac{I_{0}}{\sqrt{2}}\right)\cos\phi = V_{rms} I_{rms} \cos\phi$.

Special Cases:
(i) Only R (Pure Resistor): Voltage and current are in phase, so $\phi = 0^\circ$. $\cos(0^\circ) = 1$.
Power $P = V_{rms} I_{rms} (1) = V_{rms} I_{rms}$. (Maximum power dissipation).
(ii) Only L (Pure Inductor): Current lags voltage by $90^\circ$, so $\phi = 90^\circ$. $\cos(90^\circ) = 0$.
Power $P = V_{rms} I_{rms} (0) = 0$. (Wattless current).

Solution:
Magnifying Power of Astronomical Telescope: It is the ratio of the angle subtended at the eye by the image to the angle subtended by the object at the unaided eye. $M = \frac{\beta}{\alpha}$.
Derivation (Final image at least distance of distinct vision, D):
The objective lens forms a real inverted image $A'B'$ of a distant object at its focus, so $OB' = f_{o}$. Angle $\alpha \approx \tan\alpha = \frac{A'B'}{f_{o}}$.
The eyepiece acts as a simple microscope. The image $A'B'$ acts as an object at distance $u_{e}$. Angle $\beta \approx \tan\beta = \frac{A'B'}{-u_{e}}$.
Magnifying power $M = \frac{\beta}{\alpha} = \frac{A'B' / (-u_{e})}{A'B' / f_{o}} = -\frac{f_{o}}{u_{e}}$.
For the eyepiece, applying the lens formula: $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$.
By sign convention, $v_{e} = -D$ and object distance is $-u_{e}$.
$\frac{1}{-D} - \frac{1}{-u_{e}} = \frac{1}{f_{e}} \implies \frac{1}{u_{e}} = \frac{1}{f_{e}} + \frac{1}{D} = \frac{1}{f_{e}}\left(1 + \frac{f_{e}}{D}\right)$.
Substitute into M: $M = -f_{o} \left[ \frac{1}{f_{e}}\left(1 + \frac{f_{e}}{D}\right) \right] = -\frac{f_{o}}{f_{e}} \left(1 + \frac{f_{e}}{D}\right)$. Solution (Alternative):
Refraction at Concave Spherical Surface (Rarer $\mu_1$ to Denser $\mu_2$):
Consider a point object O on the principal axis. A ray OA strikes the concave surface. Since it goes rarer to denser, it bends towards the normal CA, diverging away from the axis. It appears to come from virtual image I.
Let angles with principal axis be $\alpha$ (object), $\beta$ (image), $\gamma$ (normal).
In $\Delta OAC$, exterior angle $\gamma = \alpha + i \implies i = \gamma - \alpha$.
In $\Delta IAC$, exterior angle $\gamma = \beta + r \implies r = \gamma - \beta$.
By Snell's Law for small angles, $\mu_{1}i = \mu_{2}r$.
$\mu_{1}(\gamma - \alpha) = \mu_{2}(\gamma - \beta) \implies \mu_{2}\beta - \mu_{1}\alpha = (\mu_{2} - \mu_{1})\gamma$.
Using $\text{angle} = \text{arc/radius}$ (small aperture): $\alpha = \frac{AP}{PO}$, $\beta = \frac{AP}{PI}$, $\gamma = \frac{AP}{PC}$.
$\mu_{2}(\frac{AP}{PI}) - \mu_{1}(\frac{AP}{PO}) = (\mu_{2} - \mu_{1})(\frac{AP}{PC})$.
By sign convention: $PO = -u$, $PI = -v$, $PC = -R$.
$\mu_{2}(\frac{1}{-v}) - \mu_{1}(\frac{1}{-u}) = (\mu_{2} - \mu_{1})(\frac{1}{-R})$.
Multiplying by -1 yields: $-\frac{\mu_{1}}{u} + \frac{\mu_{2}}{v} = \frac{\mu_{2} - \mu_{1}}{R}$.