HINTS & SOLUTIONS

Answer: (d) $NC^{-1}$
Explanation: Electric field intensity ($E$) is defined as the force experienced per unit positive test charge. $E = F/q$. The SI unit of force is Newton (N) and charge is Coulomb (C), hence $N/C$ or $NC^{-1}$.

Answer: (b) $2.8 \times 10^{-7} kWh$
Solution: We know $1 \text{ kWh} = 3.6 \times 10^6 \text{ Joules}$.
Therefore, $1 \text{ Joule} = \frac{1}{3.6 \times 10^6} \text{ kWh} \approx 2.77 \times 10^{-7} \text{ kWh} \approx 2.8 \times 10^{-7} \text{ kWh}$.

Answer: (b) helical
Explanation: When a charged particle enters a magnetic field at an angle other than $0^\circ, 90^\circ,$ or $180^\circ$, its velocity resolves into two components: the perpendicular component causes circular motion, and the parallel component causes linear translation. The combined path is a helix.

Answer: (b) $5\sqrt{3/2} A$
Solution: Peak current $I_0 = I_{rms} \sqrt{2} = 5\sqrt{2} \text{ A}$.
Angular frequency $\omega = 2\pi f = 2\pi(50) = 100\pi \text{ rad/s}$.
Instantaneous current $I = I_0 \sin(\omega t) = 5\sqrt{2} \sin\left(100\pi \times \frac{1}{300}\right)$
$I = 5\sqrt{2} \sin\left(\frac{\pi}{3}\right) = 5\sqrt{2} \left(\frac{\sqrt{3}}{2}\right) = 5\frac{\sqrt{6}}{2} = 5\sqrt{\frac{6}{4}} = 5\sqrt{\frac{3}{2}} \text{ A}$.

Answer: (c) frequency
Explanation: Frequency is a fundamental characteristic of the source producing the wave. When a wave travels from one medium to another, its velocity and wavelength change, but its frequency remains constant.

Answer: (d) red
Explanation: The velocity of light in a medium is given by $v = c/\mu$. Since the refractive index ($\mu$) of glass is minimum for red light and maximum for violet light, the velocity of red light is the highest inside the glass. Hence, it traverses the slab fastest and emerges first.

Answer: (d) zero
Explanation: A photon is a quantum of light that always travels at the speed of light ($c$). According to the theory of relativity, a particle traveling at the speed of light must have a rest mass of zero.

Answer: (c) $\gamma$-photon
Explanation: The nuclear reaction is $_3Li^7 + _1H^1 \rightarrow _4Be^8 + \text{Energy}$. Both atomic number ($3+1=4$) and mass number ($7+1=8$) are already balanced by the $_4Be^8$ nucleus. The excess energy is emitted as a $\gamma$-photon.

Answer: (b) non-ohmic resistance
Explanation: A PN junction diode does not obey Ohm's Law (V is not directly proportional to I). Its V-I characteristic curve is exponential, not a straight line passing through the origin, making it a non-ohmic device.

Answer: (b) (B) (A) (D) (E) (C)
Explanation: A basic communication system flow is: Information Source (B) $\rightarrow$ Transmitter (A) $\rightarrow$ Channel (D) $\rightarrow$ Receiver (E) $\rightarrow$ User of Information (C).

Answer: (c) 60°
Solution: For an equilateral prism, $A = 60^\circ$. Given $n = \sqrt{3}$.
Using prism formula: $n = \frac{\sin(\frac{A+\delta_m}{2})}{\sin(\frac{A}{2})}$
$\sqrt{3} = \frac{\sin(\frac{60^\circ+\delta_m}{2})}{\sin(30^\circ)}$
$\sqrt{3} \times 0.5 = \sin(\frac{60^\circ+\delta_m}{2}) \implies \frac{\sqrt{3}}{2} = \sin(\frac{60^\circ+\delta_m}{2})$
Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, we have $\frac{60^\circ+\delta_m}{2} = 60^\circ \implies 60^\circ+\delta_m = 120^\circ \implies \delta_m = 60^\circ$.

Answer: (a) Light is scattered by droplets
Explanation: Fog consists of dense, tiny water droplets suspended in the air. When light travels through fog, it undergoes multiple scattering by these droplets in random directions, destroying image formation and reducing visibility.

Solution:
Polar Dielectrics: Dielectrics in which the center of mass of positive charges does not coincide with the center of mass of negative charges. They possess a permanent, built-in electric dipole moment. (Examples: $H_2O, HCl$).
Non-Polar Dielectrics: Dielectrics in which the center of mass of positive and negative charges coincide. They have zero permanent dipole moment. A temporary dipole moment is induced only when an external electric field is applied. (Examples: $O_2, H_2$).

Solution:
Capacitive reactance is given by $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.
This indicates that $X_C$ is inversely proportional to the frequency ($f$).
Graph Description: The graph plotted between $X_C$ (y-axis) and $f$ (x-axis) is a rectangular hyperbola in the first quadrant. As frequency increases, the reactance rapidly drops towards zero.

Solution:
EM Spectrum in Ascending Order of Wavelength:
Gamma rays $<$ X-rays $<$ Ultraviolet (UV) $<$ Visible light $<$ Infrared (IR) $<$ Microwaves $<$ Radio waves. Solution (Alternative):
Properties of E.M. Waves:

• They do not require any material medium for propagation.
• They are transverse in nature (electric and magnetic fields oscillate perpendicular to propagation direction).
• They travel in a vacuum with the speed of light ($c = 3 \times 10^8 \text{ m/s}$).
• They carry energy and momentum, exerting radiation pressure on surfaces they strike.

Solution:
Photon: A photon is a fundamental particle of light. It is an elementary quantum (or packet) of electromagnetic energy.
Properties:

• The rest mass of a photon is zero.
• The energy of a photon is proportional to its frequency ($E = h\nu$).
• They are electrically neutral and are not deflected by electric or magnetic fields.

Solution:
Half-Wave Rectifier Working:
It consists of an AC source, a step-down transformer, a single p-n junction diode, and a load resistor ($R_L$).
Positive Half Cycle: The p-side of the diode becomes positive relative to the n-side. The diode is forward-biased, offers low resistance, and conducts current through $R_L$.
Negative Half Cycle: The p-side becomes negative relative to the n-side. The diode is reverse-biased, offers extremely high resistance, and blocks the current. Thus, current flows through $R_L$ in only one direction, yielding a pulsating DC output. Solution (Alternative):
AND Gate:
Boolean Expression: $Y = A \cdot B$
Logic Symbol: Represents a D-shaped symbol with two inputs (A, B) on the flat side and one output (Y) on the curved side.
Truth Table:

A	B	Y = A.B
0	0	0
0	1	0
1	0	0
1	1	1

Solution:
Mirror Formula Derivation (Concave Mirror, Real Image):
Consider an object AB placed beyond the center of curvature C. A real, inverted image A'B' forms between C and F.
By similar triangles $\Delta A'B'C$ and $\Delta ABC$: $\frac{A'B'}{AB} = \frac{CA'}{CA}$
By similar triangles $\Delta A'B'P$ and $\Delta ABP$: $\frac{A'B'}{AB} = \frac{PA'}{PA}$
Equating the ratios: $\frac{CA'}{CA} = \frac{PA'}{PA}$
Writing distances from pole P: $CA' = CP - PA'$ and $CA = PA - CP$.
So, $\frac{CP - PA'}{PA - CP} = \frac{PA'}{PA}$
Applying cartesian sign convention: $PA = -u$, $PA' = -v$, $CP = -R = -2f$.
$\frac{-R - (-v)}{-u - (-R)} = \frac{-v}{-u} \implies \frac{-R + v}{-u + R} = \frac{v}{u}$
$-uR + uv = -uv + vR \implies 2uv = vR + uR$.
Dividing both sides by $uvR$: $\frac{2}{R} = \frac{1}{u} + \frac{1}{v}$. Since $R = 2f$, we get $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Solution (Alternative):
Polarization: Polarization is a phenomenon restricted strictly to transverse waves. In transverse light waves, vibrations happen perpendicular to the direction of propagation in multiple planes, which can be restricted to a single plane (polarized). Sound waves in air are longitudinal waves where particle vibrations occur parallel to the direction of wave propagation, leaving no perpendicular planes to restrict. Hence, sound cannot be polarized.

Solution:
Metre Bridge: It is a practical application of the Wheatstone bridge principle used to measure unknown resistance accurately.
Working: It consists of a 1-meter long constantan/manganin wire stretched along a scale. An unknown resistance $S$ is connected in one gap of the metallic strips, and a known resistance box $R$ in the other gap. A galvanometer is connected between the central terminal and a sliding jockey.
The jockey is slid over the wire until the galvanometer shows zero deflection (null point) at length $l$ from the end near $R$.
At the balance point, the bridge forms four resistances: $P$ (resistance of length $l$), $Q$ (resistance of length $100-l$), $R$, and $S$.
By Wheatstone bridge principle: $\frac{P}{Q} = \frac{R}{S}$
Let $r$ be resistance per unit length. $P = l \times r$ and $Q = (100 - l) \times r$.
$\frac{l \cdot r}{(100-l) \cdot r} = \frac{R}{S} \implies \frac{l}{100-l} = \frac{R}{S}$
Unknown resistance $S = R \left(\frac{100 - l}{l}\right)$.

Solution:
(a) Kirchhoff's Laws:
1. Junction Rule (KCL): The algebraic sum of currents meeting at any junction in an electrical circuit is zero ($\sum I = 0$). It is based on the conservation of charge.
2. Loop Rule (KVL): The algebraic sum of changes in potential around any closed loop involving resistors and cells in the circuit is zero ($\sum \Delta V = 0$). It is based on the conservation of energy.

(b) Net Resistance calculation:
Note: Since the image "2024a_q20.JPG" providing the specific circuit diagram is unavailable in this text format, I cannot compute the exact numerical resistance. Please provide the resistor values and their structural arrangement (e.g., standard Wheatstone bridge, series-parallel combo) so I can solve it accurately for you.

Solution:
(a) Radial Magnetic Field: In a moving coil galvanometer, the magnetic poles are made concave and a soft iron core is placed inside the coil to produce a radial magnetic field. Its importance is that the plane of the coil is always parallel to the magnetic field regardless of its orientation. This ensures the deflecting torque is always maximum and constant ($\tau = NIAB \sin 90^\circ = NIAB$), leading to a linear scale where deflection is directly proportional to current ($\theta \propto I$).
(b) Force on Alpha Particle: The magnetic Lorentz force on a moving charge is $\vec{F} = q(\vec{v} \times \vec{B})$. Since the velocity $\vec{v}$ is parallel to the magnetic field $\vec{B}$, the angle $\theta$ between them is $0^\circ$.
$F = qvB \sin(0^\circ) = 0$. The force acting on the particle is zero. Solution (Alternative):
Torque on Bar Magnet:
Consider a bar magnet of length $2l$ and pole strength $m$ placed at an angle $\theta$ in a uniform magnetic field $B$.
Force on North pole $= +mB$ (along the field).
Force on South pole $= -mB$ (opposite to the field).
Since the forces are equal, opposite, and act along different lines of action, they form a couple. The torque ($\tau$) of this couple is:
$\tau = \text{Force} \times \text{perpendicular distance between forces}$.
From geometry, perpendicular distance $= 2l \sin\theta$.
$\tau = (mB) \times (2l \sin\theta) = (m \times 2l) B \sin\theta$.
Since magnetic dipole moment $M = m \times 2l$, we get $\tau = MB \sin\theta$. In vector form, $\vec{\tau} = \vec{M} \times \vec{B}$.

Solution:
Magnetic Field on Axis of Circular Coil:
Consider a circular coil of radius $a$ carrying current $I$. Let P be a point on its axis at distance $x$ from the center.
Take a small current element $dl$ at the top of the coil. The distance from $dl$ to P is $r = \sqrt{a^2 + x^2}$.
By Biot-Savart Law, the magnetic field at P due to $dl$ is $dB = \frac{\mu_0 I dl \sin 90^\circ}{4\pi r^2}$.
This field $dB$ resolves into two components: $dB \cos\phi$ perpendicular to the axis, and $dB \sin\phi$ along the axis.
Due to diametrically opposite elements, perpendicular components ($dB \cos\phi$) cancel out. Axial components add up.
Total magnetic field $B = \int dB \sin\phi$.
From geometry, $\sin\phi = \frac{a}{r} = \frac{a}{\sqrt{a^2 + x^2}}$.
$B = \int \frac{\mu_0 I dl}{4\pi (a^2 + x^2)} \frac{a}{\sqrt{a^2 + x^2}} = \frac{\mu_0 I a}{4\pi (a^2 + x^2)^{3/2}} \int dl$.
Since $\int dl = 2\pi a$ (circumference),
$B = \frac{\mu_0 I a}{4\pi (a^2 + x^2)^{3/2}} (2\pi a) = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$.

Solution:
Dispersion of Light: The phenomenon of splitting of white light into its constituent component colors (spectrum) when it passes through a transparent medium like a glass prism.
Prism Formula Derivation:
For a prism, the angle of deviation $\delta = i + e - A$.
At the position of minimum deviation ($\delta = \delta_m$), the ray passes symmetrically through the prism. Thus: 1. Angle of incidence equals angle of emergence ($i = e$). 2. Angle of refraction at both faces is equal ($r_1 = r_2 = r$).
Since $A = r_1 + r_2$, we get $A = r + r = 2r \implies r = \frac{A}{2}$.
Substituting $i = e$ in deviation formula: $\delta_m = i + i - A = 2i - A \implies i = \frac{A + \delta_m}{2}$.
According to Snell's Law, the refractive index of the prism is $n = \frac{\sin i}{\sin r}$.
Substituting values of $i$ and $r$: $n = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$.

Solution:
Nuclear Size: Experiments show that the volume of a nucleus is directly proportional to its mass number $A$. If $R$ is the radius, $\frac{4}{3}\pi R^3 \propto A \implies R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \text{ fm}$.
Nuclear Density: It is the ratio of mass of nucleus to its volume. Let $m$ be the average mass of a nucleon. Total mass $\approx m A$.
Volume $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$.
Density $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{mA}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}$.
Since $m$ and $R_0$ are constants, the term $A$ cancels out. Thus, nuclear density is independent of the mass number and is same ($\approx 2.3 \times 10^{17} \text{ kg/m}^3$) for all nuclei. Solution (Alternative Numerical):
Given for $^{56}_{26}Fe$: Protons $Z = 26$, Mass number $A = 56$, Neutrons $N = 56 - 26 = 30$.
Mass of 26 protons = $26 \times 1.007825 = 26.203450 \text{ amu}$.
Mass of 30 neutrons = $30 \times 1.008665 = 30.259950 \text{ amu}$.
Total constituent mass = $26.203450 + 30.259950 = 56.463400 \text{ amu}$.
Actual nuclear mass = $55.934949 \text{ amu}$.
Mass defect $\Delta m = 56.463400 - 55.934949 = 0.528451 \text{ amu}$.
Total Binding Energy = $\Delta m \times 931.5 \text{ MeV} = 0.528451 \times 931.5 = 492.25 \text{ MeV}$.
Binding Energy per nucleon = $\frac{\text{Total BE}}{A} = \frac{492.25}{56} \approx 8.79 \text{ MeV/nucleon}$.

Solution:
NPN Transistor as CE Amplifier:
In Common Emitter (CE) configuration, the emitter is common to both input and output circuits. The emitter-base junction is forward-biased with a low voltage battery ($V_{BB}$), and the collector-base junction is heavily reverse-biased by $V_{CC}$.
Working: A weak alternating input signal is superimposed on the forward bias $V_{BB}$.
During the positive half cycle of input, the forward bias increases, causing base current ($I_B$) and collector current ($I_C$) to increase heavily (since $I_C = \beta I_B$). This increases the voltage drop across the load resistor $R_L$.
Output voltage $V_{CE} = V_{CC} - I_C R_L$. As $I_C R_L$ increases, $V_{CE}$ decreases (becomes less positive/more negative). Thus, a positive input results in a negative output, creating a $180^\circ$ phase shift.
A small variation in input voltage causes a large variation in output voltage, resulting in amplification. Solution (Alternative):
Energy Bands in Solids:
In isolated atoms, electrons occupy discrete energy levels. In a solid crystal, billions of atoms are brought closely together. The outer electron orbits overlap and interact due to Pauli's exclusion principle, causing individual energy levels to split into closely spaced sub-levels, forming continuous "bands".
1. Valence Band: The highest energy band completely or partially filled with valence electrons. Electrons here are bound to their respective atoms.
2. Conduction Band: The energy band immediately above the valence band. If electrons jump here, they are free to move and conduct electricity.
3. Forbidden Energy Gap ($E_g$): The energy gap between the valence and conduction bands where no electron states exist. It determines if the material is a conductor ($E_g=0$), semiconductor ($E_g \approx 1\text{eV}$), or insulator ($E_g > 3\text{eV}$).

Solution:
Ground Wave Propagation:
When radio waves propagate gliding along the surface of the earth, it is called ground wave propagation. It is utilized for low frequencies (up to 2 MHz) like AM radio. As waves glide, they induce currents in the ground, dissipating energy rapidly due to the earth's resistance. Hence, it is suitable only for short-distance communication.
Space Wave Propagation:
When radio waves travel in a straight line directly from the transmitting antenna to the receiving antenna through the troposphere, it is called space wave or Line-of-Sight (LOS) propagation. It is used for ultra-high frequencies (UHF) and VHF above 40 MHz (like Television and Radar). Due to the earth's curvature, the range is limited by the heights of the transmitting and receiving antennas.

Solution:
Electric Flux: The total number of electric field lines crossing a given area normally. $\phi_E = \oint \vec{E} \cdot d\vec{A}$. SI unit is $\text{N m}^2 \text{C}^{-1}$ or $\text{V m}$.
Electric Field due to Infinite Plane Sheet:
Consider an infinite plane sheet of uniform surface charge density $\sigma$. To find the electric field at distance $r$, imagine a cylindrical Gaussian surface of cross-sectional area $A$ passing through the sheet, extending length $r$ on both sides.
Electric field $\vec{E}$ points outward normally from the sheet. Thus, the flux through the curved surface is zero (since $\vec{E} \perp d\vec{A}$).
Flux only passes through the two flat circular ends. Total flux $\phi_E = EA + EA = 2EA$.
By Gauss's Law, $\phi_E = \frac{q_{enclosed}}{\epsilon_0}$. The charge enclosed inside the cylinder is $q = \sigma A$.
Therefore, $2EA = \frac{\sigma A}{\epsilon_0} \implies E = \frac{\sigma}{2\epsilon_0}$. (Independent of distance $r$). Solution (Alternative):
Electric Dipole Moment ($\vec{p}$): A vector quantity measuring the strength of an electric dipole, equal to the product of magnitude of either charge and the separation distance. $p = q(2a)$. Direction is from negative to positive charge.
Field on Equatorial Line:
Consider a dipole of charges $+q$ and $-q$ separated by $2a$. Point P is on the perpendicular bisector at distance $r$ from the center.
Distance of P from both charges is $\sqrt{r^2 + a^2}$.
Magnitude of field due to $+q$ is $E_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{(r^2 + a^2)}$.
Magnitude of field due to $-q$ is $E_- = \frac{1}{4\pi\epsilon_0} \frac{q}{(r^2 + a^2)}$. Thus, $E_+ = E_-$.
Resolving these vectors, vertical components ($E\sin\theta$) cancel out. Horizontal components ($E\cos\theta$) add up parallel to the dipole axis.
Net field $E = E_+\cos\theta + E_-\cos\theta = 2E_+\cos\theta$.
From geometry, $\cos\theta = \frac{a}{\sqrt{r^2 + a^2}}$.
$E = 2 \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2} \right) \left( \frac{a}{\sqrt{r^2 + a^2}} \right) = \frac{1}{4\pi\epsilon_0} \frac{q(2a)}{(r^2 + a^2)^{3/2}}$.
Since $p = q(2a)$, $E = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}}$. Direction is opposite to dipole moment.

Solution:
Impedance ($Z$): The total effective opposition offered by a series combination of an inductor (L), capacitor (C), and resistor (R) to the flow of alternating current.
Derivation:
Let AC current $I = I_0 \sin(\omega t)$ flow through the series LCR circuit.
Voltage across resistor $V_R = I R$ (in phase with I).
Voltage across inductor $V_L = I X_L$ (leads I by $90^\circ$).
Voltage across capacitor $V_C = I X_C$ (lags I by $90^\circ$).
Using a phasor diagram, $V_L$ and $V_C$ are $180^\circ$ out of phase. Assuming $V_L > V_C$, the net reactive voltage is $(V_L - V_C)$ leading $V_R$ by $90^\circ$.
By Pythagoras theorem, resulting applied voltage $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
$V = \sqrt{(IR)^2 + (I X_L - I X_C)^2} = I \sqrt{R^2 + (X_L - X_C)^2}$.
Impedance $Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
Condition for Resonance:
Resonance occurs when the circuit allows maximum current to flow, which happens when Impedance is minimum. This condition is achieved when $X_L = X_C$, so $Z = R$.
$\omega L = \frac{1}{\omega C} \implies \omega_0 = \frac{1}{\sqrt{LC}}$.

Solution:
Lens Maker's Formula for Convex Lens:
Assumptions: Thin lens, paraxial rays, point object on principal axis.
Sign Convention: Distances measured in direction of incident light are positive; opposite are negative.
Refraction at 1st spherical surface (radius $R_1$): Object at $O$, virtual image at $I_1$.
$\frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}$ ... (Eq 1)
Refraction at 2nd surface (radius $R_2$): $I_1$ acts as virtual object, final image at $I$.
$\frac{n_1}{v} - \frac{n_2}{v_1} = \frac{n_1 - n_2}{R_2} = \frac{-(n_2 - n_1)}{R_2}$ ... (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 $n_1$, taking $\mu = n_2/n_1$. If object is at infinity ($u = -\infty$), image forms at focus ($v = f$):
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Solution (Alternative):
Compound Microscope: An optical instrument used to obtain highly magnified images of tiny objects, consisting of two convex lenses: an objective and an eyepiece.
Ray Diagram & Working: The object is placed just beyond the focus of the objective, forming a real, inverted, magnified image $A'B'$. This image falls within the focal length of the eyepiece, which acts as a simple magnifier to form a highly magnified, virtual final image $A''B''$ at the least distance of distinct vision (D).
Magnifying Power ($M$): The ratio of the angle subtended at the eye by the image ($\beta$) to the angle subtended by the object ($\alpha$) when both are at distance D. $M \approx \tan\beta / \tan\alpha$.
$M = m_o \times m_e$.
Linear magnification of objective $m_o = \frac{v_o}{u_o}$.
Since the eyepiece acts as a simple magnifier forming an image at D, $m_e = \left(1 + \frac{D}{f_e}\right)$.
Total magnification $M = \left(\frac{v_o}{u_o}\right) \left(1 + \frac{D}{f_e}\right)$.
Approximation: If object is very close to $f_o$ ($u_o \approx f_o$) and image $A'B'$ is close to the eyepiece ($v_o \approx L$, tube length), then $M = \frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)$.