Step 1: Identify RMS Values.

For the voltage, the peak value is: \[ V_m = 100 \quad \Rightarrow \quad V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{100}{\sqrt{2}}. \]
For the current, given the peak current: \[ I_m = 100 \, mA = 0.1 \, A \quad \Rightarrow \quad I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{0.1}{\sqrt{2}}. \]

Step 2: Calculate Power Factor.

The phase difference between voltage and current is: \[ \phi = \frac{\pi}{3} \quad \Rightarrow \quad \cos \phi = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}. \]

Step 3: Compute Average Power Consumed.

The average power \( P \) consumed is given by: \[ P = V_{rms} \times I_{rms} \times \cos \phi = \frac{100}{\sqrt{2}} \times \frac{0.1}{\sqrt{2}} \times \frac{1}{2}. \]
Simplifying, \[ P = \frac{100 \times 0.1}{2 \times 2} = \frac{10}{4} = 2.5 \, W. \]

Step 4: Conclusion.

Hence, the average power consumed by the circuit is \( 2.5 \, W \). Quick Tip: In AC circuits, average power is given by \(P = V_{rms} \cdot I_{rms} \cdot \cos\phi\), where \(\phi\) is the phase difference.

Step 1: Recall the Definition.

Isotones are atoms of different elements that have the same number of neutrons but differ in the number of protons.

Step 2: Calculate Number of Neutrons.

- For \( ^{14}_{6}C \): Neutrons \(= 14 - 6 = 8\).

- For \( ^{16}_{8}O \): Neutrons \(= 16 - 8 = 8\).

Since both have 8 neutrons, they qualify as isotones.


Checking other pairs:

- \(^{14}_{6}C\) has 8 neutrons, \(^{14}_{7}N\) has 7 neutrons (not isotones).

- \(^{14}_{6}C\) has 8 neutrons, \(^{17}_{8}O\) has 9 neutrons (not isotones).

- \(^{14}_{6}C\) has 8 neutrons, \(^{13}_{7}N\) has 6 neutrons (not isotones).


Step 3: Conclusion.

Therefore, the correct isotone pair is \(^{14}_{6}C\) and \(^{16}_{8}O\). Quick Tip: Remember: isotopes = same protons, isotones = same neutrons, isobars = same mass number.

Step 1: Relation Between \( E_0 \) and \( B_0 \).

For an electromagnetic wave, the peak electric field \( E_0 \) and peak magnetic field \( B_0 \) are related by: \[ \frac{E_0}{B_0} = c \]
where \( c = 3 \times 10^8 \, m/s \) is the speed of light.

Step 2: Substitute Given Values.

Using the given magnetic field amplitude \( B_0 = 3 \times 10^{-10} \, T \), we get: \[ E_0 = c \times B_0 = (3 \times 10^8) \times (3 \times 10^{-10}) = 9 \, V/m. \]

Step 3: Conclusion.

Hence, the amplitude of the electric field associated with the wave is \( 9 \, V/m \). Quick Tip: In an electromagnetic wave, \(E_0 = c \, B_0\) always holds true.

Step 1: Formula for Drift Velocity.

The drift velocity \( v_d \) is given by: \[ v_d = \frac{I}{n e A} \]
where \( A = \pi r^2 \) is the cross-sectional area of the wire.

Step 2: Compare the Two Cases.

Initial case: \[ v = \frac{i{n e \pi r^2}. \]
New case (radius \( = 2r \), current \( = 2i \)): \[ v' = \frac{2i{n e \pi (2r)^2} = \frac{2i}{4 \, n e \pi r^2} = \frac{1}{2} \cdot \frac{i}{n e \pi r^2}. \]

Step 3: Conclusion.
\[ v' = \frac{v}{2}. \]
Thus, the correct answer is (iii) \(\dfrac{v}{2}\). Quick Tip: Drift velocity is inversely proportional to the cross-sectional area of the wire.

Step 1: Formula for Deviation at Minimum Deviation.

The deviation angle \(\delta\) for a thin prism is given by: \[ \delta = (n - 1) A \]
where \( A \) is the prism angle and \( n \) is the refractive index of the material.

Step 2: Calculate Deviations for Each Refractive Index.

For \( n = 1.6 \): \[ \delta_1 = (1.6 - 1) \times 15^\circ = 0.6 \times 15 = 9^\circ. \]
For \( n = 1.5 \): \[ \delta_2 = (1.5 - 1) \times 15^\circ = 0.5 \times 15 = 7.5^\circ. \]
For \( n = 1.4 \): \[ \delta_3 = (1.4 - 1) \times 15^\circ = 0.4 \times 15 = 6^\circ. \]

Step 3: Conclusion.

Since \[ \delta_1 > \delta_2 > \delta_3, \]
the correct option is (i). Quick Tip: For a prism of the same angle, deviation increases with the refractive index.

Step 1: Recall Concept of Apparent Dip.

If the actual dip is \(\theta\), and the dip needle shows an apparent dip \(\theta'\) in a plane inclined at an angle \(\alpha\) to the magnetic meridian, then the relationship is: \[ \tan \theta' = \frac{\tan \theta}{\cos \alpha}. \]

Step 2: Take the Ratio.
\[ \frac{\tan \theta'}{\tan \theta} = \frac{1}{\cos \alpha} = \sec \alpha. \]

Step 3: Conclusion.

Therefore, the correct answer is (i) \(\sec \alpha\). Quick Tip: Remember: \(\tan \theta' = \dfrac{\tan \theta}{\cos \alpha}\) for apparent dip. If \(\alpha = 0\), then \(\theta' = \theta\).

Step 1: Recall Magnification Principle.

A compound microscope is made up of two lenses – the objective lens with magnification \( m_1 \) and the eyepiece lens with magnification \( m_2 \).

Step 2: Combine Magnifications.

The overall magnifying power \( M \) of the microscope is given by the product of the magnifications of the two lenses: \[ M = m_1 \times m_2. \]

Step 3: Conclusion.

Therefore, the magnifying power is \( M = m_1 m_2 \). Quick Tip: Compound microscopes multiply magnifications of the objective and eyepiece lenses.

Step 1: Understanding the Shunt.

A galvanometer is designed to measure only small currents. To enable it to measure larger currents, a low resistance called a shunt is connected in parallel with the galvanometer.

Step 2: Function of the Shunt.

The shunt provides an alternate path for most of the current, allowing it to bypass the galvanometer and thereby protecting it from excessive current.

Step 3: Conclusion.

Therefore, a shunt is a low resistance connected in parallel to extend the current measuring range of a galvanometer. Quick Tip: Shunts protect galvanometers and allow them to measure higher currents.

Step 1: Recall Kirchhoff’s Current Law (KCL).

At any junction in an electrical circuit, the sum of currents entering the junction is equal to the sum of currents leaving it.

Step 2: Mathematical Expression.
\[ \sum I_{in} = \sum I_{out} \quad \Rightarrow \quad \sum I = 0. \]

Step 3: Conclusion.

This statement is known as Kirchhoff’s First Law. Quick Tip: Kirchhoff’s First Law is based on conservation of charge.

Step 1: Definition.

When the current in a coil changes, an emf is induced in the coil itself, called the self-induced emf.

Step 2: Formula.
\[ e = -L \frac{di}{dt}, \]
where \(L\) is the coefficient of self-induction.

Step 3: Conclusion.

Therefore, \[ L = \frac{e}{- \frac{di}{dt}}. \] Quick Tip: Self-induction is the electrical inertia of a coil.

Step 1: Recall de Broglie relation.

As per de Broglie hypothesis, a particle in motion has a wavelength given by: \[ \lambda = \frac{h}{p}. \]

Step 2: Express momentum.

Since momentum \( p = mv \), we have: \[ \lambda = \frac{h}{mv}. \]

Step 3: Conclusion.

Thus, the de Broglie wavelength is inversely proportional to the momentum of the particle. Quick Tip: Greater momentum \(\Rightarrow\) shorter de Broglie wavelength.

Step 1: Recall definition of dipole.

A dipole consists of charges \( +q \) and \( -q \) separated by a distance \( 2a \). The dipole moment is \[ p = q \times 2a. \]

Step 2: Potential on axial line.

At a point \( P \) at distance \( r \) from the center along the axial line, the potential is given by: \[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{r^2}. \]

Step 3: Conclusion.

Therefore, the potential due to a dipole on its axial line decreases as \( \frac{1}{r^2} \) with distance. Quick Tip: On axial line: \(V = \dfrac{p}{4\pi\epsilon_0 r^2}\), On equatorial line: \(V = 0\) (due to symmetry).

Step 1: Formula for magnetic dipole moment.
\[ m = N \cdot I \cdot A \]
where \(N =\) number of turns, \(I =\) current, \(A =\) area.

Step 2: Substitution.

Given: \[ A = 5 \, cm^2 = 5 \times 10^{-4} \, m^2, \quad N = 100, \quad I = 0.2 \, A. \] \[ m = 100 \times 0.2 \times 5 \times 10^{-4}. \] \[ m = 1.0 \times 10^{-3} \, Am^2. \]

Step 3: Formula for torque.
\[ \tau = m B \sin \theta \]
For maximum torque, \(\sin \theta = 1\): \[ \tau_{max} = m B. \]

Step 4: Substitution.
\[ \tau_{max} = (1.0 \times 10^{-3})(1.5) = 1.5 \times 10^{-3} \, Nm. \]

Step 5: Conclusion.

1. \(m = 1 \times 10^{-3} \, Am^2\)

2. \(\tau_{max} = 1.5 \times 10^{-3} \, Nm\) Quick Tip: Magnetic dipole moment depends on \(NIA\), while torque depends on alignment with the magnetic field.

Step 1: Write expression for volume of nucleus.
\[ V = \frac{4}{3} \pi R^3 \] \[ R = R_0 A^{1/3} \quad \Rightarrow \quad V = \frac{4}{3} \pi (R_0 A^{1/3})^3. \] \[ V = \frac{4}{3} \pi R_0^3 A. \]

Step 2: Write expression for mass of nucleus.

Mass of nucleus \(\approx A m_p\) (where \(m_p\) is mass of one nucleon, nearly proton/neutron mass).

Step 3: Density of nucleus.
\[ \rho = \frac{Mass}{Volume} = \frac{A m_p}{\frac{4}{3} \pi R_0^3 A}. \] \[ \rho = \frac{m_p}{\frac{4}{3} \pi R_0^3}. \]

Step 4: Conclusion.

Since \(A\) cancels out, \(\rho\) is independent of mass number \(A\). Quick Tip: Nuclear density is constant (\(\sim 2.3 \times 10^{17} \, kg/m^3\)) for all nuclei, showing tightly packed nucleons.

Step 1: Symbol of AND gate.

The AND gate has two or more inputs and one output. Its symbol is:
\[ (Insert AND gate diagram: two inputs A, B and one output Y) \]

Step 2: Logical expression.

The output of AND gate is: \[ Y = A \cdot B \]

Step 3: Construct truth table.

\[ \begin{array}{|c|c|c|} \hline A & B & Y = A \cdot B
\hline 0 & 0 & 0
0 & 1 & 0
1 & 0 & 0
1 & 1 & 1
\hline \end{array} \]

Step 4: Conclusion.

The AND gate output is HIGH (1) only if both inputs are HIGH. Quick Tip: Remember: AND means multiplication in Boolean algebra.

Step 1: Effect of stretching on resistance.

Resistance of a wire: \[ R = \rho \frac{L}{A}. \]
If the wire is stretched to 3 times its original length, then \(L' = 3L\) and volume remains constant. \[ A' = \frac{A}{3}. \]
So, \[ R' = \rho \frac{3L}{A/3} = 9 \cdot \rho \frac{L}{A} = 9R. \]

Step 2: New resistance of wire.

Given original resistance = \(10 \, \Omega\). \[ R' = 9 \times 10 = 90 \, \Omega. \]

Step 3: Cutting into 3 equal parts.

When cut into 3 equal parts: \[ R_{part} = \frac{90}{3} = 30 \, \Omega \, each. \]

Step 4: Analyze circuit.

- The top two resistors (each \(30 \, \Omega\)) are in series: \[ R_{top} = 30 + 30 = 60 \, \Omega. \]
- The bottom resistor is \(30 \, \Omega\).
- These two branches are in parallel.

Step 5: Equivalent resistance.
\[ \frac{1}{R_{AB}} = \frac{1}{R_{top}} + \frac{1}{R_{bottom}} = \frac{1}{60} + \frac{1}{30}. \] \[ \frac{1}{R_{AB}} = \frac{1 + 2}{60} = \frac{3}{60} = \frac{1}{20}. \] \[ R_{AB} = 20 \, \Omega. \]

⚠ Correction: Let’s carefully recalc —

Wait: \(R_{top} = 60 \, \Omega\), \(R_{bottom} = 30 \, \Omega\).

So, \[ R_{eq} = \frac{(60)(30)}{60 + 30} = \frac{1800}{90} = 20 \, \Omega. \]

Final answer: \(R_{AB} = 20 \, \Omega\).

Step 6: Conclusion.

The total resistance between A and B is \(20 \, \Omega\). Quick Tip: Stretching a wire increases its resistance proportional to the square of stretching factor.

Step 1: Bohr’s quantization condition.

According to Bohr’s model, angular momentum of electron in \(n^{th}\) orbit is: \[ m v r = n \hbar, \]
where \(m\) = electron mass, \(v\) = velocity, \(r\) = radius.

Step 2: Radius of \(n^{th}\) orbit.

From Bohr’s theory: \[ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z}. \]
So, \[ r_n \propto n^2. \]

Step 3: Velocity of electron.

Velocity is given by: \[ v_n = \frac{Z e^2}{2 \epsilon_0 h} \cdot \frac{1}{n}. \]
So, \[ v_n \propto \frac{1}{n}. \]

Step 4: Period of revolution.

Period is time taken for one complete revolution: \[ T = \frac{2 \pi r_n}{v_n}. \]
Substituting dependencies: \[ T \propto \frac{n^2}{1/n} = n^3. \]

Step 5: Conclusion.

Thus, the period of revolution of electrons in Bohr’s model is proportional to \(n^3\). Quick Tip: In Bohr’s atom: \(r \propto n^2\), \(v \propto 1/n\), hence \(T \propto n^3\).

Step 1: Condition for diffraction minima.

For single slit: \[ a \sin \theta = m \lambda_1, \quad m = 1, 2, 3, ... \]
For first minima: \[ a \sin \theta = \lambda_1. \]

Step 2: Condition for diffraction maxima.

For interference maxima of another wavelength \(\lambda_2\): \[ a \sin \theta = n \lambda_2, \quad n = 1, 2, 3, ... \]
For first maxima: \(n=2\): \[ a \sin \theta = 2 \lambda_2. \]

Step 3: Equating conditions.

Since first minima of \(\lambda_1\) coincides with first secondary maximum of \(\lambda_2\): \[ \lambda_1 = 2 \lambda_2. \] \[ \lambda_2 = \frac{\lambda_1}{2}. \]

Step 4: Substitution.
\[ \lambda_2 = \frac{660}{2} = 330 \, nm. \]

Step 5: Conclusion.

Hence, the required wavelength is \(\lambda_2 = 330 \, nm\). Quick Tip: In single slit diffraction: minima \(\propto m \lambda\), while maxima can overlap with minima of another wavelength.

Step 1: Photoelectric equation.

The maximum kinetic energy of emitted electron is: \[ K_{\max} = e V_s = h \nu - \phi \]
where \(\phi\) = work function, \(h \nu = \dfrac{hc}{\lambda}\).

Step 2: For \(\lambda = 100 \, nm\).

Energy of photon: \[ E_1 = \frac{1240}{100} \, eV = 12.4 \, eV. \]
Kinetic energy: \[ K_{\max,1} = E_1 - \phi = 12.4 - 4.7 = 7.7 \, eV. \]
This matches the given stopping potential (7.7 V).

Step 3: For \(\lambda = 200 \, nm\).

Energy of photon: \[ E_2 = \frac{1240}{200} = 6.2 \, eV. \]
Kinetic energy: \[ K_{\max,2} = E_2 - \phi = 6.2 - 4.7 = 1.5 \, eV. \]

Step 4: Stopping potential.
\[ V_s = \frac{K_{\max,2}}{e} = 1.5 \, V. \]

Step 5: Conclusion.

The stopping potential for \(\lambda = 200 \, nm\) is \(1.5 \, V\). Quick Tip: Stopping potential depends only on photon energy minus work function.

Step 1: Biot–Savart law.

Magnetic field due to a small element \(dl\) at point \(P\): \[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl \, \sin \theta}{r^2}. \]

Step 2: For circular loop.

At centre of coil: \(r = R\), \(\sin \theta = 1\), so \[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl}{R^2}. \]

Step 3: Integrate over loop.

Total length of coil = \(2 \pi R\): \[ B = \frac{\mu_0}{4\pi} \frac{I (2\pi R)}{R^2}. \] \[ B = \frac{\mu_0 I}{2R}. \]

Step 4: For N turns.
\[ B = \frac{\mu_0 N I}{2R}. \]

Step 5: Conclusion.

The magnetic field at the centre of circular coil is: \[ B = \frac{\mu_0 N I}{2R}. \] Quick Tip: For a coil of radius \(R\) with \(N\) turns, field at centre increases with \(N\) and decreases with \(R\).

Step 1: Relation of potential with charge.

For a sphere of radius \(R\), \[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R}. \]
If both spheres are at same potential: \[ \frac{Q_1}{R_1} = \frac{Q_2}{R_2}. \]

Step 2: Express \(Q\) in terms of surface density.
\[ Q = \sigma \cdot 4\pi R^2. \]

So, \[ \frac{\sigma_1 \cdot 4\pi R_1^2}{R_1} = \frac{\sigma_2 \cdot 4\pi R_2^2}{R_2}. \] \[ \sigma_1 R_1 = \sigma_2 R_2. \]

Step 3: Substitution.
\(R_1 = 3 \, cm\), \(R_2 = 2 \, cm\). \[ \sigma_1 \cdot 3 = \sigma_2 \cdot 2. \] \[ \sigma_1 : \sigma_2 = 2 : 3. \]

Step 4: Conclusion.

The ratio of surface charge densities is \(2 : 3\). Quick Tip: For equal potentials, \(\sigma \propto \dfrac{1}{R}\).

Step 1: Recall formula for capacitance.

For a parallel plate capacitor with dielectric of thickness \(d\) and constant \(K\): \[ C = \frac{\epsilon_0 K A}{d}. \]

Step 2: Observation of system.

Here, the capacitor contains two dielectric slabs of thicknesses \(d_1, d_2\) and constants \(K_1, K_2\), placed in series (stacked one after another between plates).

Step 3: Equivalent capacitance for series connection.

For series capacitors: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}. \]

Step 4: Individual capacitances.
\[ C_1 = \frac{\epsilon_0 K_1 A}{d_1}, \quad C_2 = \frac{\epsilon_0 K_2 A}{d_2}. \]

Step 5: Substitute in series formula.
\[ \frac{1}{C} = \frac{d_1}{\epsilon_0 K_1 A} + \frac{d_2}{\epsilon_0 K_2 A}. \]

Step 6: Simplify.
\[ \frac{1}{C} = \frac{1}{\epsilon_0 A}\left(\frac{d_1}{K_1} + \frac{d_2}{K_2}\right). \] \[ C = \frac{\epsilon_0 A}{\dfrac{d_1}{K_1} + \dfrac{d_2}{K_2}}. \]

Step 7: Conclusion.

Thus, the capacitance of the system is: \[ C = \frac{\epsilon_0 A}{\dfrac{d_1}{K_1} + \dfrac{d_2}{K_2}}. \] Quick Tip: For multiple dielectrics in series, treat them as series capacitors. The effective plate separation is replaced by \(\dfrac{d_1}{K_1} + \dfrac{d_2}{K_2}\).

Step 1: Formula for motional emf.

When a conductor of length \(l\) moves with velocity \(v\) perpendicular to magnetic field \(B\): \[ E = B l v. \]

Step 2: Substitution.
\[ E = 4 \times 1 \times 2 = 8 \, V. \]

So, the induced emf across PQ is \(8 \, V\).

Step 3: Relation between charge and potential.

For capacitor: \[ Q = C V. \]

Step 4: Substitution.
\[ Q = (10 \times 10^{-6})(8) = 80 \times 10^{-6} \, C. \] \[ Q = 80 \, \mu C. \]

Step 5: Direction of charge flow.

Using Fleming’s right-hand rule:
- Magnetic field (\(B\)) is into the page.
- Velocity (\(v\)) is towards right.
- Force on positive charge (\(q\)) is upwards (towards P).

Thus, end \(P\) becomes positive and end \(Q\) becomes negative.
Hence, plate \(A\) connected to \(P\) is positive.

Step 6: Conclusion.

(i) Induced emf \(= 8 \, V\)

(ii) Charge on capacitor \(= 80 \, \mu C\)

(iii) Plate \(A\) is positive. Quick Tip: In motional emf: \(E = B l v\). Always use Fleming’s right-hand rule to decide polarity.

Step 1: Principle of potentiometer.

The potential difference across any length of the potentiometer wire is directly proportional to its balancing length.
\[ V \propto l \]

Step 2: First observation (no external resistance).

- The cell is connected directly to the potentiometer.
- Let the balancing length be \(l_1\).
- Then the emf of the cell is: \[ E \propto l_1. \]

Step 3: Second observation (with external resistance R).

- Now the cell is connected across a resistance \(R\).
- The potential difference across the cell (terminal voltage) is: \[ V = \frac{E R}{R + r}. \]
- Let the new balancing length be \(l_2\).
- Then, \[ V \propto l_2. \]

Step 4: Ratio of lengths.
\[ \frac{E}{V} = \frac{l_1}{l_2}. \]

Substitute \(V = \dfrac{E R}{R + r}\): \[ \frac{E}{\dfrac{E R}{R + r}} = \frac{l_1}{l_2}. \] \[ \frac{R + r}{R} = \frac{l_1}{l_2}. \]

Step 5: Solve for \(r\).
\[ R + r = R \cdot \frac{l_1}{l_2}. \] \[ r = R \left( \frac{l_1}{l_2} - 1 \right). \] \[ r = R \cdot \frac{l_1 - l_2}{l_2}. \]

Step 6: Conclusion.

The internal resistance of the cell is: \[ r = R \cdot \frac{l_1 - l_2}{l_2}. \] Quick Tip: Using a potentiometer avoids error due to current draw, as it measures emf and potential difference by null deflection method.

(i) Viewed through the plane surface.

Step 1: Apparent depth formula.

For a point inside a medium of refractive index \(\mu\), when viewed through a plane surface: \[ Apparent depth = \frac{Real depth}{\mu}. \]

Step 2: Substitution.

Real depth \(= 8 \, cm - 4 \, cm = 4 \, cm\) from plane surface. \[ Apparent depth = \frac{4}{1.6} = 2.5 \, cm. \]
So the image appears at distance: \[ 8 - 2.5 = 5.5 \, cm \quad from center, \quad or 6.4 \, cm from plane surface. \]

Conclusion (i): Image is at \(6.4 \, cm\) from plane surface.

---

(ii) Viewed through the spherical surface.

Step 1: Refraction at spherical surface.

Formula: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}. \]
Here, \(\mu_1 = 1.6\), \(\mu_2 = 1\), \(u = 4 \, cm\), \(R = 8 \, cm\).

Step 2: Substitution.
\[ \frac{1}{v} - \frac{1.6}{4} = \frac{1 - 1.6}{8}. \] \[ \frac{1}{v} - 0.4 = \frac{-0.6}{8}. \] \[ \frac{1}{v} = 0.4 - 0.075 = 0.325. \] \[ v \approx 3.08 \, cm. \]

Step 3: Sign convention.

Since the result is positive inside the medium, the final image will appear as a virtual image on the same side, extended behind surface. Equivalent distance = \(-16 \, cm\).

Conclusion (ii): The image appears as a virtual image at \(-16 \, cm\) behind the spherical surface.

--- Quick Tip: For plane surfaces: Apparent depth = Real depth / μ. For spherical surfaces: Use \(\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}\).

Step 1: Principle.

A transformer works on the principle of mutual induction, i.e., when alternating current flows through the primary coil, it produces a changing magnetic flux, which induces an emf in the secondary coil.

Step 2: Step-up transformer working.

- In a step-up transformer, the number of turns in the secondary coil (\(N_s\)) is greater than that in the primary coil (\(N_p\)).
- The emf ratio is: \[ \frac{E_s}{E_p} = \frac{N_s}{N_p}. \]
- Hence, voltage across secondary increases, while current decreases correspondingly to conserve power.

Step 3: Circuit diagram.

(Insert simple diagram: AC source connected to primary coil; secondary coil connected to load, with \(N_s > N_p\)).

Step 4: Energy losses in transformer.

Two common reasons:
1. Eddy current loss: Currents induced in the iron core produce heating. Laminating the core reduces this.
2. Hysteresis loss: Repeated magnetization and demagnetization of the iron core consumes energy. Using soft iron with small hysteresis loop minimizes this.

Step 5: Conclusion.

Thus, a transformer is based on mutual induction. In a step-up transformer, voltage is increased by using more turns in the secondary. Energy losses occur due to eddy currents and hysteresis. Quick Tip: Step-up transformer: \(N_s > N_p\), increases voltage. Step-down transformer: \(N_s < N_p\), decreases voltage.

Step 1: Identify given values.
\[ R_1 = 10 \, \Omega, \quad R_2 = 20 \, \Omega, \quad R_3 = 10 \, \Omega. \]
Two batteries: \(30 \, V\) (with \(R_1\)), \(10 \, V\) (with \(R_3\)).

Step 2: Apply Kirchhoff’s current law (KCL).

At the common junction: \[ I_1 = I_2 + I_3. \]

Step 3: Apply Kirchhoff’s voltage law (KVL).

Loop 1 (with \(R_1\) and \(R_2\)): \[ 30 - 10I_1 - 20I_2 = 0. \] \[ 10I_1 + 20I_2 = 30. \quad (1) \]

Loop 2 (with \(R_2\) and \(R_3\)): \[ 10 - 10I_3 - 20I_2 = 0. \] \[ 10I_3 + 20I_2 = 10. \quad (2) \]

Step 4: Use current relation.
\(I_1 = I_2 + I_3\). Substituting into (1): \[ 10(I_2 + I_3) + 20I_2 = 30. \] \[ 10I_2 + 10I_3 + 20I_2 = 30. \] \[ 30I_2 + 10I_3 = 30. \quad (3) \]

Step 5: Solve equations (2) and (3).

From (2): \[ 10I_3 + 20I_2 = 10. \]
From (3): \[ 30I_2 + 10I_3 = 30. \]

Subtract equations: \[ (30I_2 + 10I_3) - (20I_2 + 10I_3) = 30 - 10. \] \[ 10I_2 = 20 \quad \Rightarrow \quad I_2 = 2 \, A. \]

Substitute in (2): \[ 10I_3 + 20(2) = 10. \] \[ 10I_3 + 40 = 10 \quad \Rightarrow \quad I_3 = -3 \, A. \]

Negative sign means actual direction is opposite to assumed.

So, \[ I_1 = I_2 + I_3 = 2 + (-3) = -1 \, A. \]

Thus, \[ I_1 = -1 \, A, \quad I_2 = 2 \, A, \quad I_3 = -3 \, A. \]

Step 6: Final Answer.

Hence, actual directions of \(I_1\) and \(I_3\) are opposite to assumed.
Magnitudes: \[ I_1 = 1 \, A, \quad I_2 = 2 \, A, \quad I_3 = 3 \, A. \]

--- Quick Tip: If current comes out negative, it means the actual direction is opposite to the assumed direction in KVL/KCL equations.

Step 1: Balanced bridge condition.
\[ \frac{R_1}{R_2} = \frac{l_1}{l_2}. \]

Step 2: First case (without extra resistor).
\(l_1 = 40 \, cm, \quad l_2 = 60 \, cm\). \[ \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3}. \quad (1) \]

Step 3: Second case (with 10 \(\Omega\) in series with \(R_1\)).

Balanced point at \(40 \, cm\) from B \(\Rightarrow\) \(l_1 = 60 \, cm, l_2 = 40 \, cm\). \[ \frac{R_1 + 10}{R_2} = \frac{60}{40} = \frac{3}{2}. \quad (2) \]

Step 4: Solve equations (1) and (2).

From (1): \[ R_1 = \frac{2}{3} R_2. \quad (3) \]

Substitute in (2): \[ \frac{\frac{2}{3}R_2 + 10}{R_2} = \frac{3}{2}. \] \[ \frac{2R_2}{3R_2} + \frac{10}{R_2} = \frac{3}{2}. \] \[ \frac{2}{3} + \frac{10}{R_2} = \frac{3}{2}. \] \[ \frac{10}{R_2} = \frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}. \] \[ R_2 = \frac{10 \times 6}{5} = 12 \, \Omega. \]

Correction check: Let's recalc carefully:
\[ \frac{\frac{2}{3}R_2 + 10}{R_2} = \frac{3}{2}. \] \[ \frac{2R_2}{3R_2} + \frac{10}{R_2} = \frac{3}{2}. \] \[ \frac{2}{3} + \frac{10}{R_2} = \frac{3}{2}. \] \[ \frac{10}{R_2} = \frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}. \] \[ R_2 = \frac{10 \times 6}{5} = 12 \, \Omega. \]

Then, from (3): \[ R_1 = \frac{2}{3} \times 12 = 8 \, \Omega. \]

Step 5: Final Answer.
\[ R_1 = 8 \, \Omega, \quad R_2 = 12 \, \Omega. \]

--- Quick Tip: In meter bridge problems, balanced length ratio equals resistance ratio: \(\dfrac{R_1}{R_2} = \dfrac{l_1}{l_2}\).

Step 1: Principle.

A transistor oscillator works on the principle of positive feedback. The output of an amplifier is fed back to its input in phase to maintain continuous oscillations.

Step 2: Essential components.

- A transistor amplifier.
- Feedback network (usually LC tank circuit).
- DC power supply.

Step 3: Working process.

1. When supply is given, noise voltage in the circuit provides initial signal.
2. The LC tank circuit produces oscillations of its natural frequency.
3. The transistor amplifies these oscillations.
4. A part of output is fed back in phase with the input, reinforcing it.
5. Continuous undamped oscillations are maintained.

Step 4: Barkhausen criterion.

For sustained oscillations: \[ Loop gain = A \beta = 1. \]

Step 5: Circuit diagram.

(Insert diagram of transistor oscillator: LC tank circuit at input, transistor amplifier, feedback loop).

Step 6: Conclusion.

Thus, a transistor oscillator converts DC power into continuous AC output of a desired frequency.


Quick Tip: Oscillator = Amplifier + Positive Feedback + LC/RC tank circuit.

Step 1: Principle.

A p-n junction diode allows current only during forward bias. In full-wave rectifier, two diodes conduct alternately for each half cycle of AC input.

Step 2: Circuit arrangement.

- A center-tapped transformer provides AC input.
- Two diodes \(D_1\) and \(D_2\) are connected across the secondary.
- Load resistance \(R_L\) is connected across the output.

Step 3: Working process.

1. During the positive half cycle of input AC:
- Upper half of secondary is positive.
- \(D_1\) conducts (forward biased), \(D_2\) is reverse biased.
- Current flows through \(R_L\) in one direction.

2. During the negative half cycle of input AC:
- Lower half of secondary is positive.
- \(D_2\) conducts (forward biased), \(D_1\) is reverse biased.
- Current again flows through \(R_L\) in the same direction.

Step 4: Output.

Both halves of input AC are converted into pulsating DC. The frequency of output is double the input frequency.

Step 5: Circuit diagram.

(Insert standard full-wave rectifier diagram with center-tapped transformer and two diodes).

Step 6: Conclusion.

Thus, the full-wave rectifier efficiently converts AC into DC, utilizing both halves of input.

Quick Tip: Half-wave rectifier uses only one half cycle, while full-wave rectifier uses both halves of AC.

Step 1: Recall condition for interference.

In Young’s double slit experiment, interference fringes are formed due to path difference between the two waves from slits.

Step 2: Condition for dark fringes.

Dark fringe occurs when: \[ \Delta = \left( n - \tfrac{1}{2} \right) \lambda, \quad n = 1, 2, 3 \dots \]

Step 3: Relation between path difference and fringe position.

Path difference at distance \(y\) from central maximum is: \[ \Delta = \frac{y d}{D}, \]
where \(d\) = distance between slits, \(D\) = distance between slits and screen.

Step 4: Substitute condition.
\[ \frac{y_n d}{D} = \left( n - \tfrac{1}{2} \right) \lambda. \]

Step 5: Solve for \(y_n\).
\[ y_n = \left( n - \tfrac{1}{2} \right) \frac{\lambda D}{d}. \]

Step 6: Conclusion.

Thus, the distance of the \(n^{th}\) dark fringe from the central fringe is: \[ y_n = \left( n - \tfrac{1}{2} \right) \frac{\lambda D}{d}. \] Quick Tip: In Young’s double slit experiment: - Bright fringe: \(y_n = n \frac{\lambda D}{d}\) - Dark fringe: \(y_n = (n - \tfrac{1}{2}) \frac{\lambda D}{d}\)

Step 1: Statement of Huygens’ principle.

Every point on a wavefront is a source of secondary wavelets, which spread out in all directions with speed of light in that medium. The new wavefront is the envelope of all secondary wavelets.

Step 2: Application to refraction.

Consider a plane wavefront incident on a refracting surface separating two media:
- Speed in medium 1: \(v_1\), refractive index \(n_1\).
- Speed in medium 2: \(v_2\), refractive index \(n_2\).

Step 3: Geometry of refraction.

From Huygens’ construction: \[ \frac{\sin i}{\sin r} = \frac{v_1}{v_2}. \]

Step 4: Use relation between speed and refractive index.

Since \(v \propto \tfrac{1}{n}\), \[ \frac{\sin i}{\sin r} = \frac{n_2}{n_1}. \]

Step 5: Conclusion.

This is Snell’s law of refraction, derived from Huygens’ wave theory. Quick Tip: Huygens’ principle explains reflection, refraction, diffraction, but not polarization.

Step 1: Identify given values.
\[ L = 20 \, mH = 20 \times 10^{-3} \, H, \quad C = 2 \, \mu F = 2 \times 10^{-6} \, F, \quad R = 1000 \, \Omega. \]

Step 2: Inductive reactance.
\[ X_L = 2 \pi f L. \]

Step 3: Capacitive reactance.
\[ X_C = \frac{1}{2 \pi f C}. \]

Step 4: Resonance condition.

At resonance, \(X_L = X_C\). So, \[ 2 \pi f L = \frac{1}{2 \pi f C}. \] \[ f^2 = \frac{1}{(2 \pi)^2 L C}. \] \[ f = \frac{1}{2 \pi \sqrt{LC}}. \]

Step 5: Substitution.
\[ f = \frac{1}{2 \pi \sqrt{20 \times 10^{-3} \times 2 \times 10^{-6}}}. \] \[ f = \frac{1}{2 \pi \sqrt{40 \times 10^{-9}}}. \] \[ f = \frac{1}{2 \pi \times 2 \times 10^{-4.5}} \approx 1000 \, Hz. \]

Step 6: Reactances at resonance.
\[ X_L = 2 \pi f L = 2 \pi (1000)(20 \times 10^{-3}) = 125.6 \, \Omega. \] \[ X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (1000)(2 \times 10^{-6})} = 125.6 \, \Omega. \]

Step 7: Impedance at resonance.

At resonance, \(X_L = X_C\), so impedance is purely resistive: \[ Z = R = 1000 \, \Omega. \]

Step 8: Conclusion.

(i) \(X_L = 125.6 \, \Omega\), \(X_C = 125.6 \, \Omega\)

(ii) \(f = 1000 \, Hz\)

(iii) \(Z = 1000 \, \Omega\) Quick Tip: At resonance, \(X_L = X_C\) and the circuit behaves like a pure resistor.

Step 1: Faraday’s laws.

1. Whenever magnetic flux linked with a circuit changes, an emf is induced in it.
2. The magnitude of induced emf is equal to the rate of change of flux: \[ e = - \frac{d\Phi}{dt}. \]

Step 2: Initial flux (circular loop).

Area of circle: \[ A_1 = \pi r^2 = \pi (0.1)^2 = 0.0314 \, m^2. \]
Flux: \[ \Phi_1 = B A_1 = 100 \times 0.0314 = 3.14 \, Wb. \]

Step 3: Final flux (square loop).

Perimeter of circle = perimeter of square. \[ 2 \pi r = 4a \quad \Rightarrow \quad a = \frac{\pi r}{2}. \] \[ a = \frac{3.14 \times 0.1}{2} = 0.157 \, m. \]
Area of square: \[ A_2 = a^2 = (0.157)^2 \approx 0.0247 \, m^2. \]
Flux: \[ \Phi_2 = B A_2 = 100 \times 0.0247 = 2.47 \, Wb. \]

Step 4: Change in flux.
\[ \Delta \Phi = \Phi_1 - \Phi_2 = 3.14 - 2.47 = 0.67 \, Wb. \]

Step 5: Average emf.
\[ e = \frac{\Delta \Phi}{\Delta t} = \frac{0.67}{0.1} = 6.7 \, V. \]

Correction: Wait! Magnetic field is \(100 T\), so flux values must be:
\[ \Phi_1 = 100 \times 0.0314 = 3.14 \, Wb, \quad \Phi_2 = 100 \times 0.0247 = 2.47 \, Wb. \] \[ \Delta \Phi = 0.67 \, Wb. \] \[ e = \frac{0.67}{0.1} = 6.7 \, V. \]

Step 6: Conclusion.

The average induced emf is \(6.7 \, V\). Quick Tip: Always compare areas when shape changes in magnetic field — emf depends on change in flux, not initial shape.