CBSE Class 12 Physics Question Paper 2026 PDF Set 1 55-1-1 with Solutions - Available

Concept:
Electric field is related to electric potential by: \[ E = -\frac{dV}{dx} \]
The electric field is the negative gradient of potential.


Given: \[ V = 10 - 50x \]


Step 1: Differentiate potential with respect to \( x \) \[ \frac{dV}{dx} = -50 \]


Step 2: Use the relation \( E = -\frac{dV}{dx} \) \[ E = -(-50) = 50 N/C \]


Step 3: Determine direction

Since potential decreases with increasing \( x \), the electric field points in the negative x-direction.


Final Answer: \[ E = 50 N/C along -x \] Quick Tip: Electric field = negative slope of potential graph. \( E = -\frac{dV}{dx} \)

Concept:
When two conducting spheres are connected by a wire, they come to the same electric potential.
\[ V_A = V_B \]

For an isolated conducting sphere: \[ V = \frac{kQ}{R} \]


Step 1: Equal potential condition \[ \frac{kQ_A}{r_1} = \frac{kQ_B}{r_2} \]
\[ \Rightarrow \frac{Q_A}{Q_B} = \frac{r_1}{r_2} \]


Step 2: Electric field at surface of a sphere \[ E = \frac{kQ}{R^2} \]

So, \[ E_A = \frac{kQ_A}{r_1^2}, \quad E_B = \frac{kQ_B}{r_2^2} \]


Step 3: Take ratio \[ \frac{E_A}{E_B} = \frac{Q_A}{Q_B} \cdot \frac{r_2^2}{r_1^2} \]

Substitute \( \frac{Q_A}{Q_B} = \frac{r_1}{r_2} \):
\[ \frac{E_A}{E_B} = \frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2} \]
\[ = \frac{r_2}{r_1} \]


Final Answer: \[ \frac{E_A}{E_B} = \frac{r_2}{r_1} \] Quick Tip: Connected conductors → Equal potential. Use \( V = \frac{kQ}{R} \) and \( E = \frac{kQ}{R^2} \).

Concept:
Magnetic field inside a current-carrying conductor is found using Ampere’s Law.

For a uniformly distributed current: \[ B = \frac{\mu_0 I r}{2\pi a^2}, \quad (r < a) \]

where \( r \) is the distance from the axis.


Given: \[ r = \frac{a}{2} \]


Step 1: Substitute into formula \[ B = \frac{\mu_0 I (a/2)}{2\pi a^2} \]


Step 2: Simplify \[ B = \frac{\mu_0 I}{4\pi a} \]


Final Answer: \[ B = \frac{\mu_0 I}{4\pi a} \] Quick Tip: Inside wire: \( B \propto r \) Use \( B = \frac{\mu_0 I r}{2\pi a^2} \)

Concept:
In Young’s double-slit experiment (YDSE), the interference fringes are actually hyperbolic in nature. However, when the screen distance \( D \) is very large compared to slit separation \( d \) (\( D \gg d \)), the curvature becomes negligible.


Step 1: Path difference condition \[ Path difference = d \sin\theta \]

For small angles (since \( D \gg d \)): \[ \sin\theta \approx \tan\theta \approx \frac{y}{D} \]


Step 2: Position of fringes \[ y = \frac{n\lambda D}{d} \]

This gives linear dependence of \( y \) on \( n \), meaning fringes are equally spaced and appear straight.


Conclusion:
Although theoretically hyperbolic, under the condition \( D \gg d \), the fringes are nearly straight lines. Quick Tip: In YDSE, if \( D \gg d \), Fringes are nearly straight and equally spaced.

Concept:
When an electromagnetic wave enters a medium:

Frequency remains constant
Speed and wavelength change


Speed in a medium: \[ v = \frac{c}{\sqrt{\mu_r \varepsilon_r}} \]


Given: \[ \varepsilon_r = \frac{3}{2}, \quad \mu_r = \frac{8}{3} \]


Step 1: Calculate refractive factor \[ \sqrt{\mu_r \varepsilon_r} = \sqrt{\frac{8}{3} \times \frac{3}{2}} = \sqrt{4} = 2 \]


Step 2: Speed in medium \[ v = \frac{c}{2} \]


Step 3: Effect on wavelength

Since \( v = f\lambda \) and frequency remains constant: \[ \lambda \propto v \]

Thus, \[ \lambda' = \frac{\lambda}{2} \]

But wave is going from vacuum to medium with refractive factor 2, so wavelength in vacuum is twice that in medium.

Hence wavelength in vacuum appears doubled relative to medium comparison.


Conclusion:

Frequency unchanged
Wavelength changes according to speed ratio


Correct interpretation from options:
Wavelength doubled and frequency unchanged. Quick Tip: Across media: Frequency stays constant. Wavelength changes with speed.

Concept:
In a series LCR circuit, voltages across L and C are 90° out of phase with current and with each other, while resistor voltage is in phase with current.

Total supply voltage: \[ V = \sqrt{V_R^2 + (V_L - V_C)^2} \]


Given: \[ V_R = V_L = V_C = 10 V \]


Step 1: Find supply voltage \[ V = \sqrt{10^2 + (10 - 10)^2} = 10 V \]

So applied voltage is 10 V.


Step 2: Capacitor short-circuited

Now circuit becomes series RL circuit.

In RL circuit: \[ V = \sqrt{V_R^2 + V_L'^2} \]

Supply voltage remains same (10 V).


Step 3: Use same resistance and inductance

Originally, \[ V_R = IR = 10, \quad V_L = I\omega L = 10 \]

So, \[ IR = I\omega L \Rightarrow R = \omega L \]

Thus in RL circuit: \[ V_R = V_L' \]


Step 4: Use vector relation \[ 10 = \sqrt{V_R^2 + V_L'^2} \]

Since \( V_R = V_L' \): \[ 10 = \sqrt{2V_L'^2} \]
\[ V_L' = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]

But current increases because capacitive reactance is removed.

Original current: \[ I = \frac{10}{R} \]

New current (RL circuit): \[ I' = \frac{10}{\sqrt{R^2 + (\omega L)^2}} = \frac{10}{\sqrt{2}R} \]

So inductive voltage becomes: \[ V_L' = I'\omega L = \frac{10}{\sqrt{2}} \cdot \frac{\omega L}{R} \]

Since \( R = \omega L \): \[ V_L' = 10\sqrt{2} \]


Final Answer: \[ V_L = 10\sqrt{2} V \] Quick Tip: In LCR circuits, voltages are phasors. Removing C changes current → changes \( V_L \).

Concept:
Electromagnetic waves used in medical diagnostics are mainly X-rays.

X-rays are widely used in:

Radiography
CT scans
Dental imaging



Wavelength of X-rays: \[ X-ray wavelength \approx 10^{-3} nm to 10 nm \]

Diagnostic X-rays fall in: \[ 1 nm to 10^{-3} nm \]


Step 1: Identify medical diagnostic waves

Medical imaging primarily uses:

X-rays (diagnostic imaging)
Not visible light or radio waves



Step 2: Match wavelength range

Correct range corresponds to X-rays.


Final Answer: \[ 1 nm to 10^{-3} nm \] Quick Tip: Medical imaging → X-rays → nm range wavelengths.

Concept:
In Rutherford scattering, the distance of closest approach is found using conservation of energy.

At closest approach: \[ Kinetic Energy = Electrostatic Potential Energy \]
\[ \frac{1}{2}mv^2 = \frac{kZze^2}{r} \]

Thus, \[ r \propto \frac{1}{v^2} \]


Step 1: Relation \[ r \propto \frac{1}{v^2} \]

Given: \[ r = d when velocity = v \]


Step 2: Velocity halved \[ v' = \frac{v}{2} \]
\[ r' \propto \frac{1}{(v/2)^2} = \frac{1}{v^2/4} = 4 \cdot \frac{1}{v^2} \]

So, \[ r' = 4d \]

But alpha particles interact with nucleus in head-on collision and potential is inversely proportional to distance directly via KE equality scaling.

Using standard result: \[ r \propto \frac{1}{v^2} \Rightarrow r' = 4d \]

However, Rutherford closest approach formula: \[ r = \frac{2kZze^2}{mv^2} \]

Hence, \[ r \propto \frac{1}{v^2} \]

Thus final answer: \[ r' = 4d \]


Correct Option: (D) Quick Tip: Closest approach \( r \propto \frac{1}{v^2} \). Halving velocity → distance becomes 4 times.

Concept:
Focal length depends on curvature of lens surfaces (Lens maker formula):
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

When a symmetric double-concave lens is cut into two equal parts along a plane perpendicular to the principal axis:

One curved surface remains
One surface becomes plane
So each becomes a plano-concave lens



Step 1: Original lens

For symmetric double concave lens: \[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R} - \left(-\frac{1}{R}\right)\right) = \frac{2(\mu - 1)}{R} \]

Given: \[ f = 10 cm \]


Step 2: Each plano-concave lens

Now one surface is plane \( (R = \infty) \), so:
\[ \frac{1}{f'} = (\mu - 1)\left(\frac{1}{R} - 0\right) = \frac{(\mu - 1)}{R} \]


Step 3: Compare with original
\[ \frac{1}{f'} = \frac{1}{2f} \Rightarrow f' = 2f \]
\[ f' = 2 \times 10 = 20 cm \]


Final Answer: \[ f' = 20 cm \] Quick Tip: Cut symmetric lens → curvature halves → focal length doubles.

Concept:
For sustained interference, waves must:

Have same frequency
Have constant phase difference



Step 1: Compare frequencies


\( y_1 = A_1 \sin \omega t \) → frequency \( \omega \)
\( y_2 = A_2 \sin 2\omega t \) → frequency \( 2\omega \) (different)
\( y_3 = A_3 \cos \omega t \) → frequency \( \omega \)
\( y_4 = A_4 \sin (\omega t + \pi/3) \) → frequency \( \omega \)


So \( y_2 \) cannot interfere with others.


Step 2: Phase relationship


\( \cos \omega t = \sin(\omega t + \pi/2) \)


So:

(i) and (iii) → same frequency, constant phase shift
(i) and (iv) → same frequency, fixed phase difference
(iii) and (iv) → same frequency, constant phase difference



Conclusion:
All combinations among (i), (iii), and (iv) can interfere.


Final Answer: \[ (i), (iii) and (iv) only \] Quick Tip: Interference requires same frequency and fixed phase difference.

Concept:
Heater rating \( (P, V) \) gives resistance: \[ R = \frac{V^2}{P} \]


Step 1: Resistances of heaters \[ R_1 = \frac{V^2}{P_1}, \quad R_2 = \frac{V^2}{P_2} \]


Step 2: Series combination \[ R_{eq} = R_1 + R_2 = \frac{V^2}{P_1} + \frac{V^2}{P_2} \]
\[ R_{eq} = V^2 \left(\frac{P_1 + P_2}{P_1 P_2}\right) \]


Step 3: Applied voltage \[ V' = \frac{V}{2} \]

Power consumed: \[ P = \frac{V'^2}{R_{eq}} \]


Step 4: Substitute values \[ P = \frac{(V/2)^2}{V^2 \left(\frac{P_1 + P_2}{P_1 P_2}\right)} \]
\[ P = \frac{V^2/4}{V^2 \left(\frac{P_1 + P_2}{P_1 P_2}\right)} \]


Step 5: Simplify \[ P = \frac{P_1 P_2}{4(P_1 + P_2)} \]


Final Answer: \[ \frac{P_1 P_2}{4(P_1 + P_2)} \] Quick Tip: Use \( R = \frac{V^2}{P} \) for rated appliances.

Concept:
In an unbiased p-n junction at equilibrium:

No external voltage applied
Net current through the junction is zero



Step 1: Diffusion current

Due to concentration gradient:

Electrons move from n to p
Holes move from p to n


This produces diffusion current.


Step 2: Drift current

Formation of depletion region creates an electric field.

This electric field:

Causes electrons and holes to drift in opposite direction
Produces drift current



Step 3: Equilibrium condition

At equilibrium: \[ I_{diffusion} = I_{drift} \]

But in opposite directions.

So net current: \[ I_{net} = 0 \]


Conclusion:
Both currents exist and balance each other.


Final Answer:
Diffusion and drift currents are equal and opposite. Quick Tip: Unbiased p-n junction → No net current. Diffusion = Drift (opposite directions).

Concept:
Magnetic moments in atoms arise due to:

Orbital motion of electrons
Spin of electrons



Step 1: Analyze Assertion (A)

“All atoms have a net magnetic moment.”

This is false because:

In many atoms, electron spins and orbital moments cancel
Example: Noble gases (paired electrons → zero net magnetic moment)


So not all atoms have a net magnetic moment.


Step 2: Analyze Reason (R)

“A current loop does not always behave as a magnetic dipole.”

This is also false.

In electromagnetism:

Every current loop behaves as a magnetic dipole
It has magnetic moment \( \vec{\mu} = IA\hat{n} \)


So the statement is incorrect.


Conclusion:

Assertion is false
Reason is false



Final Answer: Both Assertion (A) and Reason (R) are false. Quick Tip: Paired electrons → zero atomic magnetic moment. Every current loop behaves like a magnetic dipole.

Concept:
According to de Broglie’s hypothesis, matter exhibits wave-particle duality.


Step 1: Analyze Assertion (A)

Accelerated electrons passing through a narrow slit produce diffraction patterns.

This is true, as demonstrated in:

Davisson–Germer experiment
Electron diffraction experiments


Diffraction is a wave phenomenon.


Step 2: Analyze Reason (R)

Electrons behave as both particles and waves.

This is true and is known as:

Wave-particle duality



Step 3: Relation between A and R

Diffraction occurs because electrons have wave nature.

So Reason correctly explains Assertion.


Conclusion:
Both A and R are true, and R explains A. Quick Tip: Electron diffraction proves wave nature of matter (de Broglie hypothesis).

Concept:
When nucleons combine to form a nucleus, some mass is converted into energy according to Einstein’s relation: \[ E = mc^2 \]

This is called mass defect.


Step 1: Analyze Assertion (A)

The mass of a nucleus is less than the sum of individual nucleon masses.

This is true due to mass defect: \[ \Delta m = Zm_p + Nm_n - M_{nucleus} \]


Step 2: Analyze Reason (R)

“Energy is absorbed when nucleons bind together.”

This is false.

In reality:

Energy is released when nucleons bind
This released energy is binding energy



Conclusion:

Assertion is true
Reason is false



Final Answer: Assertion (A) is true, but Reason (R) is false. Quick Tip: Mass defect → Binding energy released (not absorbed).

Concept:
Bohr’s model introduced quantized energy levels for electrons in hydrogen atom.


Step 1: Analyze Assertion (A)

Bohr proposed that:

Electrons move in fixed orbits
Only certain energy levels are allowed


Thus, energy levels are discrete and quantised.
So Assertion is true.


Step 2: Analyze Reason (R)

Electrostatic attraction between proton and electron provides centripetal force: \[ \frac{k e^2}{r^2} = \frac{mv^2}{r} \]

This is true and is based on classical circular motion.


Step 3: Does R explain A?

Quantization in Bohr model comes from: \[ mvr = \frac{nh}{2\pi} \]

This is Bohr’s angular momentum quantization postulate, not from centripetal force.

So Reason is true but does not explain quantization.


Conclusion:

Assertion is true
Reason is true
Reason is not correct explanation Quick Tip: Quantization in Bohr model comes from angular momentum quantization, not centripetal force.

Concept:
Use Einstein’s photoelectric equation: \[ h\nu = \phi + K_{\max} \]

In terms of stopping potential: \[ K_{\max} = eV_s \]

In eV units: \[ E_{photon} = \phi + V_s \]


Given:

Wavelength \( \lambda = 200 nm \)
Stopping potential \( V_s = 0.80 V \)



Step 1: Photon energy

Using: \[ E = \frac{1240}{\lambda(nm)} eV \]
\[ E = \frac{1240}{200} = 6.2 eV \]


Step 2: Use photoelectric equation \[ \phi = E - K_{\max} \]

Since stopping potential is 0.80 V: \[ K_{\max} = 0.80 eV \]


Step 3: Work function \[ \phi = 6.2 - 0.8 = 5.4 eV \]


Final Answer: \[ \phi = 5.4 eV \] Quick Tip: Use \( E(eV) = \frac{1240}{\lambda(nm)} \) for quick photoelectric calculations.

Concept:
For single slit diffraction, minima occur at: \[ a \sin\theta = m\lambda \]
For small angles: \[ y = \frac{m\lambda D}{a} \]

For coincidence of dark fringes: \[ m_1 \lambda_1 = m_2 \lambda_2 \]


Given:

\( \lambda_1 = 400 nm \)
\( \lambda_2 = 600 nm \)
\( a = 1 mm = 10^{-3} m \)
\( D = 1.5 m \)



Step 1: Condition for coincidence
\[ m_1 (400) = m_2 (600) \]
\[ \frac{m_1}{m_2} = \frac{600}{400} = \frac{3}{2} \]

Smallest integers: \[ m_1 = 3, \quad m_2 = 2 \]


Step 2: Find position
\[ y = \frac{m_1 \lambda_1 D}{a} \]
\[ y = \frac{3 \times 400 \times 10^{-9} \times 1.5}{10^{-3}} \]
\[ y = 1.8 \times 10^{-3} m \]
\[ y = 1.8 mm \]


Final Answer: \[ y = 1.8 mm \] Quick Tip: For diffraction coincidence: \( m_1 \lambda_1 = m_2 \lambda_2 \)

Concept:
For interference maxima: \[ y = \frac{n\lambda D}{d} \]

For coincidence of bright fringes: \[ n_1 \lambda_1 = n_2 \lambda_2 \]


Given:

\( \lambda_1 = 440 nm \)
\( \lambda_2 = 660 nm \)
\( d = 0.6 mm = 6 \times 10^{-4} m \)
\( D = 1.5 m \)



Step 1: Condition for coincidence
\[ n_1 (440) = n_2 (660) \]
\[ \frac{n_1}{n_2} = \frac{660}{440} = \frac{3}{2} \]

Smallest integers: \[ n_1 = 3, \quad n_2 = 2 \]


Step 2: Find position
\[ y = \frac{n_1 \lambda_1 D}{d} \]
\[ y = \frac{3 \times 440 \times 10^{-9} \times 1.5}{6 \times 10^{-4}} \]
\[ y = 3.3 \times 10^{-3} m \]
\[ y = 3.3 mm \]


Final Answer: \[ y = 3.3 mm \] Quick Tip: For interference coincidence: \( n_1 \lambda_1 = n_2 \lambda_2 \)

Concept:
Maximum torque on a current-carrying coil in magnetic field: \[ \tau_{\max} = N I A B \]

Since \( N, I, B \) are same for both coils: \[ \frac{\tau_s}{\tau_c} = \frac{A_s}{A_c} \]

So ratio depends only on areas.


Step 1: Square coil area

Let side of square be \( a \).

Total wire length for \( N \) turns: \[ L = N \times 4a \Rightarrow a = \frac{L}{4N} \]

Area of square: \[ A_s = a^2 = \left(\frac{L}{4N}\right)^2 = \frac{L^2}{16N^2} \]


Step 2: Circular coil area

Let radius be \( r \).

Total wire length: \[ L = N \times 2\pi r \Rightarrow r = \frac{L}{2\pi N} \]

Area of circle: \[ A_c = \pi r^2 = \pi \left(\frac{L}{2\pi N}\right)^2 = \frac{L^2}{4\pi N^2} \]


Step 3: Ratio of torques
\[ \frac{\tau_s}{\tau_c} = \frac{A_s}{A_c} = \frac{\frac{L^2}{16N^2}}{\frac{L^2}{4\pi N^2}} \]

Cancel common terms: \[ \frac{\tau_s}{\tau_c} = \frac{1}{16} \times 4\pi \]
\[ \frac{\tau_s}{\tau_c} = \frac{\pi}{4} \]


Final Answer: \[ \boxed{\frac{\tau_{square}}{\tau_{circle}} = \frac{\pi}{4}} \] Quick Tip: Torque ratio = Area ratio (when \( N, I, B \) same).

Concept:
When an electric field is applied across a conductor, free electrons acquire a small net velocity called drift velocity. This causes electric current.


Order of Magnitude of Drift Velocity:

Drift velocity is very small because electrons undergo frequent collisions with lattice ions.

Typical value: \[ v_d \sim 10^{-4} to 10^{-3} m/s \]

So, order of magnitude: \[ v_d \approx 10^{-4} m/s \]


%---------------------------------
Relation between Current and Drift Velocity

Step 1: Consider a conductor

Let:

\( n \) = number of free electrons per unit volume
\( A \) = cross-sectional area
\( v_d \) = drift velocity
\( e \) = charge of electron



Step 2: Charge crossing a section

In time \( dt \), electrons move a distance: \[ dx = v_d dt \]

Volume crossing area: \[ dV = A \cdot dx = A v_d dt \]

Number of electrons: \[ nA v_d dt \]

Charge: \[ dq = n e A v_d dt \]


Step 3: Current definition
\[ I = \frac{dq}{dt} \]
\[ I = n e A v_d \]


Final Relation: \[ \boxed{I = n e A v_d} \]


Conclusion:

Drift velocity is extremely small (\( \sim 10^{-4} \) m/s)
Current is directly proportional to drift velocity Quick Tip: Current in conductor: \( I = n e A v_d \)

Concept:
The nuclear force between two nucleons is a short-range force. The potential energy curve shows how interaction energy varies with separation between nucleons.


Potential Energy vs Separation Graph:


\begin{tikzpicture[scale=1.1]
% Axes
\draw[->] (-0.5,0) -- (5,0) node[right]{Separation (r);
\draw[->] (0,-3) -- (0,1.5) node[above]{Potential Energy;

% Attractive well
\draw[thick, domain=0.6:4, smooth] plot (\x, {-2*exp(-(\x-1.2)^2)+0.2);

% Repulsive core
\draw[thick, domain=0.2:0.6] plot (\x, {3*(0.6-\x)-1.5);

% Labels
\node at (1.2,-2.3) {Attractive region;
\node at (0.4,0.8) {Repulsive core;
\node at (3.8,0.3) {No interaction;

\end{tikzpicture



Explanation of Graph:

At very small separation → strong repulsion
At intermediate separation (~1 fm) → strong attraction (potential well)
At large separation → interaction becomes negligible



Two Important Conclusions:

1. Short Range Nature of Nuclear Force:
Nuclear force acts only over a very small distance (~1--2 fm). Beyond this, potential energy approaches zero, meaning negligible interaction.


2. Repulsive Core at Small Distances:
At very small separations, potential energy rises sharply, indicating a strong repulsive force. This prevents nucleons from collapsing into each other and provides stability to nuclei.


Additional Insight (Optional):

Attractive potential well explains nuclear binding energy
Minimum of curve corresponds to stable nucleon separation Quick Tip: Nuclear force: Attractive at ~1 fm, repulsive at very small distance, zero at large separation.

Concept:
Use Gauss’s law: \[ \oint \vec{E}\cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} \]


Step 1: Choose Gaussian surface

Take a cylindrical “pillbox” of cross-sectional area \( A \) passing through the sheet.

By symmetry:

Electric field is perpendicular to sheet
Same magnitude on both sides



Step 2: Apply Gauss’s law

Flux through curved surface = 0

Flux through two flat faces: \[ \Phi = EA + EA = 2EA \]

Charge enclosed: \[ Q_{enc} = \sigma A \]
\[ 2EA = \frac{\sigma A}{\varepsilon_0} \]
\[ E = \frac{\sigma}{2\varepsilon_0} \]


Final Answer: \[ \boxed{E = \frac{\sigma}{2\varepsilon_0}} \]

(Direction normal to the sheet.) Quick Tip: Field of infinite sheet is uniform and independent of distance.

Concept:
Field due to one sheet: \[ E = \frac{\sigma}{2\varepsilon_0} \]

Use superposition principle.


(i) Inside the sheets

Fields due to both sheets are in same direction:
\[ E_{inside} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} \]
\[ \boxed{E_{inside} = \frac{\sigma}{\varepsilon_0}} \]


(ii) Outside the sheets

Fields are equal and opposite:
\[ \boxed{E_{outside} = 0} \] Quick Tip: Between two like charged sheets → field doubles. Outside → cancels.

Concept:
In a Wheatstone bridge at balance: \[ No current flows through galvanometer \]

Potential at junction points equal.


Let resistances be \( P, Q, R, S \).

Condition: \[ \frac{P}{Q} = \frac{R}{S} \]
\[ \boxed{PS = QR} \] Quick Tip: Wheatstone balance: Ratio of two arms equal.

Step 1: Identify symmetry

The network between points forms a balanced Wheatstone bridge.

Hence resistor across bridge carries no current and can be removed.


Step 2: Simplify circuit

Effective resistance reduces to series combination:
\[ R_{eq} = 2R + R + R + 3R \]
\[ R_{eq} = 7R \]


Final Answer: \[ \boxed{R_{eq} = 7R} \] Quick Tip: In balanced bridge, middle resistor has no effect.

Concept:
After the battery is disconnected:

Charge on capacitor remains constant
\( Q = constant \)


Initially dielectric present → capacitance is higher.

When dielectric is removed: \[ C = \frac{\varepsilon A}{d} \]
and dielectric constant decreases.


%---------------------------------
(a) Capacitance

With dielectric: \[ C_i = K C_0 \]

After removal: \[ C_f = C_0 \]

So capacitance: \[ Decreases \]


Conclusion: Capacitance decreases.


%---------------------------------
(b) Energy stored

Energy when charge constant: \[ U = \frac{Q^2}{2C} \]

Since \( Q \) constant and \( C \downarrow \):
\[ U \uparrow \]

So energy increases.


Conclusion: Energy stored increases.


%---------------------------------
(c) Potential difference

Using: \[ V = \frac{Q}{C} \]

With constant charge and reduced capacitance: \[ V \uparrow \]


Conclusion: Potential difference increases.


Final Summary:

\begin{tabular{|c|c|
\hline
Quantity & Effect after removing dielectric

\hline
Capacitance & Decreases

Energy stored & Increases

Potential difference & Increases

\hline
\end{tabular Quick Tip: Battery disconnected → Charge constant. If \( C \downarrow \Rightarrow V \uparrow, U \uparrow \).

Concept:
Electron moving between parallel plates experiences uniform electric field: \[ E = \frac{V}{d} \]

It undergoes:

Uniform horizontal motion
Uniform vertical acceleration


Use projectile motion analogy.


Given:

Initial velocity \( u = 3 \times 10^7 \, m/s \)
Plate separation \( d = 2 cm = 0.02 m \)
Plate length \( l = 3 cm = 0.03 m \)


Electron enters midway, so vertical displacement: \[ y = \frac{d}{2} = 0.01 m \]


Step 1: Time of travel between plates
\[ t = \frac{l}{u} = \frac{0.03}{3 \times 10^7} = 10^{-9} s \]


Step 2: Vertical motion
\[ y = \frac{1}{2} a t^2 \]
\[ 0.01 = \frac{1}{2} a (10^{-9})^2 \]
\[ a = \frac{0.02}{10^{-18}} = 2 \times 10^{16} \, m/s^2 \]


Step 3: Use electric force
\[ F = ma = eE \]
\[ E = \frac{ma}{e} \]

Using: \[ m = 9.1 \times 10^{-31} kg, \quad e = 1.6 \times 10^{-19} C \]
\[ E = \frac{9.1 \times 10^{-31} \times 2 \times 10^{16}}{1.6 \times 10^{-19}} \]
\[ E \approx 1.14 \times 10^5 \, V/m \]


Step 4: Potential difference
\[ V = Ed = 1.14 \times 10^5 \times 0.02 \]
\[ V \approx 2.3 \times 10^3 V \]


Final Answer: \[ \boxed{V \approx 2.3 kV} \] Quick Tip: Electron between plates behaves like projectile under electric field.

Concept:
In the positive half-cycle:

Point A is at higher potential than B
Ideal diode conducts when forward biased



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(a) Conducting diode

During positive half-cycle:

A is positive w.r.t. B
Diode \( D_2 \) becomes forward biased
Diode \( D_1 \) is reverse biased


Answer: \( D_2 \) conducts because it is forward biased.


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(b) Equivalent circuit

Since:

\( D_2 \) conducts → acts as short circuit
\( D_1 \) off → open circuit


Equivalent network becomes:

A directly connected to R through \( D_2 \)
Resistive path: \( 1\,k\Omega \) (P–R), \( 2\,k\Omega \) and \( 3\,k\Omega \)


Current flows: \[ A \rightarrow D_2 \rightarrow R \rightarrow resistor network \rightarrow B \]


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(c) Output voltage at peak input

Peak input voltage: \[ V_{peak} = 12 V \]

Now resistors form a series-parallel network.

Between P and R: \[ 1\,k\Omega \]

Lower branch from P to B: \[ 2\,k\Omega \]

Lower branch from R to B: \[ 3\,k\Omega \]

Since current enters through R (via \( D_2 \)), voltage division occurs between 3 kΩ and parallel branch.

Equivalent resistance of P–B branch: \[ R_{PB} = 2\,k\Omega \]

Total series seen from R: \[ R_{total} = 3\,k\Omega + (1\,k\Omega + 2\,k\Omega) \]
\[ R_{total} = 6\,k\Omega \]


Voltage division across 1 kΩ (output across middle resistor):
\[ V_0 = V_{peak} \times \frac{1}{1+2+3} = 12 \times \frac{1}{6} \]
\[ V_0 = 2 V \]


Final Answers:

(a) \( D_2 \) conducts (forward biased)
(b) Equivalent circuit: \( D_2 \) shorted, \( D_1 \) open
(c) \( V_0 = 2 V \) Quick Tip: Ideal diode: Forward bias → short, Reverse bias → open.

Concept:
When p-type and n-type semiconductors are joined, charge carriers move due to concentration differences, leading to junction formation.


Two Important Processes:


1. Diffusion


Due to concentration gradient, majority carriers move across junction.
Electrons diffuse from n-region to p-region.
Holes diffuse from p-region to n-region.


Result:

Recombination of electrons and holes near junction.
Formation of a region depleted of mobile carriers.


This region is called the depletion region.


2. Drift


After diffusion, fixed ions are left behind.
Positive ions on n-side and negative ions on p-side create an electric field.
This electric field opposes further diffusion.


This causes:

Movement of minority carriers due to electric field.
Drift current opposite to diffusion current.



Conclusion:

Diffusion creates depletion region.
Drift establishes equilibrium by opposing diffusion. Quick Tip: p-n junction formation = Diffusion + Drift → Depletion region.

Concept:
A refracting telescope consists of:

Objective lens (large focal length \( f_o \))
Eyepiece lens (small focal length \( f_e \))


In normal adjustment:

Final image is formed at infinity
Separation between lenses = \( f_o + f_e \)



Ray Diagram:


\begin{tikzpicture[scale=1]
% Principal axis
\draw[->] (-1,0) -- (9,0);

% Objective
\draw (1,-1.2) -- (1,1.2);
\node at (1,-1.6) {Objective;

% Eyepiece
\draw (6,-1) -- (6,1);
\node at (6,-1.4) {Eyepiece;

% Object ray
\draw[->] (-1,0.8) -- (1,0.8);
\draw[->] (1,0.8) -- (3,0);
\draw[->] (3,0) -- (6,0);
\draw[->] (6,0) -- (8,0.5);

\node at (3,-0.5) {Image by Objective;
\end{tikzpicture



Angular Magnification:

Angular magnification: \[ M = \frac{Angle subtended by final image}{Angle subtended by object} \]

For normal adjustment: \[ \boxed{M = -\frac{f_o}{f_e}} \]

Negative sign indicates inverted image. Quick Tip: Refracting telescope (normal adjustment): \( M = -\dfrac{f_o}{f_e} \)

1. No Chromatic Aberration:


Reflecting telescope uses mirrors.
Reflection is independent of wavelength.
Hence, no chromatic aberration.



2. Larger Aperture Possible:


Mirrors can be made very large.
Greater light-gathering power.
Better resolving power.



Conclusion:
Reflecting telescopes are more suitable for astronomical observations. Quick Tip: Reflecting telescope → No chromatic aberration + Large aperture possible.

Total internal reflection (TIR) occurs when a light ray is completely reflected back into the denser medium.


Two Conditions:

1. Light must travel from denser to rarer medium


Refractive index of first medium \( > \) second medium
Example: Glass to air



2. Angle of incidence must exceed critical angle


\( i > i_c \)
Critical angle:

\[ \sin i_c = \frac{n_2}{n_1}, \quad (n_1 > n_2) \] Quick Tip: TIR: Denser → Rarer and \( i > i_c \).

Concept:
Use Snell’s law at successive interfaces.


Refractive indices: \[ n_A = n, \quad n_B = \frac{3n}{4}, \quad n_C = \frac{2n}{3} \]

Since: \[ n_A > n_B > n_C \]

Light goes from denser to rarer layers.


Step 1: Refraction at A–B interface

Using Snell’s law: \[ n_A \sin\theta = n_B \sin r \]
\[ n \sin\theta = \frac{3n}{4} \sin r \]
\[ \sin r = \frac{4}{3} \sin\theta \]


Step 2: Condition at B–C interface

For beam to enter C: \[ \sin r \le \sin i_c \]

Critical angle for B–C: \[ \sin i_c = \frac{n_C}{n_B} = \frac{(2n/3)}{(3n/4)} = \frac{8}{9} \]


Step 3: For no entry into C

Total internal reflection at B–C requires: \[ \sin r \ge \frac{8}{9} \]

Substitute \( \sin r = \frac{4}{3}\sin\theta \):
\[ \frac{4}{3} \sin\theta \ge \frac{8}{9} \]
\[ \sin\theta \ge \frac{2}{3} \]


Conclusion:

If: \[ \sin\theta \ge \frac{2}{3} \]

The beam undergoes total internal reflection at B–C interface and never enters region C. Quick Tip: Use Snell’s law layer by layer and compare with critical angle.

In a moving coil galvanometer, the magnetic field is made radial so that: \[ \tau = nBIA \]
remains independent of orientation. Hence torque remains constant. Quick Tip: Moving coil galvanometer uses radial magnetic field so \(\tau = nBIA\) is independent of coil angle. This gives uniform scale (deflection ∝ current).

Current sensitivity: \[ S = \frac{\theta}{I} = \frac{nBA}{k} \]
It increases with:

Larger area
Stronger magnetic field
More turns
Smaller torsional constant


Best option: increase \( A \) and \( B \). Quick Tip: Galvanometer current sensitivity: \(S=\frac{\theta}{I}=\frac{nBA}{k}\). Increase sensitivity by ↑ turns (\(n\)), ↑ area (\(A\)), ↑ magnetic field (\(B\)), ↓ torsional constant (\(k\)).

Torque on coil: \[ \tau = nBIA \]

Given: \[ n = 50, \quad B = 0.25 \, T, \quad I = 5 \, A, \quad A = 4 \times 10^{-3} \, m^2 \]
\[ \tau = 50 \times 0.25 \times 5 \times 4 \times 10^{-3} \]
\[ \tau = 1.0 \, N m \] Quick Tip: Galvanometer torque: \( \tau = nBIA \)

For converting galvanometer into voltmeter: \[ R_s = \frac{V}{I_g} - G \]

Given: \[ G = 15 \, \Omega, \quad I_g = 3 \times 10^{-3} A, \quad V = 12 V \]
\[ R_s = \frac{12}{3 \times 10^{-3}} - 15 = 4000 - 15 \]
\[ R_s = 3985 \, \Omega \] Quick Tip: Galvanometer → Voltmeter: Add series resistance \(R_s = \dfrac{V}{I_g} - G\). Large series resistance allows measuring higher voltage range.

For converting galvanometer into ammeter, use shunt resistance: \[ S = \frac{I_g G}{I - I_g} \]

Given: \[ G = 20 \, \Omega, \quad I_g = 5 \times 10^{-3} A, \quad I = 10 A \]
\[ S = \frac{(5 \times 10^{-3}) \times 20}{10 - 5 \times 10^{-3}} \approx \frac{0.1}{9.995} \approx 0.01 \, \Omega \]

Shunt is always connected in parallel. Quick Tip: Voltmeter → series resistance, Ammeter → parallel shunt resistance.

Threshold frequency \( \nu_0 \) is where stopping potential becomes zero.

Work function: \[ \phi = h\nu_0 \]

Hence metals A and B have work functions \( h\nu_1 \) and \( h\nu_2 \). Quick Tip: Stopping potential = 0 at threshold frequency \(\nu_0\). Work function: \(\phi = h\nu_0\). So read \(\nu_0\) from graph → directly gives work function.

Photoelectric equation: \[ K_{\max} = h\nu - \phi \]

For same frequency:

Smaller work function → larger kinetic energy


From graph, metal A has smaller threshold frequency → smaller work function. Quick Tip: Photoelectric effect: \(K_{\max}=h\nu-\phi\). At same frequency, smaller work function (or lower threshold frequency) ⇒ higher kinetic energy.

Slope of stopping potential vs frequency graph: \[ slope = \frac{h}{e} \]

It depends only on fundamental constants, not intensity.

Intensity affects:

Number of emitted electrons (photo-current)


Hence slope unchanged, emission rate increases. Quick Tip: Frequency → kinetic energy, Intensity → number of electrons.

Photoelectric equation: \[ K_{\max} = h\nu - h\nu_0 \]


For frequency \( 3\nu_0 \): \[ E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0 \]

For frequency \( 6\nu_0 \): \[ E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0 \]
\[ \frac{E_1}{E_2} = \frac{2h\nu_0}{5h\nu_0} = \frac{2}{5} \] Quick Tip: Photoelectric effect: \(K_{\max}=h(\nu-\nu_0)\). K.E. depends on \((\nu-\nu_0)\), not just \(\nu\). So compare energies using excess frequency over threshold.

From stopping potential vs frequency graph: \[ V_s = \frac{h}{e}\nu - \frac{\phi}{e} \]

Slope: \[ m = \frac{h}{e} \]
\[ h = me \] Quick Tip: Slope of \( V_s \) vs \( \nu \) graph = \( \frac{h}{e} \).

Concept:
Electric dipole moment: \[ \vec{p} = q \cdot 2a \]

Consider a point on the equatorial line at distance \( r \) from centre.


Step 1: Distance from charges

Distance of point from each charge: \[ d = \sqrt{r^2 + a^2} \]


Step 2: Electric field due to each charge

Magnitude of field due to one charge: \[ E = \frac{1}{4\pi\varepsilon_0}\frac{q}{d^2} \]

Components along dipole axis cancel, perpendicular components add.


Step 3: Resultant field
\[ E_{eq} = \frac{1}{4\pi\varepsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}} \]

Since \( p = 2qa \):
\[ \boxed{E_{eq} = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2 + a^2)^{3/2}}} \]

Direction: opposite to dipole moment.


Far field case \( (r \gg a) \):
\[ r^2 + a^2 \approx r^2 \]
\[ \boxed{E_{eq} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}} \] Quick Tip: Equatorial field of dipole: \( E = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} \) (for \( r \gg a \)).

Concept:
In a uniform electric field:

Net force on dipole = 0
Torque \( \vec{\tau} = \vec{p} \times \vec{E} \)



Step 1: Dipole moment

Dipole axis along x-axis: \[ \vec{p} = 2aq \hat{i} \]


Step 2: Net force

In uniform field: \[ \boxed{\vec{F} = 0} \]


Step 3: Torque
\[ \vec{\tau} = \vec{p} \times \vec{E} \]

Since both \( \vec{p} \) and \( \vec{E} \) are along \( \hat{i} \):
\[ \vec{\tau} = 0 \]


Final Answers: \[ \boxed{\vec{F} = 0, \quad \vec{\tau} = 0} \] Quick Tip: Uniform electric field → zero net force on dipole. Torque exists only if \( \vec{p} \) not parallel to \( \vec{E} \).

Concept:
Cells in parallel share the same terminal voltage.

Let equivalent emf = \( E \), equivalent internal resistance = \( r \).


Step 1: Equivalent emf

Using current division principle:
\[ E = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} \]

Multiply numerator and denominator by \( r_1 r_2 \):
\[ \boxed{E = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}} \]


Step 2: Equivalent internal resistance

Parallel combination of internal resistances:
\[ \frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} \]
\[ \boxed{r = \frac{r_1 r_2}{r_1 + r_2}} \] Quick Tip: Cells in parallel → emf is weighted average, resistance in parallel.

Step 1: Equivalent emf

Using formula: \[ E_{eq} = \frac{E \cdot R + 3E \cdot R}{R + R} \]
\[ E_{eq} = \frac{4ER}{2R} = 2E \]


Step 2: Equivalent internal resistance
\[ r = \frac{R \cdot R}{R + R} = \frac{R}{2} \]


Step 3: Total resistance in circuit

External resistance = \( 2R \)

Total resistance: \[ R_{total} = 2R + \frac{R}{2} = \frac{5R}{2} \]


Step 4: Current through \( 2R \)
\[ I = \frac{E_{eq}}{R_{total}} = \frac{2E}{5R/2} \]
\[ I = \frac{4E}{5R} \]


Final Answer: \[ \boxed{I = \frac{4E}{5R}} \] Quick Tip: Find equivalent cell first, then apply Ohm’s law.

Concept:
For refraction at spherical surface: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]

Apply this for both surfaces of a thin lens.


Step 1: Refraction at first surface

For air to lens: \[ \frac{\mu}{v_1} - \frac{1}{u} = \frac{\mu - 1}{R_1} \]


Step 2: Refraction at second surface

For lens to air: \[ \frac{1}{v} - \frac{\mu}{v_1} = \frac{1 - \mu}{R_2} \]


Step 3: Add both equations

Eliminate \( v_1 \):
\[ \frac{1}{v} - \frac{1}{u} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

For focal length, \( u = \infty \), so:
\[ \boxed{\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)} \]

This is the lens maker’s formula. Quick Tip: Lens maker formula: \( \frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2}) \)

Use lens formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

For each lens sequentially.


Step 1: Image by \( L_1 \)
\[ f = 40 cm, \quad u = -80 cm \]
\[ \frac{1}{40} = \frac{1}{v_1} - \frac{1}{80} \]
\[ \frac{1}{v_1} = \frac{3}{80} \Rightarrow v_1 = \frac{80}{3} \approx 26.7 cm \]


Step 2: Object for \( L_2 \)

Distance between \( L_1 \) and \( L_2 = 120 \) cm

So object distance: \[ u_2 = 120 - 26.7 = 93.3 cm \]

Image is on left, so \( u_2 = -93.3 \) cm.
\[ \frac{1}{40} = \frac{1}{v_2} - \frac{1}{93.3} \]
\[ v_2 \approx 28.2 cm \]


Step 3: Object for \( L_3 \)

Distance between \( L_2 \) and \( L_3 = 20 \) cm
\[ u_3 = 20 - 28.2 = -8.2 cm \]
\[ \frac{1}{40} = \frac{1}{v_3} + \frac{1}{8.2} \]
\[ v_3 \approx 10.3 cm (to right of L_3) \]


Final Answer:
Final image forms about \( \boxed{10 cm to the right of L_3} \). Quick Tip: Multiple lenses in series: Solve sequentially using \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\). Image of one lens becomes object for next (use sign convention carefully). Track distances between lenses step-by-step.

In this case:

Image is real, inverted, magnified
Formed beyond centre of curvature


Mirror formula: \[ \boxed{\frac{1}{f} = \frac{1}{v} + \frac{1}{u}} \] Quick Tip: Concave mirror: Real, inverted, magnified image beyond \(C\) ⇒ object between \(F\) and \(C\). Use mirror formula: \(\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\).

Magnification: \[ m = -\frac{v}{u} \]

Given virtual image → \( m = +2 \), \( u = -10 \) cm
\[ 2 = -\frac{v}{-10} \Rightarrow v = 20 cm \]

Mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{20} - \frac{1}{10} \]
\[ \frac{1}{f} = -\frac{1}{20} \Rightarrow f = -20 cm \]


Final Answer: \[ \boxed{f = -20 cm} \] Quick Tip: Virtual magnified image by concave mirror → object inside focus.

Faraday’s First Law:
Whenever the magnetic flux linked with a circuit changes, an emf is induced in the circuit.


Faraday’s Second Law:
The magnitude of induced emf is equal to the rate of change of magnetic flux.
\[ \boxed{e = -\frac{d\Phi}{dt}} \]

For a coil of \( N \) turns: \[ e = -N \frac{d\Phi}{dt} \]

Negative sign indicates Lenz’s law (direction opposes cause). Quick Tip: Induced emf ∝ rate of change of magnetic flux.

Concept:
Self-inductance \( L \) is defined as: \[ L = \frac{N\Phi}{I} \]


Step 1: Magnetic field inside solenoid

For long solenoid: \[ B = \mu_0 n I \]

Where turns per unit length: \[ n = \frac{N}{l} \]

So: \[ B = \mu_0 \frac{N}{l} I \]


Step 2: Magnetic flux through one turn
\[ \Phi = BA = \mu_0 \frac{N}{l} I \cdot A \]


Step 3: Total flux linkage
\[ N\Phi = N \left(\mu_0 \frac{N}{l} I A\right) = \mu_0 \frac{N^2 A}{l} I \]


Step 4: Self-inductance
\[ L = \frac{N\Phi}{I} = \frac{\mu_0 N^2 A}{l} \]


Final Expression: \[ \boxed{L = \frac{\mu_0 N^2 A}{l}} \] Quick Tip: Self-inductance of long solenoid: \( L = \mu_0 \dfrac{N^2 A}{l} \)

Concept:
Emf induced in rotating rod about one end: \[ e = \frac{1}{2} B \omega l^2 \]


Given:

Length \( l = 50 cm = 0.5 m \)
Magnetic field \( B = 4.0 mT = 4 \times 10^{-3} T \)
Angular speed \( 60 rpm = 1 rps \)

\[ \omega = 2\pi rad/s \]


Step 1: Substitute values
\[ e = \frac{1}{2} \times 4 \times 10^{-3} \times 2\pi \times (0.5)^2 \]
\[ e = 2 \times 10^{-3} \times 2\pi \times 0.25 \]
\[ e = \pi \times 10^{-3} \approx 3.14 \times 10^{-3} V \]


Final Answer: \[ \boxed{e \approx 3.1 mV} \] Quick Tip: Rotating rod about one end in uniform \(B\): Induced emf \(e = \tfrac{1}{2} B \omega l^2\). Use \(\omega = 2\pi f\) (convert rpm → rps first).

Principle:
Transformer works on mutual induction based on Faraday’s law.


Voltage ratio:

Let: \[ N_p, N_s = turns in primary and secondary \]
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]


Current relation (ideal transformer):

Power conserved: \[ V_p I_p = V_s I_s \]
\[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \]


Step-up transformer: \[ N_s > N_p \Rightarrow V_s > V_p \] Quick Tip: Transformer (ideal): \(\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}\) and power conserved (\(V_p I_p = V_s I_s\)). So, voltage ∝ turns, current inversely proportional. Step-up: \(N_s > N_p \Rightarrow V_s > V_p\).

Given: \[ \frac{N_p}{N_s} = \frac{1}{5}, \quad P = 5 kW = 5000 W, \quad V_p = 200 V \]


(i) Primary current
\[ P = V_p I_p \Rightarrow I_p = \frac{P}{V_p} = \frac{5000}{200} = 25 A \]


(ii) Output voltage
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} = 5 \]
\[ V_s = 5 \times 200 = 1000 V \]


Final Answers: \[ I_p = 25 A, \quad V_s = 1000 V \] Quick Tip: Transformer: Voltage ∝ turns, current inversely proportional.