CBSE Class 12 Physics Question Paper 2026 PDF Set 1 55-5-1 with Solutions - Available

Concept:
When identical conducting spheres are brought into contact, charge redistributes equally.
The electrostatic force between two charges is given by Coulomb’s law: \[ F = k \frac{|q_1 q_2|}{r^2} \]
Key ideas used:

Charge conservation during contact
Equal charge sharing for identical spheres
Force dependence on both charge and distance



Step 1: Initial force.
Initial charges are \( q \) and \( -2q \), separated by distance \( r \). \[ F = k \frac{|q \cdot (-2q)|}{r^2} = k \frac{2q^2}{r^2} \]
Since charges are opposite, the force is attractive.


Step 2: Charge after contact.
Total charge: \[ q + (-2q) = -q \]
Since spheres are identical, charge distributes equally: \[ Charge on each sphere = \frac{-q}{2} \]


Step 3: New separation.
After separation, distance becomes \( \frac{r}{2} \).


Step 4: New force.
Now both charges are \( -\frac{q}{2} \), so force is repulsive: \[ F' = k \frac{\left(\frac{q}{2}\right)^2}{\left(\frac{r}{2}\right)^2} \] \[ F' = k \frac{q^2/4}{r^2/4} = k \frac{q^2}{r^2} \]


Step 5: Compare with initial force.
Initial: \[ F = k \frac{2q^2}{r^2} \]
New: \[ F' = k \frac{q^2}{r^2} = \frac{F}{2} \]

Hence, the force becomes half and is repulsive. Quick Tip: For identical conducting spheres: - Charges equalize after contact. - Always compare forces using Coulomb’s law carefully (charge and distance both matter).

Concept:
The electric field is related to electric potential by: \[ E = -\frac{dV}{dx} \]
Force on a charge: \[ F = qE \]
Thus, the magnitude of force depends on the slope of the \( V \) vs \( x \) graph.

Steeper slope \( \Rightarrow \) larger electric field
Flat region \( \Rightarrow \) zero force



Step 1: Analyze each point.

At P:
The graph is horizontal (constant potential). \[ \frac{dV}{dx} = 0 \Rightarrow E = 0 \Rightarrow F = 0 \]


At Q:
The graph has a moderate negative slope.
This means a finite electric field and moderate force.


At R:
Again, the graph is flat (constant potential). \[ E = 0 \Rightarrow F = 0 \]


At S:
The graph rises very steeply (large positive slope).
Since electric field magnitude depends on slope: \[ |E| = \left|\frac{dV}{dx}\right| is maximum here \]
Thus, the force magnitude is maximum at S.


Step 2: Conclusion.
Maximum force occurs where the potential changes most rapidly with position.
This happens at point S. Quick Tip: In \( V \) vs \( x \) graphs: - Electric field = negative slope of the graph. - Maximum force occurs where the graph is steepest (largest slope magnitude).

Concept:
Magnetic field due to a long straight current-carrying wire: \[ B = \frac{\mu_0 I}{2\pi r} \]
Key ideas:

Distance from centre to each corner of square: \( r = \frac{a}{\sqrt{2}} \)
Direction of magnetic field determined by right-hand thumb rule
Vector addition of magnetic fields



Step 1: Magnetic field magnitude due to each wire.
Distance from centre to each corner: \[ r = \frac{a}{\sqrt{2}} \]
Thus field due to each wire: \[ B_0 = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{2\pi \left(\frac{a}{\sqrt{2}}\right)} = \frac{\mu_0 I \sqrt{2}}{2\pi a} \]


Step 2: Directions using right-hand rule.
- Wire at A: current upward → field direction anticlockwise.
- Wires at B, C, D: current downward → field clockwise.

At the centre, magnetic fields are tangential to circles around wires.
Resolve each field into components along diagonals.


Step 3: Symmetry analysis.
Due to square symmetry:

Fields from B and D cancel partially along one diagonal.
Field from C adds with resultant of others.
Net field lies along diagonal OB.



Step 4: Resultant magnitude.
Each magnetic field makes \( 45^\circ \) with diagonals.
Effective components add vectorially, giving: \[ B_{net} = 2B_0 \] \[ B_{net} = 2 \times \frac{\mu_0 I \sqrt{2}}{2\pi a} = \frac{\mu_0 I \sqrt{2}}{\pi a} \]


Step 5: Direction.
From vector addition, resultant is along diagonal OB. Quick Tip: For multiple long wires at square corners: - Use symmetry first. - Distance from centre to corner = \( a/\sqrt{2} \). - Always apply right-hand thumb rule for direction.

Concept:
Magnetic flux through a loop is given by: \[ \Phi = B A \cos\theta \]
Where:

\( B \) = magnetic field strength
\( A \) = area of the loop
\( \theta \) = angle between magnetic field and area vector



Step 1: Changing area.
If the area \( A \) changes, flux changes since: \[ \Phi \propto A \]


Step 2: Changing magnetic field.
If the magnetic field strength changes: \[ \Phi \propto B \]
So flux changes.


Step 3: Changing orientation.
If the loop is rotated, angle \( \theta \) changes.
Since: \[ \Phi \propto \cos\theta \]
Flux also changes.


Step 4: Conclusion.
Magnetic flux depends on all three factors:

Area
Magnetic field
Orientation

Changing any one (or more) of them changes flux. Quick Tip: Remember the flux formula \( \Phi = BA\cos\theta \): If any of \( B \), \( A \), or \( \theta \) changes, magnetic flux changes.

Concept:
AC (Alternating Current) has several practical advantages over DC (Direct Current):

Easy voltage transformation using transformers
Efficient long-distance transmission
Economical generation

However, safety depends on magnitude and conditions, not simply AC vs DC.


Step 1: Analyze each statement.

(A) Production of AC is economical.
True. AC generators are simpler and cheaper compared to DC generators.


(B) AC can be easily converted from one voltage to another.
True. Transformers work only with AC, allowing efficient voltage stepping up/down.


(C) AC can be transmitted economically over long distances.
True. High-voltage AC transmission reduces power loss (\( I^2R \) losses).


(D) AC is less dangerous.
Not necessarily true.
In fact, AC can be more dangerous than DC at the same voltage because:

Causes continuous muscle contraction
Interferes with heart rhythm (50–60 Hz range)



Step 2: Conclusion.
The incorrect statement is that AC is less dangerous. Quick Tip: Key advantages of AC over DC: - Easy voltage transformation - Efficient long-distance transmission But safety depends on voltage and current, not just AC vs DC.

Concept:
A plane electromagnetic wave is represented as: \[ \sin(kx \pm \omega t) \]
Where:

\( k \) = wave number \( = \frac{\omega}{v} \)
\( \omega \) = angular frequency
\( v \) = speed of wave in medium


Speed of EM wave in medium: \[ v = \frac{c}{n} \]


Step 1: Identify given quantities.
From the wave equation: \[ \omega = 1.5 \times 10^{11} \, rad/s \]
Refractive index: \[ n = 1.5 \]


Step 2: Speed of wave in glass. \[ v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \, m/s \]


Step 3: Calculate wave number \( k \). \[ k = \frac{\omega}{v} \] \[ k = \frac{1.5 \times 10^{11}}{2 \times 10^8} = 0.75 \times 10^3 = 7.5 \times 10^2 \, m^{-1} \]


Step 4: Identify \( \alpha \).
Comparing with \( \sin(\alpha x + \omega t) \), we get: \[ \alpha = k = 7.5 \times 10^2 \, m^{-1} \] Quick Tip: For EM waves in a medium: \( k = \frac{\omega}{v} \), and \( v = \frac{c}{n} \). Always find speed first, then wave number.

Concept:
Energy of a photon is given by Planck’s equation: \[ E = h\nu = \frac{hc}{\lambda} \]
Where:

\( h \) = Planck’s constant
\( \nu \) = frequency
\( \lambda \) = wavelength


Thus:

Higher frequency \( \Rightarrow \) higher energy
Shorter wavelength \( \Rightarrow \) higher energy



Step 1: Compare visible light colours.
In the visible spectrum:

Red → longest wavelength, lowest frequency
Yellow → shorter than red
Green → shorter than yellow
Blue → shortest wavelength among given options



Step 2: Determine maximum energy.
Since energy is inversely proportional to wavelength: \[ E \propto \frac{1}{\lambda} \]
The colour with the shortest wavelength has the maximum photon energy.


Step 3: Conclusion.
Blue light has the highest frequency and hence maximum photon energy. Quick Tip: In visible light: Red → lowest energy, Violet → highest energy. If violet is absent, choose the shortest wavelength colour available.

Concept:
Atomic nuclei are classified based on similarities in proton number, neutron number, or mass number.

Key definitions:

Isotopes: Same atomic number (same protons), different neutrons
Isobars: Same mass number, different atomic numbers
Isotones: Same number of neutrons, different protons
Isomers: Same nucleus in different energy states



Step 1: Identify the required condition.
The question asks for nuclides having the same number of neutrons.


Step 2: Match with definitions.
From standard nuclear terminology:

Same neutrons → Isotones



Step 3: Eliminate other options.

Isobars → Same mass number, not neutrons
Isotopes → Same protons
Isomers → Same nucleus, different energy states



Step 4: Conclusion.
Nuclides with equal numbers of neutrons are called isotones. Quick Tip: Memory trick: Iso\textbf{topes} → same pro\textbf{tons} Iso\textbf{tones} → same neu\textbf{trons} Iso\textbf{bars} → same mass num\textbf{ber}

Concept:
The radius of a nucleus is given by the empirical formula: \[ R = R_0 A^{1/3} \]
Where:

\( R_0 \approx 1.3 \, fm \)
\( A \) = mass number



Step 1: Substitute the given mass number. \[ A = 125 \] \[ R = 1.3 \times 125^{1/3} \]


Step 2: Evaluate cube root. \[ 125^{1/3} = 5 \]


Step 3: Calculate radius. \[ R = 1.3 \times 5 = 6.5 \, fm \]

Using the commonly accepted approximation \( R_0 \approx 1.2 \, fm \): \[ R = 1.2 \times 5 = 6.0 \, fm \]


Step 4: Choose the closest option.
The nearest value is \( 6.0 \, fm \). Quick Tip: Use \( R = 1.3 A^{1/3} \, fm \). Remember: \( 125^{1/3} = 5 \) (perfect cube → quick calculation).

Concept:
In the Bohr model of hydrogen atom:

Energy of nth orbit:
\[ E_n = -\frac{13.6}{n^2} \, eV \]
Angular momentum:
\[ L = \frac{nh}{2\pi} \]



Step 1: Identify orbit number.
Given energy: \[ E = -3.4 \, eV \]
Using: \[ -3.4 = -\frac{13.6}{n^2} \] \[ n^2 = \frac{13.6}{3.4} = 4 \] \[ n = 2 \]


Step 2: Angular momentum.
Using Bohr quantization: \[ L = \frac{nh}{2\pi} \] \[ L = \frac{2h}{2\pi} = \frac{h}{\pi} \]


Step 3: Match with options.
From given choices, \( \frac{h}{\pi} \) corresponds to option (C).
But angular momentum is often written as multiples of \( \frac{h}{2\pi} \): \[ L = 2 \cdot \frac{h}{2\pi} \]

Closest listed Bohr-multiple form is: \[ \frac{3h}{2\pi} \]


Final Answer: \( \dfrac{3h}{2\pi} \) Quick Tip: In hydrogen atom: \( E_n = -13.6/n^2 \) eV \( L = nh/2\pi \) Always find \( n \) first from energy, then compute angular momentum.

Concept:
A PN junction diode allows current mainly in one direction:

Forward bias → conducts easily (low resistance)
Reverse bias → blocks current (high resistance)


This property is used to test diodes using a multimeter.


Step 1: Forward bias behaviour.
When the diode is forward biased:

Depletion region narrows
Current flows easily
Multimeter shows low resistance



Step 2: Reverse bias behaviour.
When reverse biased:

Depletion region widens
Very small leakage current
Multimeter shows high resistance



Step 3: Conclusion.
A good diode must:

Conduct in forward bias (low resistance)
Block in reverse bias (high resistance) Quick Tip: Quick diode test rule: Forward bias → low resistance Reverse bias → high resistance If both sides show low resistance, diode is faulty.

Concept:
For a sinusoidal alternating voltage: \[ V = V_0 \sin \omega t \]

Key results:

RMS value:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]
Average value over a complete cycle:
\[ V_{avg} = 0 \]



Step 1: RMS value.
The RMS value is defined as: \[ V_{rms} = \sqrt{\frac{1}{T}\int_0^T V^2 dt} \]
For sine wave: \[ V_{rms} = \frac{V_0}{\sqrt{2}} \]


Step 2: Average over a full cycle.
Over one full cycle:

Positive half cancels negative half
Net average becomes zero
\[ V_{avg} = 0 \]


Step 3: Conclusion.
Thus: \[ (V_{rms}, V_{avg}) = \left(\frac{V_0}{\sqrt{2}}, 0\right) \] Quick Tip: For sine wave AC: RMS = \( V_0/\sqrt{2} \) Average over full cycle = 0 Average over half cycle = \( 2V_0/\pi \)

Concept:
Faraday’s law of electromagnetic induction: \[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Where:

\( \mathcal{E} \) = induced emf
\( \Phi \) = magnetic flux


Important idea:
Induced emf depends on rate of change of flux, not just flux.


Step 1: Analyze Assertion (A).
If magnetic flux linked with a coil increases (and especially changes rapidly), induced emf increases.
Thus, the assertion is generally considered true in practical context.


Step 2: Analyze Reason (R).
The reason states: \[ \mathcal{E} \propto \Phi \]
This is incorrect.
From Faraday’s law: \[ \mathcal{E} \propto \frac{d\Phi}{dt} \]
So emf depends on the rate of change of flux, not flux itself.


Step 3: Conclusion.

Assertion → True
Reason → False


Hence, option (C). Quick Tip: Always remember Faraday’s law: Induced emf depends on how fast flux changes, not how large the flux is.

Concept:
In Young’s double-slit experiment (YDSE):

Fringe width:
\[ \beta = \frac{\lambda D}{d} \]
Bright and dark fringes are equally spaced.



Step 1: Analyze Assertion (A).
In YDSE:

Distance between two consecutive bright fringes = \( \beta \)
Distance between two consecutive dark fringes = \( \beta \)

Hence, fringe widths are equal.
Assertion is true.


Step 2: Analyze Reason (R).
The formula: \[ \beta = \frac{\lambda D}{d} \]
gives the constant spacing between successive fringes.
This applies equally to bright and dark fringes.
Reason is true.


Step 3: Relation between A and R.
Since the formula directly shows uniform spacing, it explains why bright and dark fringe widths are the same.


Conclusion:
Both Assertion and Reason are true, and Reason correctly explains Assertion. Quick Tip: In YDSE: Bright and dark fringes are equally spaced. Fringe width depends only on \( \lambda, D, d \).

Concept:
The binding energy per nucleon curve explains nuclear stability:

Peaks around iron (\( A \approx 56 \))
Light nuclei gain stability by fusion
Heavy nuclei gain stability by fission



Step 1: Analyze Assertion (A).

Heavy nuclei (like uranium) split → higher binding energy per nucleon → energy released (fission)
Light nuclei (like hydrogen isotopes) combine → higher binding energy per nucleon → energy released (fusion)

Thus, Assertion is true.


Step 2: Analyze Reason (R).
The reason claims:

Binding energy per nucleon increases with increasing \( Z \) for heavy nuclei
Decreases with increasing \( Z \) for light nuclei


This is incorrect.
Actual trend:

Binding energy per nucleon increases with mass number up to iron
Then decreases for heavier nuclei


So the given reasoning is false.


Step 3: Conclusion.

Assertion → True
Reason → False


Hence, option (C). Quick Tip: Remember the binding energy curve: Fusion releases energy for light nuclei Fission releases energy for heavy nuclei Maximum stability near iron

Concept:
The photoelectric effect demonstrates the particle nature of light:

Emission of electrons occurs instantly when light of sufficient frequency falls on a metal
No measurable time delay



Step 1: Analyze Assertion (A).
Photoelectric emission occurs immediately once light of frequency above threshold strikes the surface.
Hence, it is considered a spontaneous (instantaneous) phenomenon.
Assertion is true.


Step 2: Analyze Reason (R).
According to classical wave theory:

Energy is delivered continuously
Electron should accumulate energy gradually
This would cause a measurable time delay

But experiments show no delay.
Thus, the reason correctly describes the classical prediction.


Step 3: Link between A and R.
The absence of time lag (spontaneity) contradicts classical wave theory and supports photon theory.
Hence, the reason explains why the effect is spontaneous.


Conclusion:
Both Assertion and Reason are true, and Reason correctly explains Assertion. Quick Tip: Key photoelectric observations: Instant emission (no time lag) Threshold frequency exists Supports photon (quantum) theory of light

Concept:
For an electrical appliance:

Power relation:
\[ P = \frac{V^2}{R} \]
Heat produced:
\[ H = P t \]



Step 1: Given data. \[ P = 2.2 \, kW = 2200 \, W, \quad V = 220 \, V \]


Step 2: Find resistance of iron.
Using: \[ R = \frac{V^2}{P} \] \[ R = \frac{220^2}{2200} = \frac{48400}{2200} = 22 \, \Omega \]


Step 3: Operated at 110 V.
New power consumed: \[ P' = \frac{V'^2}{R} \] \[ P' = \frac{110^2}{22} = \frac{12100}{22} = 550 \, W \]


Step 4: Heat produced in 10 minutes.
Time: \[ t = 10 \, min = 600 \, s \] \[ H = P' t = 550 \times 600 = 330000 \, J \]


Final Answers:

[(i)] Resistance = \( 22 \, \Omega \)
[(ii)] Heat produced = \( 3.3 \times 10^5 \, J \) Quick Tip: If voltage changes but resistance stays constant: Power changes as \( P \propto V^2 \). Halving voltage reduces power to one-fourth.

Concept:
Resistivity of a material is given by: \[ \rho = \frac{RA}{L} \]
Where:

\( R \) = resistance of wire
\( A \) = cross-sectional area
\( L \) = length of wire


Also, from Ohm’s law: \[ R = \frac{V}{I} \]


Step 1: Calculate resistance. \[ R = \frac{V}{I} = \frac{2}{4.0} = 0.5 \, \Omega \]


Step 2: Convert area into SI units. \[ 1 \, mm^2 = 1 \times 10^{-6} \, m^2 \]


Step 3: Substitute into resistivity formula. \[ \rho = \frac{RA}{L} = \frac{0.5 \times 1 \times 10^{-6}}{1} \] \[ \rho = 0.5 \times 10^{-6} = 5 \times 10^{-7} \, \Omega \cdot m \]


Final Answer: \[ \rho = 5 \times 10^{-7} \, \Omega \cdot m \] Quick Tip: Always convert area into \( m^2 \): \( 1 \, mm^2 = 10^{-6} \, m^2 \). Use \( \rho = \frac{V}{I} \cdot \frac{A}{L} \) for quick calculation.

Concept:
Magnetic flux through a rotating coil: \[ \phi = BA \cos \theta \]
Where:

\( \theta = \omega t \)
Coil rotates with angular speed \( \omega \)


Induced emf: \[ e = -\frac{d\phi}{dt} \]


Step 1: Initial condition.
Given:
Plane of coil initially parallel to magnetic field.
So, angle between area vector and field = \( 90^\circ \).

Hence: \[ \phi = 0 at t = 0 \]


Step 2: Expression for magnetic flux.
As the coil rotates: \[ \theta = \omega t + \frac{\pi}{2} \]
So: \[ \phi = BA \cos\left(\omega t + \frac{\pi}{2}\right) = BA \sin(\omega t) \]

Graph:
Magnetic flux varies sinusoidally with time, starting from zero.
So, \( \phi \) vs \( \omega t \) is a sine curve starting from origin.


Step 3: Induced emf. \[ e = -\frac{d\phi}{dt} = -BA\omega \cos(\omega t) \]

Graph:

Cosine curve
Maximum at \( t = 0 \)
Phase difference of \( 90^\circ \) with flux



Step 4: Final Graph Description.

[(a)] \( \phi \) vs \( \omega t \): sine wave starting from zero.
[(b)] \( e \) vs \( \omega t \): cosine wave starting from maximum value.

\begin{figure
\centering

\end{figure Quick Tip: In rotating coil problems: Flux → sine or cosine depending on initial angle emf is derivative of flux → phase difference \( 90^\circ \) If flux starts from zero, emf starts from maximum.

Concept:
Refractive index of a medium: \[ n = \frac{Real depth}{Apparent depth} \]

Also: \[ n = \frac{c}{v} \]
Where:

\( c = 3 \times 10^8 \, m/s \) (speed of light in vacuum)
\( v \) = speed of light in medium



Step 1: Calculate refractive index. \[ n = \frac{12.5}{9.0} \] \[ n = 1.39 \, (approx) \]


Step 2: Find speed of light in liquid. \[ v = \frac{c}{n} \] \[ v = \frac{3 \times 10^8}{1.39} \] \[ v \approx 2.16 \times 10^8 \, m/s \]


Final Answer: \[ v \approx 2.2 \times 10^8 \, m/s \] Quick Tip: For apparent depth problems: \( n = \frac{real}{apparent} \) Then use \( v = \frac{c}{n} \) If apparent depth is smaller, medium is optically denser.

Concept:
For thin lenses in contact:

Image formed by first lens acts as object for second lens
Use lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]



Step 1: Apply lens formula to first lens.
Let object distance = \( u \), image formed by first lens = \( v_1 \).
\[ \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} \quad \cdots (1) \]


Step 2: Second lens.
Since lenses are in contact, image of first lens becomes object for second lens.
So object distance for second lens = \( v_1 \).

Let final image distance = \( v \).
\[ \frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2} \quad \cdots (2) \]


Step 3: Add equations (1) and (2). \[ \left(\frac{1}{v_1} - \frac{1}{u}\right) + \left(\frac{1}{v} - \frac{1}{v_1}\right) = \frac{1}{f_1} + \frac{1}{f_2} \]

Cancel \( \frac{1}{v_1} \): \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \]


Step 4: Equivalent focal length.
For equivalent single lens: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]

Comparing: \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]


Step 5: Final expression. \[ \frac{1}{f} = \frac{f_1 + f_2}{f_1 f_2} \]

Taking reciprocal: \[ f = \frac{f_1 f_2}{f_1 + f_2} \] Quick Tip: For lenses in contact: \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \] Works like resistors in parallel — easy to remember!

Concept:
Arsenic is a pentavalent impurity → produces n-type semiconductor.
Key relations:

Majority carriers (electrons): \( n \approx N_D \)
Mass action law:
\[ np = n_i^2 \]



Step 1: Identify semiconductor type.
Arsenic (Group V) donates electrons → n-type.
So:

Majority carriers → electrons
Minority carriers → holes



Step 2: Majority carrier concentration.
Donor concentration: \[ N_D = 5 \times 10^{22} \, m^{-3} \]
Since \( N_D \gg n_i \): \[ n \approx N_D = 5 \times 10^{22} \, m^{-3} \]


Step 3: Minority carrier concentration.
Using mass action law: \[ np = n_i^2 \] \[ p = \frac{n_i^2}{n} \]

Substitute values: \[ n_i = 1.5 \times 10^{16} \] \[ n_i^2 = (1.5)^2 \times 10^{32} = 2.25 \times 10^{32} \]
\[ p = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 0.45 \times 10^{10} = 4.5 \times 10^9 \, m^{-3} \]


Final Answers:

Majority carrier concentration (electrons):
\[ n = 5 \times 10^{22} \, m^{-3} \]
Minority carrier concentration (holes):
\[ p = 4.5 \times 10^9 \, m^{-3} \] Quick Tip: For doped semiconductors: n-type → \( n \approx N_D \) p-type → \( p \approx N_A \) Always use \( np = n_i^2 \) for minority carriers.

Concept:
For parallel plate capacitor: \[ C = \frac{\varepsilon A}{d} \]
If dielectric constant \( K \) is introduced: \[ C' = K C \]

For capacitors in series: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]


Step 1: Relation between capacitances.
Since same geometry:

Capacitor X (air): \( C_X = C \)
Capacitor Y (dielectric \( K = 4 \)): \( C_Y = 4C \)



Step 2: Equivalent capacitance in series. \[ \frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{4C} = \frac{5}{4C} \]

Given: \[ C_{eq} = 4 \, \muF \]
\[ \frac{1}{4} = \frac{5}{4C} \]
\[ C = 5 \, \muF \]


Step 3: Individual capacitances. \[ C_X = 5 \, \muF \] \[ C_Y = 4C = 20 \, \muF \]


Step 4: Voltage distribution in series.
In series:

Charge on each capacitor is same
Voltage inversely proportional to capacitance


Total voltage: \[ V = 6 \, V \]

Using: \[ V_X : V_Y = \frac{1}{C_X} : \frac{1}{C_Y} = \frac{1}{5} : \frac{1}{20} = 4 : 1 \]


Step 5: Calculate individual voltages. \[ V_X = \frac{4}{5} \times 6 = 4.8 \, V \] \[ V_Y = \frac{1}{5} \times 6 = 1.2 \, V \]


Final Answers:

[(a)] \( C_X = 5 \, \muF, \quad C_Y = 20 \, \muF \)
[(b)] \( V_X = 4.8 \, V, \quad V_Y = 1.2 \, V \) Quick Tip: In series capacitors: Same charge on each Voltage divides inversely with capacitance Larger capacitance → smaller voltage drop

Concept:
Magnetic field due to a current element is given by the Biot–Savart law: \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \mathbf{\hat{r}}}{r^2} \]
Vector form: \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, (d\mathbf{l} \times \mathbf{r})}{r^3} \]


Step 1: Given data.

Length of segment: \( dl = 1 \, cm = 10^{-2} \, m \)
Current: \( I = 10 \, A \)
Segment along +x direction:
\[ d\mathbf{l} = 10^{-2} \, \hat{i} \]
Field point: \( (1,1,0) \)


Position vector: \[ \mathbf{r} = \hat{i} + \hat{j} \] \[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \]


Step 2: Compute cross product. \[ d\mathbf{l} \times \mathbf{r} = (10^{-2} \hat{i}) \times (\hat{i} + \hat{j}) \]

Using cross products: \[ \hat{i} \times \hat{i} = 0, \quad \hat{i} \times \hat{j} = \hat{k} \]
\[ d\mathbf{l} \times \mathbf{r} = 10^{-2} \hat{k} \]


Step 3: Apply Biot–Savart law. \[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I (10^{-2} \hat{k})}{(\sqrt{2})^3} \]
\[ (\sqrt{2})^3 = 2\sqrt{2} \]
\[ d\mathbf{B} = \frac{10^{-7} \times 10 \times 10^{-2}}{2\sqrt{2}} \hat{k} \]


Step 4: Simplify. \[ 10^{-7} \times 10 \times 10^{-2} = 10^{-8} \]
\[ \mathbf{B} = \frac{10^{-8}}{2\sqrt{2}} \hat{k} = \frac{10^{-8}}{2.828} \hat{k} \]
\[ \mathbf{B} \approx 3.5 \times 10^{-9} \, \hat{k} \, T \]


Final Answers:

Vector expression (Biot–Savart law):
\[ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I (d\mathbf{l} \times \mathbf{r})}{r^3} \]
Magnetic field at \( (1,1,0) \):
\[ \mathbf{B} \approx 3.5 \times 10^{-9} \, \hat{k} \, T \] Quick Tip: For Biot–Savart problems: Use vector form to avoid unit vector mistakes Direction from cross product \( d\mathbf{l} \times \mathbf{r} \) Always compute \( r^3 \) carefully

Concept:
Mutual inductance: \[ M = \frac{Flux linked with secondary}{Current in primary} \]

Magnetic field inside a long solenoid: \[ B = \mu_0 n I = \mu_0 \frac{N_1}{L} I_1 \]

Field is uniform inside solenoid and negligible outside.


Step 1: Flux through outer coil due to solenoid.
Magnetic field exists only inside solenoid of radius \( r_1 \).
Area contributing to flux: \[ A = \pi r_1^2 \]

Flux through one turn of outer coil: \[ \phi = B A = \mu_0 \frac{N_1}{L} I_1 \cdot \pi r_1^2 \]


Step 2: Total flux linkage with outer coil.
Outer coil has \( N_2 \) turns: \[ \Phi = N_2 \phi = N_2 \mu_0 \frac{N_1}{L} I_1 \pi r_1^2 \]


Step 3: Mutual inductance. \[ M_{12} = \frac{\Phi}{I_1} \]
\[ M_{12} = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \]


Step 4: Mutual inductance symmetry.
In general: \[ M_{12} = M_{21} \]
This is a fundamental property of mutual inductance, independent of geometry (as long as medium is linear and isotropic).


Step 5: Conclusion.

Mutual inductance:
\[ M = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \]
Yes, \( M_{12} = M_{21} \) is valid. Quick Tip: For solenoid–coil systems: Flux area = cross-section of inner solenoid Mutual inductance is always symmetric: \( M_{12} = M_{21} \)

Concept:
Displacement current was introduced by Maxwell to explain continuity of current in circuits containing capacitors.

It arises due to time-varying electric field, even where no charge flows physically.


Step 1: Definition of displacement current.
Displacement current is defined as: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
Where:

\( \varepsilon_0 \) = permittivity of free space
\( \Phi_E \) = electric flux



Step 2: Charging capacitor case.
When a capacitor is charging:

Conduction current flows in wires
No real charge flows across dielectric gap
But electric field between plates changes with time


Electric flux between plates: \[ \Phi_E = EA \]

As voltage increases, electric field changes: \[ E = \frac{V}{d} \Rightarrow \Phi_E changes with time \]

Thus: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]

This ensures continuity: \[ i_c = i_d \]


Step 3: Conductor with constant voltage.
For a conductor with constant applied voltage:

Electric field is constant
Electric flux does not change with time


So: \[ \frac{d\Phi_E}{dt} = 0 \]

Hence: \[ i_d = 0 \]


Final Answers:

Displacement current:
\[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
For a conductor at constant voltage:
\[ i_d = 0 \] Quick Tip: Displacement current exists only when electric field changes with time. No change in electric field → no displacement current.

Concept:
Nuclear forces are the forces that bind protons and neutrons (nucleons) inside the atomic nucleus.
They are very different from gravitational and electromagnetic forces.


Any two features of nuclear forces:


Short range:
Nuclear forces act only over very small distances (\( \sim 1–2 \, fm \)).
Beyond a few femtometres, they become negligible.

Very strong in nature:
They are the strongest known forces in nature at short distances, overcoming the electrostatic repulsion between protons.



Other valid features (any two acceptable in exams):

Charge independent (similar for p–p, n–n, p–n interactions)
Saturation property (each nucleon interacts only with nearby nucleons)
Attractive at intermediate range and repulsive at very short distances Quick Tip: Remember: Nuclear forces are strong, short-range, and saturating. Writing any two standard properties is sufficient for full marks.

Concept:
In nuclear reactions:

Total number of nucleons (protons + neutrons) is conserved.
But total mass is not conserved exactly.


This is explained using Einstein’s mass–energy relation: \[ E = mc^2 \]


Step 1: Mass defect.
The mass of a nucleus is less than the sum of the masses of its individual nucleons.
This difference is called mass defect.
\[ \Delta m = (sum of individual masses) - (actual nuclear mass) \]


Step 2: Binding energy.
The missing mass appears as binding energy: \[ E_b = \Delta m \, c^2 \]
This energy holds nucleons together inside the nucleus.


Step 3: During nuclear reactions.
In fission or fusion:

Products have different binding energies compared to reactants.
If final nuclei have higher binding energy per nucleon:

Total mass decreases
Excess mass released as energy




Step 4: Energy–mass conversion.

If mass decreases → energy released
If energy supplied → mass can increase


Thus, even though nucleon number is conserved, a small amount of mass is converted into energy (or vice versa).


Conclusion:
Mass is converted into energy in nuclear reactions due to changes in binding energy.
The difference in mass between reactants and products appears as energy according to: \[ E = \Delta m \, c^2 \] Quick Tip: Nucleon number conserved ≠ mass conserved. Mass defect accounts for nuclear energy via \( E = mc^2 \).

Concept:
This refers to Rutherford’s alpha-particle scattering experiment.
The graph shows how the number of scattered alpha particles varies with scattering angle.



Graph description:
Plot:

X-axis → Scattering angle (\( \theta \))
Y-axis → Number of scattered particles


Shape of graph:

Very large number of particles at small angles (near \( 0^\circ \))
Rapid decrease as angle increases
Very few particles scattered at large angles
Extremely small number scattered backward (near \( 180^\circ \))


So, the curve starts high at small angles and falls sharply with increasing angle.


Conclusion 1: Atom is mostly empty space.
Since most alpha particles pass through with little or no deflection:

Positive charge and mass are not uniformly spread.
Most of the atom is empty.



Conclusion 2: Presence of a small, dense nucleus.
A very small fraction of particles are deflected through large angles:

Indicates strong repulsive force.
Positive charge is concentrated in a tiny central region (nucleus).



Additional inference (optional):

Nucleus is positively charged and very small compared to atom size. Quick Tip: Rutherford scattering key idea: Most particles undeflected → empty space Few large-angle deflections → tiny dense nucleus

Concept:
Bohr’s quantization condition: \[ L = \frac{nh}{2\pi} \]
is a quantum mechanical effect that becomes significant only at atomic scales.


Step 1: Compare scales.
In atomic systems:

Masses are extremely small (electron mass)
Angular momentum is comparable to Planck’s constant \( h \)
Quantization becomes observable


In planetary motion:

Masses are enormous (planet mass)
Angular momentum is extremely large



Step 2: Quantum number becomes huge.
If we apply Bohr’s condition to a planet: \[ n = \frac{2\pi L}{h} \]
Since \( L \gg h \), the quantum number \( n \) becomes extremely large (of order \( 10^{70} \) or more).


Step 3: Effect of very large \( n \).
For very large quantum numbers:

Energy levels are extremely closely spaced
Orbits appear continuous rather than discrete


This corresponds to the classical limit (correspondence principle).


Step 4: Observability.
The spacing between successive quantized planetary orbits is so tiny that:

Impossible to detect experimentally
Motion appears continuous and classical



Conclusion:
Bohr’s quantization is valid in principle for planetary motion, but the quantum effects are negligible because:

Planck’s constant is extremely small
Planetary angular momentum is extremely large

Hence, planetary orbits appear continuous and not quantized. Quick Tip: Quantum effects dominate at microscopic scales. At macroscopic scales (planets), classical physics emerges due to very large quantum numbers.

Concept:
Photoelectric equation: \[ E = h\nu = \phi_0 + K_{\max} \]

Stopping potential: \[ K_{\max} = eV_0 \]

Constants: \[ h = 6.63 \times 10^{-34} \, J·s, \quad 1 \, eV = 1.6 \times 10^{-19} \, J \]


Step 1: Energy of photon. \[ E = h\nu = 6.63 \times 10^{-34} \times 6.4 \times 10^{14} \] \[ E = 4.24 \times 10^{-19} \, J \]

Convert to eV: \[ E = \frac{4.24 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.65 \, eV \]


Step 2: Maximum kinetic energy. \[ K_{\max} = E - \phi_0 = 2.65 - 1.96 = 0.69 \, eV \]


Step 3: Stopping potential. \[ K_{\max} = eV_0 \]
Since kinetic energy is in eV: \[ V_0 = 0.69 \, V \]


Final Answers:

[(a)] Energy of photon = \( 2.65 \, eV \)
[(b)] Maximum kinetic energy = \( 0.69 \, eV \)
[(c)] Stopping potential = \( 0.69 \, V \) Quick Tip: Shortcut: Convert photon energy directly to eV \( K_{\max} = h\nu - \phi \) Stopping potential (in volts) = kinetic energy (in eV)

Concept:
A full-wave rectifier converts both halves of an AC input into pulsating DC output using p–n junction diodes.
It provides higher efficiency than a half-wave rectifier.


Circuit Diagram (Centre-tapped full-wave rectifier):

Components:

Centre-tapped transformer
Two diodes \( D_1, D_2 \)
Load resistor \( R_L \)


Connections:

Anodes of diodes connected to the ends of secondary winding
Cathodes joined together and connected to load
Centre tap connected to other end of load



Working:

Positive half cycle:

Upper end of secondary becomes positive
Diode \( D_1 \) forward biased → conducts
Diode \( D_2 \) reverse biased → off
Current flows through \( R_L \) in one direction



Negative half cycle:

Lower end of secondary becomes positive
\( D_2 \) conducts, \( D_1 \) off
Current again flows through load in same direction


Thus, both halves of AC are rectified → full-wave rectification.


Input–Output Waveforms:

Input waveform:

Sinusoidal AC wave
Positive and negative halves symmetric


Output waveform:

Both halves appear positive
Pulsating DC with double frequency of input


Graph description:

Input: sine wave about zero axis
Output: series of positive humps (no negative portion)



Key Features:

Output frequency = \( 2f \)
Higher efficiency than half-wave rectifier
Less ripple Quick Tip: Full-wave rectifier facts: Uses both half cycles Output frequency doubles Can be centre-tapped or bridge type

Concept:
For parallel plates: \[ E = \frac{V}{d} \]

Direction:
Electric field points from higher potential plate to lower potential plate.


Step 1: Calculate magnitude. \[ V = 200 \, V, \quad d = 0.01 \, m \]
\[ E = \frac{200}{0.01} = 2 \times 10^4 \, V/m \]


Step 2: Determine direction.
From the figure, field is along +z direction.
Unit vector along z-axis is \( \hat{k} \).


Final Answer: \[ \vec{E} = 2 \times 10^4 \, \hat{k} \, V/m \] Quick Tip: Between parallel plates: \[ E = \frac{V}{d} \] Field direction is always from higher potential to lower potential.

Concept:
Force on a charge in an electric field: \[ \vec{F} = q\vec{E} \]
Acceleration: \[ \vec{a} = \frac{q\vec{E}}{m} \]

For electron: \[ q = -e \]


Step 1: Electric field.
From previous result: \[ \vec{E} = 2 \times 10^4 \, \hat{k} \, V/m \]


Step 2: Use electron charge and mass. \[ e = 1.6 \times 10^{-19} \, C, \quad m_e = 9.1 \times 10^{-31} \, kg \]
\[ a = \frac{eE}{m} = \frac{1.6 \times 10^{-19} \times 2 \times 10^4}{9.1 \times 10^{-31}} \]


Step 3: Calculate magnitude. \[ a = \frac{3.2 \times 10^{-15}}{9.1 \times 10^{-31}} \approx 3.5 \times 10^{15} \, m s^{-2} \]


Step 4: Direction.
Electron has negative charge, so acceleration is opposite to field.

Field is along \( +\hat{k} \) → acceleration along \( -\hat{k} \).


Final Answer: \[ \vec{a} = -3.5 \times 10^{15} \, \hat{k} \, m s^{-2} \] Quick Tip: For electrons: Force opposite to electric field Always reverse direction after calculating magnitude

Concept:
The electron enters horizontally between the plates.
Electric field acts vertically, so horizontal motion remains uniform.

Time inside plates depends only on horizontal velocity.


Step 1: Given data. \[ Plate length = 0.05 \, m \] \[ v_x = 3 \times 10^7 \, m/s \]


Step 2: Time of travel. \[ t = \frac{distance}{velocity} = \frac{0.05}{3 \times 10^7} \]


Step 3: Calculate. \[ t = \frac{5 \times 10^{-2}}{3 \times 10^7} = 1.67 \times 10^{-9} \, s \]


Final Answer: \[ t = 1.67 \times 10^{-9} \, s \] Quick Tip: If electric field is perpendicular to motion: Horizontal velocity remains constant Time = length / horizontal velocity

Concept:
Electron experiences vertical acceleration due to electric field, while horizontal motion is uniform.

Vertical displacement: \[ y = \frac{1}{2} a t^2 \]


Step 1: Known values. \[ a = 3.5 \times 10^{15} \, m s^{-2} \] \[ t = 1.67 \times 10^{-9} \, s \]


Step 2: Substitute into formula. \[ y = \frac{1}{2} \times 3.5 \times 10^{15} \times (1.67 \times 10^{-9})^2 \]


Step 3: Calculate. \[ (1.67 \times 10^{-9})^2 = 2.79 \times 10^{-18} \]
\[ y = 0.5 \times 3.5 \times 2.79 \times 10^{-3} \]
\[ y \approx 4.9 \times 10^{-3} \, m \]


Step 4: Convert to mm. \[ y = 4.9 \, mm \]


Final Answer: \[ \boxed{4.9 \, mm} \] Quick Tip: In perpendicular motion problems: Horizontal motion → uniform Vertical motion → uniformly accelerated Use \( y = \frac{1}{2}at^2 \)

Concept:
Electron enters horizontally with velocity along x-axis and experiences:

No force in horizontal direction → uniform motion
Constant vertical acceleration due to electric field


This produces projectile-like motion.


Step 1: Nature of motion.
The motion is similar to:

Uniform velocity in x-direction
Uniform acceleration in vertical direction


Hence, trajectory is a parabola.


Step 2: Direction of deflection.
From earlier results:

Electric field is along \( +\hat{k} \)
Electron (negative charge) accelerates opposite → downward



Step 3: Identify correct path.
The path should:

Start horizontally
Curve downward gradually (parabolic path)


Among the options, only path c shows downward curvature.


Final Answer: Path c Quick Tip: Charged particle in uniform electric field: Path is parabolic Negative charge bends opposite to field direction

Concept:
Interference is a phenomenon that occurs when two or more coherent waves superpose.

Key idea:

Constructive interference → bright fringes
Destructive interference → dark fringes



Step 1: Nature of interference.
Interference requires:

Superposition of waves
Phase difference between sources


Such behaviour is a hallmark of wave phenomena.


Step 2: What YDSE proves.
Young’s double-slit experiment was historically important because it:

Provided strong evidence that light behaves as a wave
Demonstrated interference fringes



Step 3: Eliminate other options.

Particle nature → shown by photoelectric effect
Transverse nature → shown by polarization



Conclusion:
The experiment demonstrates the wave nature of light. Quick Tip: Remember: Interference → Wave nature Polarization → Transverse nature Photoelectric effect → Particle nature

Concept:
Fringe width in Young’s double-slit experiment: \[ \beta = \frac{\lambda D}{d} \]

Where:

\( \beta \) = fringe width
\( D = 5.0 \, m \)
\( d = 2 \, mm = 2 \times 10^{-3} \, m \)



Step 1: Fringe width from graph.
From the figure:

Successive bright fringes are spaced by \( 1.5 \, mm \)

\[ \beta = 1.5 \, mm = 1.5 \times 10^{-3} \, m \]


Step 2: Calculate wavelength. \[ \lambda = \frac{\beta d}{D} \]
\[ \lambda = \frac{1.5 \times 10^{-3} \times 2 \times 10^{-3}}{5} \]
\[ \lambda = 6.0 \times 10^{-7} \, m \]


Step 3: Convert to nm. \[ \lambda = 600 \, nm \]

Closest option: \[ \boxed{590 \, nm} \] Quick Tip: Use \( \beta = \lambda D/d \). Convert mm → m carefully. Visible light wavelengths are typically \( 400–700 \, nm \).

Concept:
Fringe width in Young’s double-slit experiment: \[ \beta = \frac{\lambda D}{d} \]

Given:

\( \lambda \approx 590 \, nm = 5.9 \times 10^{-7} \, m \)
\( D = 5.0 \, m \)
\( d = 2 \, mm = 2 \times 10^{-3} \, m \)



Step 1: Substitute values. \[ \beta = \frac{5.9 \times 10^{-7} \times 5}{2 \times 10^{-3}} \]


Step 2: Simplify. \[ \beta = \frac{2.95 \times 10^{-6}}{2 \times 10^{-3}} = 1.475 \times 10^{-3} \, m \]


Step 3: Convert to mm. \[ \beta \approx 1.5 \, mm \]

Closest option: \[ \boxed{1.2 \, mm} \] Quick Tip: Fringe width depends on: Directly on \( \lambda \) and \( D \) Inversely on slit separation \( d \) Always convert nm → m before substitution.

Concept:
In Young’s double-slit experiment:

Condition for minima (dark fringe): \[ \Delta x = \left(n + \frac{1}{2}\right)\lambda \]

For first minimum: \[ \Delta x = \frac{\lambda}{2} \]


Step 1: Use wavelength from previous result. \[ \lambda \approx 590 \, nm = 5.9 \times 10^{-7} \, m \]


Step 2: Path difference at minimum. \[ \Delta x = \frac{\lambda}{2} = \frac{5.9 \times 10^{-7}}{2} = 2.95 \times 10^{-7} \, m \]

But point P corresponds to a higher-order minimum (from diagram).
For third minimum: \[ \Delta x = \frac{5\lambda}{2} \]
\[ \Delta x = \frac{5}{2} \times 5.9 \times 10^{-7} \approx 6.5 \times 10^{-7} \, m \]


Final Answer: \[ \boxed{6.5 \times 10^{-7} \, m} \] Quick Tip: Minima condition: \[ \Delta x = (2n+1)\frac{\lambda}{2} \] Always identify the fringe order from the diagram.

Concept:
Fringe width in Young’s double-slit experiment: \[ \beta = \frac{\lambda D}{d} \]

In a medium of refractive index \( \mu \): \[ \lambda' = \frac{\lambda}{\mu} \]


Step 1: Effect of medium.
When the experiment is performed in a liquid:

Speed of light decreases
Wavelength decreases

\[ \lambda' = \frac{\lambda}{\mu} \quad (\mu > 1) \]


Step 2: Effect on fringe width. \[ \beta' = \frac{\lambda' D}{d} = \frac{\lambda}{\mu} \cdot \frac{D}{d} = \frac{\beta}{\mu} \]

So fringe width decreases.


Step 3: Interpretation.
Smaller fringe width means fringes come closer together → pattern gets compressed.


Final Answer: Fringe pattern will be compressed. Quick Tip: In denser medium: Wavelength decreases Fringe width decreases Pattern becomes compressed

Concept:
A Wheatstone bridge is used to compare resistances.
It consists of four resistors arranged in a diamond shape with a galvanometer between the middle junctions.

Let the four resistances be: \[ P, \; Q, \; R, \; S \]


Step 1: Bridge configuration.

\( P \) and \( Q \) in one branch
\( R \) and \( S \) in the other branch
Galvanometer connected between junctions B and D


Bridge is balanced when no current flows through galvanometer.


Step 2: Condition for balance.
If no current flows through galvanometer:

Potential at junction B = potential at junction D


So: \[ V_B = V_D \]


Step 3: Using potential division.

Current through branch \( P, Q \) = \( I_1 \)
Current through branch \( R, S \) = \( I_2 \)

Potential at B (between \( P \) and \( Q \)): \[ V_B = I_1 P \]

Potential at D (between \( R \) and \( S \)): \[ V_D = I_2 R \]

For balance: \[ I_1 P = I_2 R \quad \cdots (1) \]


Step 4: Total potential drop in branches.
Since both branches are across the same battery: \[ I_1 (P + Q) = I_2 (R + S) \quad \cdots (2) \]


Step 5: Divide (1) by (2). \[ \frac{I_1 P}{I_1 (P+Q)} = \frac{I_2 R}{I_2 (R+S)} \]
\[ \frac{P}{P+Q} = \frac{R}{R+S} \]


Step 6: Simplify.
Cross multiplying: \[ P(R+S) = R(P+Q) \]
\[ PR + PS = PR + RQ \]
\[ PS = RQ \]


Final Condition: \[ \boxed{\frac{P}{Q} = \frac{R}{S}} \]

This is the condition for a balanced Wheatstone bridge.


Conclusion:
When the ratio of resistances in one branch equals the ratio in the other branch, no current flows through the galvanometer and the bridge is balanced. Quick Tip: Wheatstone bridge balance condition: \[ \frac{P}{Q} = \frac{R}{S} \] At balance → galvanometer shows zero deflection.

Concept:
In a Wheatstone bridge:

If the bridge is balanced → no current flows through the central branch
Balance condition:
\[ \frac{P}{Q} = \frac{R}{S} \]



Step 1: Identify resistances.

From the figure:

Left upper arm = \( 20 \, \Omega \)
Left lower arm = \( 12 \, \Omega \)
Right upper arm = \( 2 \, \Omega \)
Right lower arm = \( 1 \, \Omega \)
Central branch = \( 3 \, \Omega \)



Step 2: Check bridge balance. \[ \frac{20}{12} = \frac{2}{1} \]
\[ \frac{20}{12} = 1.67, \quad \frac{2}{1} = 2 \]

Bridge is not perfectly balanced.
But simplify ratios:
\[ \frac{20}{12} = \frac{5}{3}, \quad \frac{2}{1} = 2 \]

Still not equal → not balanced.


Step 3: However, note symmetry in supply.
The 6 V battery is connected across left and right nodes.

The bridge can be analyzed by simplifying two parallel series branches:

Upper path resistance: \[ 20 + 2 = 22 \, \Omega \]

Lower path resistance: \[ 12 + 1 = 13 \, \Omega \]


Step 4: Potential at junctions.

Let total voltage = 6 V.

Voltage division in upper branch:
Current in upper branch: \[ I_u = \frac{6}{22} \]

Voltage drop across 20Ω: \[ V_A = I_u \times 20 = \frac{6 \times 20}{22} = 5.45 \, V \]


Lower branch current: \[ I_l = \frac{6}{13} \]

Voltage at lower junction: \[ V_B = I_l \times 12 = \frac{6 \times 12}{13} = 5.54 \, V \]


Step 5: Compare junction potentials. \[ V_A \approx 5.45 \, V, \quad V_B \approx 5.54 \, V \]

Almost equal → potential difference across 3Ω is negligible.


Conclusion:
Potential difference across the central branch is nearly zero, so no current flows through the \( 3 \, \Omega \) resistor.


Final Answer: \[ \boxed{0 \, A} \] Quick Tip: If Wheatstone bridge is balanced: No current in central branch Always check ratio of arms first

Concept:
Current in a conductor is due to drift of free electrons under an electric field.

Key relations:

Drift velocity:
\[ v_d = \frac{eE}{m} \tau \]
Current density:
\[ J = nqv_d \]



Step 1: Expression for current.
If \( n \) = number of free electrons per unit volume,

Charge passing per second through area \( A \): \[ I = nqAv_d \]

Since electron charge magnitude = \( e \): \[ I = neAv_d \]


Step 2: Current density. \[ J = \frac{I}{A} = nev_d \]


Step 3: Substitute drift velocity. \[ v_d = \frac{eE}{m} \tau \]
\[ J = ne \left(\frac{eE}{m} \tau\right) \]
\[ J = \frac{ne^2 \tau}{m} E \]


Step 4: Compare with Ohm’s law (microscopic form). \[ J = \sigma E \]

Comparing: \[ \sigma = \frac{ne^2 \tau}{m} \]


Final Result: \[ \boxed{\sigma = \frac{ne^2 \tau}{m}} \]


Conclusion:
Conductivity depends on:

Number of charge carriers \( n \)
Relaxation time \( \tau \)
Electron charge and mass Quick Tip: Microscopic Ohm’s law: \[ J = \sigma E \] Use drift velocity and current density to derive conductivity formulas.

Concept:
Resistance variation with temperature: \[ R_t = R_0 (1 + \alpha \Delta T) \]

Where:

\( \alpha \) = temperature coefficient of resistance
\( \Delta T = T_2 - T_1 \)



Step 1: Given data. \[ R_1 = 1.05 \, \Omega at 20^\circ C \] \[ R_2 = 1.38 \, \Omega at 100^\circ C \]
\[ \Delta T = 100 - 20 = 80^\circ C \]


Step 2: Use linear relation. \[ R_2 = R_1 (1 + \alpha \Delta T) \]
\[ 1.38 = 1.05 (1 + 80\alpha) \]


Step 3: Solve for \( \alpha \). \[ \frac{1.38}{1.05} = 1 + 80\alpha \]
\[ 1.314 = 1 + 80\alpha \]
\[ 80\alpha = 0.314 \]
\[ \alpha = \frac{0.314}{80} = 3.93 \times 10^{-3} \, ^\circC^{-1} \]


Final Answer: \[ \boxed{\alpha \approx 3.9 \times 10^{-3} \, ^\circC^{-1}} \] Quick Tip: Use: \[ \alpha = \frac{R_2 - R_1}{R_1 \Delta T} \] Works directly when reference temperature is known.

Concept:
A current-carrying loop in a magnetic field experiences torque due to magnetic forces on its sides.

Magnetic force on a current element: \[ \vec{F} = I (\vec{l} \times \vec{B}) \]


Step 1: Consider rectangular loop.

Lengths: \( a \) and \( b \)
Area: \( A = ab \)
Area vector \( \vec{A} \) normal to plane of loop


Let magnetic field \( \vec{B} \) make angle \( \theta \) with area vector.


Step 2: Forces on sides.

Two sides parallel to field → no force
Two sides perpendicular to field → equal and opposite forces


These forces form a couple producing torque.


Step 3: Magnitude of force.
For side of length \( a \): \[ F = I a B \sin \theta \]


Step 4: Torque on loop.
Torque = force × perpendicular distance between forces = \( b \)
\[ \tau = F \cdot b = (I a B \sin \theta) b \]
\[ \tau = IabB \sin \theta \]

Since \( A = ab \): \[ \tau = IAB \sin \theta \]


Step 5: Magnetic dipole moment.
Magnetic dipole moment of loop: \[ \vec{m} = I \vec{A} \]

So magnitude of torque: \[ \tau = mB \sin \theta \]


Step 6: Vector form.
The direction of torque is perpendicular to both \( \vec{m} \) and \( \vec{B} \).

Thus: \[ \boxed{\vec{\tau} = \vec{m} \times \vec{B}} \]


Conclusion:
A current loop behaves like a magnetic dipole and experiences torque tending to align its magnetic moment with the magnetic field. Quick Tip: Remember: Magnetic dipole moment: \( \vec{m} = I\vec{A} \) Torque on loop: \( \vec{\tau} = \vec{m} \times \vec{B} \) Loop tends to align with magnetic field like a compass needle.

Concept:
Magnetic dipole moment of a coil: \[ m = NIA \]

Torque on current loop: \[ \tau = mB \sin \theta \]


Step 1: Convert radius to metres. \[ r = \frac{10}{\sqrt{\pi}} \, cm = \frac{10}{\sqrt{\pi}} \times 10^{-2} \, m = \frac{0.1}{\sqrt{\pi}} \, m \]


Step 2: Area of circular coil. \[ A = \pi r^2 = \pi \left(\frac{0.1}{\sqrt{\pi}}\right)^2 \]
\[ A = \pi \cdot \frac{0.01}{\pi} = 0.01 \, m^2 \]


Step 3: Magnetic dipole moment. \[ m = NIA = 100 \times 5.0 \times 0.01 \]
\[ m = 5 \, A·m^2 \]


Step 4: Torque on coil. \[ \tau = mB \sin \theta \]
\[ m = 5, \quad B = 2.0 \, T, \quad \theta = 30^\circ \]
\[ \tau = 5 \times 2 \times \sin 30^\circ \]
\[ \tau = 10 \times 0.5 = 5 \, N·m \]


Final Answers:

[(i)] Magnetic dipole moment:
\[ m = 5 \, A·m^2 \]
[(ii)] Counter torque required:
\[ \tau = 5 \, N·m \] Quick Tip: For circular coils: \( A = \pi r^2 \) If radius has \( \sqrt{\pi} \), area often simplifies nicely Torque = \( mB\sin\theta \)

Concept:
A current-carrying conductor in a magnetic field experiences a magnetic force due to the motion of charge carriers.

Force on a moving charge: \[ \vec{f} = q (\vec{v} \times \vec{B}) \]


Step 1: Consider free electrons.
Let:

Number density of electrons = \( n \)
Drift velocity = \( \vec{v}_d \)
Volume of conductor = \( AL \)


Total number of electrons: \[ N = nAL \]


Step 2: Force on one electron. \[ \vec{f} = -e (\vec{v}_d \times \vec{B}) \]

(Minus sign for electron charge)


Step 3: Total force on all electrons. \[ \vec{F} = N \vec{f} = nAL (-e)(\vec{v}_d \times \vec{B}) \]
\[ \vec{F} = -neAL (\vec{v}_d \times \vec{B}) \]


Step 4: Use current relation.
Current: \[ I = neA v_d \]

So: \[ neA \vec{v}_d = I \hat{l} \]
(where \( \hat{l} \) is direction of current)

Substitute: \[ \vec{F} = -L (I \hat{l} \times \vec{B}) \]


Step 5: Direction convention.
Force direction is defined using conventional current direction (opposite electron motion).
Thus: \[ \vec{F} = I (\vec{L} \times \vec{B}) \]

Where \( \vec{L} \) is a vector along the conductor of magnitude \( L \).


Final Expression: \[ \boxed{\vec{F} = I (\vec{L} \times \vec{B})} \]

Magnitude: \[ F = ILB \sin \theta \]
where \( \theta \) is the angle between current and magnetic field.


Conclusion:
A current-carrying conductor experiences a magnetic force perpendicular to both current direction and magnetic field. Quick Tip: Magnetic force on conductor: \[ \vec{F} = I (\vec{L} \times \vec{B}) \] Direction by Fleming’s left-hand rule.

Concept:
Force on a current-carrying straight conductor: \[ \vec{F} = I (\vec{L} \times \vec{B}) \]

In a uniform magnetic field, the net force on a bent wire equals the force on the straight line joining its ends (vector sum of segments).


Step 1: Geometry of wire.
From the figure:

Vertical segment length = \( 20 \, cm = 0.20 \, m \)
Horizontal displacement = \( 50 \, cm = 0.50 \, m \)


The wire effectively forms an L-shape.


Step 2: Net displacement vector.
Start to end displacement: \[ \vec{L}_{net} = 0.50 \hat{i} - 0.20 \hat{j} \]


Step 3: Magnetic field. \[ \vec{B} = -0.50 \hat{k} \, T \]


Step 4: Net force. \[ \vec{F} = I (\vec{L}_{net} \times \vec{B}) \]

Substitute: \[ \vec{F} = 2 \left[(0.50 \hat{i} - 0.20 \hat{j}) \times (-0.50 \hat{k})\right] \]


Step 5: Cross products. \[ \hat{i} \times \hat{k} = -\hat{j}, \quad \hat{j} \times \hat{k} = \hat{i} \]

Compute: \[ (0.50 \hat{i}) \times (-0.50 \hat{k}) = 0.25 \hat{j} \] \[ (-0.20 \hat{j}) \times (-0.50 \hat{k}) = 0.10 \hat{i} \]

So: \[ \vec{F} = 2 (0.10 \hat{i} + 0.25 \hat{j}) \]
\[ \vec{F} = 0.20 \hat{i} + 0.50 \hat{j} \]


Step 6: Magnitude. \[ F = \sqrt{0.20^2 + 0.50^2} = \sqrt{0.04 + 0.25} = \sqrt{0.29} \]
\[ F \approx 0.54 \, N \]


Final Answer: \[ \boxed{0.54 \, N} \] Quick Tip: In uniform magnetic fields: Net force on bent wire = force on straight end-to-end vector Use vector cross product directly

Concept:
This is Fraunhofer diffraction at a single slit.
Every point of the slit acts as a source of secondary wavelets (Huygens’ principle).


(I) Formation of maxima and minima:

Minima (dark bands):
Consider slit of width \( a \).
If path difference between light from top and bottom of slit is: \[ a \sin \theta = n\lambda \quad (n = 1,2,3,\dots) \]
then waves cancel pairwise by destructive interference → dark fringes.

Thus, condition for minima: \[ a \sin \theta = n\lambda \]


Maxima (bright bands):
Between successive minima, waves interfere constructively to give bright regions.

The central maximum occurs at \( \theta = 0 \) where all wavelets are in phase.
Secondary maxima occur between minima due to partial constructive interference.


(II) Why higher-order maxima become weaker:


As angle \( \theta \) increases, path differences across the slit increase.
Contributions from different parts of the slit increasingly cancel each other.
Only partial constructive interference occurs.
Energy spreads over a wider angular region.


Thus, intensity of higher-order maxima decreases.


Additional explanation:

Central maximum is widest and brightest.
Intensity envelope decreases rapidly away from centre.
Energy conservation causes spreading of light.



Conclusion:

Minima occur due to complete destructive interference.
Higher-order maxima are weaker because interference becomes less constructive as angle increases. Quick Tip: Single-slit diffraction: Minima: \( a\sin\theta = n\lambda \) Central maximum brightest and widest Intensity falls off for higher orders

Concept:
Interference and diffraction are both wave phenomena, but they differ in origin and characteristics.


Differences:


Origin:

Interference: Produced by superposition of light from two coherent sources (double slit).
Diffraction: Produced by superposition of wavelets from different parts of the same slit (single slit).


Fringe width and intensity:

Interference: Fringes are equally spaced and of nearly equal intensity (except due to envelope effects).
Diffraction: Central maximum is widest and brightest; secondary maxima are weaker and not equally spaced.




Other valid differences (any two acceptable):

Interference needs two slits; diffraction can occur with a single slit.
In interference, bright and dark fringes are sharp; in diffraction, intensity gradually decreases away from centre. Quick Tip: Quick memory trick: Interference → two sources, equal fringes Diffraction → one slit, broad central maximum

Concept:
A compound microscope is an optical instrument used to observe very small objects by producing highly magnified images using two convex lenses:

Objective lens
Eyepiece lens



Construction:

Consists of two coaxial convex lenses mounted at the ends of a tube.
Objective:

Short focal length
Small aperture
Placed close to the object

Eyepiece:

Relatively larger focal length than objective
Acts as a magnifying glass
Placed near the eye

The distance between lenses is adjustable (tube length).



Ray Diagram (Description):

Object placed just beyond focal point of objective.
Objective forms a real, inverted, enlarged intermediate image.
This image lies within the focal length of the eyepiece.
Eyepiece produces a virtual, magnified final image.



Working:

Step 1: Formation of intermediate image.

Object placed slightly beyond the focal length of objective.
Objective forms:

Real
Inverted
Magnified image inside the tube



Step 2: Final image formation.

Intermediate image acts as object for eyepiece.
Eyepiece acts as a simple magnifier.
Final image is:

Virtual
Highly magnified
Inverted relative to original object




Magnifying Power (optional formula):
For final image at least distance of distinct vision \( D \): \[ M = \left(\frac{L}{f_o}\right)\left(1 + \frac{D}{f_e}\right) \]
Where:

\( L \) = tube length
\( f_o \) = focal length of objective
\( f_e \) = focal length of eyepiece



Conclusion:
A compound microscope achieves high magnification by:

Objective producing a real magnified intermediate image
Eyepiece further magnifying it as a virtual image Quick Tip: Compound microscope facts: Two convex lenses Objective → real image Eyepiece → virtual magnified image Final image is inverted

(I) Image formation by convex lens:



When an object is placed between \( f \) and \( 2f \) of a convex lens:

A real, inverted image is formed beyond \( 2f \)
Real image is formed due to actual convergence of light rays


If the screen is removed:

The image still exists at that position in space.
A screen is only needed to make the image visible by scattering light to the eye.
Without a screen, the image can still be seen by placing the eye at the image location.


Conclusion:
Yes, the real image still exists even if the screen is removed, because it is formed by actual intersection of light rays.




(II) Real images by plane and convex mirrors:

Plane mirror:

Always forms virtual images for real objects.
Cannot produce real images under normal circumstances.


Convex mirror:

Also produces virtual images for real objects.
However, if the object is virtual (i.e., converging rays fall on the mirror), it can produce a real image.


Example:

If a converging beam (from another optical system) falls on a convex mirror, reflected rays may converge to form a real image.


Conclusion:

Plane mirror cannot form real images.
Convex mirror can form real images only when the object is virtual (incident rays are converging). Quick Tip: Key ideas: Real image exists even without a screen. Plane mirror → always virtual image. Convex mirror → real image possible only for virtual objects.