WBCHSE 2026 Class 12 Physics Question Paper with Solution PDF

Concept:
According to Coulomb's Law, the electrostatic force between two point charges is: \[ F = k \frac{q_1 q_2}{r^2} \]
where:

\( F \) = force between charges
\( r \) = distance between charges


Thus, force is inversely proportional to the square of the distance.

Step 1: Original force
\[ F \propto \frac{1}{r^2} \]

Step 2: If distance is doubled

New distance = \( 2r \)
\[ F' \propto \frac{1}{(2r)^2} = \frac{1}{4r^2} \]

Step 3: Compare forces
\[ F' = \frac{F}{4} \]

So, the force becomes one-fourth of the original value.
\[ \therefore Correct option: (c) One-fourth \] Quick Tip: In Coulomb's law, if distance becomes: Double \(\rightarrow\) Force becomes one-fourth Half \(\rightarrow\) Force becomes four times

Concept:
Magnetic flux (\( \Phi \)) measures the total magnetic field passing through a surface.
It is defined as: \[ \Phi = B \cdot A \]
where:

\( B \) = magnetic field (Tesla)
\( A \) = area (m\(^2\))


Step 1: SI unit calculation
\[ Magnetic flux unit = Tesla \times m^2 \]

This unit is called the Weber (Wb).

Final Answer: \[ \boxed{Weber (Wb)} \] Quick Tip: Magnetic flux unit = Weber (Wb) Magnetic field unit = Tesla (T)

Concept:
To convert a voltmeter into an ammeter, a shunt resistance is connected in parallel.
Key idea:

Voltmeter full-scale current \( I_v = \frac{V}{R_v} \)
Shunt carries remaining current
Same voltage across voltmeter and shunt


Step 1: Find full-scale current of voltmeter

Given: \[ V = 150\,V, R_v = 300\,\Omega \]
\[ I_v = \frac{V}{R_v} = \frac{150}{300} = 0.5\,A \]

So, voltmeter allows maximum current of \( 0.5\,A \).

Step 2: Total current required

Desired ammeter range: \[ I = 8\,A \]

Current through shunt: \[ I_s = I - I_v = 8 - 0.5 = 7.5\,A \]

Step 3: Voltage across shunt

Voltage across voltmeter (and shunt in parallel): \[ V = I_v \times R_v = 0.5 \times 300 = 150\,V \]

Step 4: Calculate shunt resistance
\[ R_s = \frac{V}{I_s} = \frac{150}{7.5} = 20\,\Omega \]

Final Answer: \[ \boxed{R_s = 20\,\Omega} \] Quick Tip: To convert a voltmeter to an ammeter: Find voltmeter full-scale current first, then use parallel shunt formula \( R_s = \frac{V}{I - I_v} \).

Concept:
Total Internal Reflection (TIR) is a phenomenon in which a light ray is completely reflected back into the denser medium instead of refracting into the rarer medium.

Definition of Critical Angle:
The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium becomes \( 90^\circ \).
\[ At critical angle: r = 90^\circ \]

Mathematical expression:
From Snell's law, \[ n_1 \sin C = n_2 \sin 90^\circ \] \[ \sin C = \frac{n_2}{n_1} \]
where:

\( n_1 \) = refractive index of denser medium
\( n_2 \) = refractive index of rarer medium



Conditions for Total Internal Reflection (TIR):


The light must travel from a denser medium to a rarer medium.
(\( n_1 > n_2 \))

The angle of incidence in the denser medium must be greater than the critical angle.
(\( i > C \))


Result:
Under these conditions, the refracted ray disappears and the entire light is reflected back into the denser medium. Quick Tip: TIR occurs only when light goes from denser to rarer medium and the angle of incidence exceeds the critical angle.

Concept:
A Wheatstone bridge is used to measure an unknown resistance by balancing two arms of a bridge circuit.
At balance condition: \[ \frac{P}{Q} = \frac{R}{X} \]
where \( X \) is the unknown resistance.

However, this method has limitations for very high and very low resistances.


Case 1: Very High Resistance


Current through the bridge becomes extremely small.
The galvanometer receives negligible current, making null detection difficult.
Leakage currents and insulation resistance introduce significant errors.


Result: Poor sensitivity and inaccurate measurement.


Case 2: Very Low Resistance


Contact resistance and resistance of connecting wires become comparable to the unknown resistance.
These additional resistances disturb the balance condition.
Small errors produce large percentage deviations.


Result: Large measurement errors.


Conclusion:

Very high resistance \(\rightarrow\) current too small, leakage errors.
Very low resistance \(\rightarrow\) contact and lead resistances cause large errors.


Hence, the Wheatstone bridge is not suitable for accurately measuring extremely high or very low resistances. Quick Tip: Wheatstone bridge works best for medium resistances. Use Kelvin bridge for low resistance and megohm meters for high resistance.

Concept:
Gauss’s theorem helps calculate electric fields for highly symmetric charge distributions such as spherical, cylindrical, or planar symmetry.


Statement of Gauss’s Theorem:

The total electric flux through a closed surface is equal to \( \frac{1}{\varepsilon_0} \) times the total charge enclosed within the surface.
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} \]

where:

\( \vec{E} \) = electric field
\( d\vec{A} \) = area vector
\( \varepsilon_0 \) = permittivity of free space



Electric field due to an infinitely long straight charged wire

Let:

Linear charge density = \( \lambda \) (charge per unit length)
Distance from wire = \( r \)


Step 1: Choose Gaussian surface

Due to cylindrical symmetry, choose a cylindrical Gaussian surface:

Radius = \( r \)
Length = \( L \)


The electric field is:

Radial
Same magnitude everywhere on curved surface



Step 2: Calculate electric flux

Flux through cylinder:

Curved surface contributes flux
Flat ends give zero flux (field parallel to surface)

\[ Flux = E \times Curved surface area \] \[ Curved area = 2\pi rL \]
\[ \Rightarrow \Phi = E(2\pi rL) \]


Step 3: Charge enclosed
\[ Q_{enc} = \lambda L \]


Step 4: Apply Gauss’s Law
\[ E(2\pi rL) = \frac{\lambda L}{\varepsilon_0} \]

Cancel \( L \):
\[ E(2\pi r) = \frac{\lambda}{\varepsilon_0} \]
\[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]


Final Result:
\[ \boxed{E = \frac{\lambda}{2\pi \varepsilon_0 r}} \]


Field decreases inversely with distance.
Field is radially outward for positive charge. Quick Tip: Use Gauss’s law when symmetry exists: Spherical \(\rightarrow\) sphere, Cylindrical \(\rightarrow\) cylinder, Planar \(\rightarrow\) Gaussian pillbox.

Concept:
The Lens Maker’s Formula relates the focal length of a lens to:

Refractive index of lens material
Radii of curvature of its two surfaces


It is derived using refraction at two spherical surfaces.


Assumptions:

Thin convex lens
Refractive index of lens = \( \mu \)
Lens in air (\( \mu_{air} = 1 \))
Radii of curvature = \( R_1 \) and \( R_2 \)



Step 1: Refraction at first spherical surface

Using refraction formula at a spherical surface: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]

For first surface:

Light travels from air to glass
\( \mu_1 = 1, \mu_2 = \mu \)
Object at distance \( u \), image formed at \( v_1 \)

\[ \frac{\mu}{v_1} - \frac{1}{u} = \frac{\mu - 1}{R_1} \cdots (1) \]


Step 2: Refraction at second spherical surface

Now the image formed by the first surface acts as the object for the second surface.

For second surface:

Light travels from glass to air
\( \mu_1 = \mu, \mu_2 = 1 \)
Object distance = \( v_1 \)
Final image distance = \( v \)

\[ \frac{1}{v} - \frac{\mu}{v_1} = \frac{1 - \mu}{R_2} \cdots (2) \]


Step 3: Add equations (1) and (2)
\[ \left( \frac{\mu}{v_1} - \frac{1}{u} \right) + \left( \frac{1}{v} - \frac{\mu}{v_1} \right) = \frac{\mu - 1}{R_1} + \frac{1 - \mu}{R_2} \]

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


Step 4: For focal length

For focal length, object at infinity: \[ u = \infty \Rightarrow \frac{1}{u} = 0, v = f \]
\[ \frac{1}{f} = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]


Lens Maker’s Formula:
\[ \boxed{ \frac{1}{f} = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) } \]


Notes:

Valid for thin lenses.
Sign convention must be used carefully.
For convex lens in air, \( R_1 > 0 \), \( R_2 < 0 \). Quick Tip: Lens Maker’s Formula gives focal length from lens geometry. Use it when refractive index and radii of curvature are known.

Concept:
Lenz’s Law describes the direction of induced current in electromagnetic induction and ensures that energy is conserved during the induction process.


Statement of Lenz’s Law:

Lenz’s Law states that:
\begin{quote
The direction of induced current in a circuit is such that it opposes the change in magnetic flux that produces it.
\end{quote

Mathematically, it is represented by the negative sign in Faraday’s law: \[ \mathcal{E} = -\frac{d\Phi}{dt} \]


Relation with Conservation of Energy:

To show that Lenz’s Law follows from conservation of energy, consider the following situation.


Case: Magnet approaching a coil

Step 1: Change in magnetic flux

When a bar magnet is moved towards a coil:

Magnetic flux linked with the coil increases.
An emf and current are induced in the coil.


Step 2: Direction of induced current

According to Lenz’s Law:

The induced current produces a magnetic field opposing the approaching magnet.
The face of the coil near the magnet behaves like a similar pole (repulsion).


Step 3: Work done by external agent

Because of opposition:

An external force is required to push the magnet towards the coil.
Mechanical work is done.


This mechanical energy is converted into:

Electrical energy in the coil
Heat due to resistance (Joule heating)



If Lenz’s Law were not true:

If the induced current aided the motion:

The magnet would accelerate automatically.
Electrical energy would be produced without external work.


This would violate the law of conservation of energy.


Conclusion:

Since the induced current always opposes the cause producing it, external work is required, ensuring that: \[ Mechanical energy \rightarrow Electrical energy \]

Thus, Lenz’s Law is a direct consequence of the Law of Conservation of Energy. Quick Tip: Lenz’s Law prevents creation of energy from nothing. Opposition of induced current ensures energy conservation.