BSEB Class 12 Physics Top 5 Sure Shot Questions 2027 PDF — Most Expected Derivations with Complete Solutions

Summary: This publication provides the complete solved solutions for the official BSEB Class 12 Physics Top 5 Sure Shot Questions (2027 Examination). Curated by PrepOne from a 10-year question frequency analysis of Bihar School Examination Board papers, covering the 5 most repeated 5-mark long-answer derivations and principles.

Recommended Practice: Download the official PDF below, practice drawing the labeled diagrams on physical paper, and evaluate your handwritten copy on the PrepOne Evaluation Engine for instant step-by-step scoring against official BSEB rubrics.

In the Bihar Board Class 12 Physics examination, Section B contains 6 long-answer questions (5 marks each), and students are required to attempt only 3 questions (yielding a 100% internal choice advantage). Because the syllabus features a defined set of foundational derivations, historical analysis indicates that these exact 5 question models repeat with the highest frequency across board sessions.

Official Sure Shot Questions PDF Download

Access and download the complete 12-page handwritten PDF notes directly from Google Drive:

Resource / Question Set	Board & Level	Action
BSEB Class 12 Physics Top 5 Sure Shot Questions (Official PrepOne Guide)	BSEB Class 12 Physics (2027)	Download PDF →

Get Full Marks on 5-Mark Physics Derivations

BSEB examiners allocate marks in steps: 1 mark for the labeled diagram, 1 mark for stating the principle/assumptions, 2 marks for intermediate algebraic steps, and 1 mark for the final formula with SI units. Scan your handwritten practice derivations on PrepOne to verify step compliance before the exam.

Download PrepOne on Google Play →

Complete Solutions to the Top 5 Sure Shot Questions

Question 1: Principle, Working & Losses of a Transformer

5 Marks — Alternating Current

"Explain the working principle of a transformer with a neat labelled diagram. Mention the types of losses in a transformer and how they can be minimised."

1. Principle & Definition

A transformer is a static electrical device that transfers electrical energy between two circuits through mutual electromagnetic induction, accompanied by a change in voltage and current without altering frequency. It operates on Faraday's Law of Electromagnetic Induction: when an alternating voltage $V_1$ is applied to the primary coil, a time-varying magnetic flux is produced in the laminated soft-iron core. This changing flux links with the secondary coil, inducing an EMF.

2. Construction & Transformation Ratio Derivation

A transformer consists of a primary coil ( $N_1$ turns), a secondary coil ( $N_2$ turns), and a laminated soft-iron core providing a low-reluctance closed flux path.

E_1 = -N_1 \frac{d\Phi}{dt}, \quad E_2 = -N_2 \frac{d\Phi}{dt}

\frac{V_2}{V_1} = \frac{N_2}{N_1} = k \quad (\text{Transformation Ratio})

$k > 1 \implies$ Step-up transformer (voltage increases, current decreases).
$k < 1 \implies$ Step-down transformer (voltage decreases, current increases).
For an ideal transformer (100% efficiency, input power = output power): $V_1 I_1 = V_2 I_2 \implies \frac{I_1}{I_2} = \frac{N_2}{N_1}$

3. Types of Losses and Minimisation Methods

Loss Type	Physical Cause	Minimisation Method
Copper Loss	Winding resistance heat dissipation: $P_{\text{Cu}} = I_1^2 R_1 + I_2^2 R_2$	Use thick copper conductors with low resistivity.
Hysteresis Loss	Energy lost during continuous cycle magnetization & demagnetization of the iron core	Use silicon steel (grain-oriented) core with narrow hysteresis loop.
Eddy Current Loss	Time-varying magnetic flux induces circulating currents: $P_e \propto B_{\max}^2 f^2 t^2$	Use a laminated core of thin insulated silicon steel sheets.
Flux Leakage	Incomplete magnetic coupling between primary and secondary windings	Co-wind coils over each other or adopt a shell-type core design.
Mechanical (Humming)	Core vibration due to magnetostriction during alternating cycles	Proper rigid mechanical clamping of the transformer core assembly.

Question 2: Biot-Savart Law & Axial Magnetic Field of a Loop

5 Marks — Moving Charges & Magnetism

"State Biot-Savart Law. Derive the expression for the magnetic field at a point on the axis of a current-carrying circular coil. Draw a neat labelled diagram."

1. Statement of Biot-Savart Law

The magnetic field $d\vec{B}$ at a point $P$ due to an infinitesimal current element $I d\vec{l}$ is directly proportional to current $I$ , length $dl$ , $\sin \theta$ , and inversely proportional to distance square $r^2$ .

d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2}, \quad dB = \frac{\mu_0}{4\pi} \frac{I dl \sin \theta}{r^2}

2. Axial Magnetic Field Derivation

Consider a circular loop of radius $R$ carrying current $I$ . Point $P$ is located at distance $x$ along the axis from the centre $O$ . Distance from any element is $r = \sqrt{R^2 + x^2}$ . Since $d\vec{l} \perp \vec{r}$ ( $\theta = 90^\circ$ , $\sin 90^\circ = 1$ ):

dB = \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2}

\cos \phi = \frac{R}{\sqrt{R^2 + x^2}}, \quad \sin \phi = \frac{x}{\sqrt{R^2 + x^2}}

The perpendicular components $dB_\perp$ cancel in pairs due to diametric symmetry. The axial components $dB_x = dB \cos \phi$ sum constructively over the circumference:

B = \oint dB \cos \phi = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \oint dl = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \cdot (2\pi R)

B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}

Special Cases: At the centre ( $x = 0$ ), $B = \frac{\mu_0 I}{2R}$ . At far distance ( $x \gg R$ ), $B \approx \frac{\mu_0}{4\pi} \frac{2M}{x^3}$ where $M = I\pi R^2$ is the magnetic dipole moment.

Question 3: Photoelectric Effect & Einstein's Equation

5 Marks — Dual Nature of Radiation

"What is the photoelectric effect? State its laws. Derive Einstein’s photoelectric equation and explain the significance of threshold frequency and work function."

1. Definition & Laws

The photoelectric effect is the emission of electrons from a metallic surface when electromagnetic radiation of sufficiently high frequency is incident on it.

Threshold Frequency: Emission occurs only if frequency $f > f_0$ , regardless of intensity.
Photocurrent Proportionality: For $f > f_0$ , rate of emission is directly proportional to incident intensity.
Kinetic Energy Independence: Maximum kinetic energy is independent of intensity and increases linearly with incident frequency.
Instantaneous Process: The time lag between incidence and emission is negligible ( $< 10^{-9}$ s).

2. Derivation of Einstein's Photoelectric Equation

By the Law of Conservation of Energy, a single incident photon of energy $hf$ transfers its entire energy to a single conduction electron:

hf = \Phi + K_{\max} = \Phi + \frac{1}{2} m v_{\max}^2

At threshold frequency $f = f_0$ , electrons are released with zero kinetic energy ( $K_{\max} = 0$ ), hence $\Phi = hf_0$ . Substituting:

hf = hf_0 + K_{\max} \implies K_{\max} = h(f - f_0)

Significance: Work function $\Phi$ is the minimum energy required to liberate an electron from the metal surface. Threshold frequency $f_0 = \Phi / h$ is an intrinsic characteristic property of each metal.

Question 4: Capacitance with Dielectric Slab

5 Marks — Electrostatics

"Derive the expression for the capacitance of a parallel plate capacitor. What is the effect of inserting a dielectric slab between the plates?"

1. Derivation of Capacitance in Vacuum

Consider two parallel conducting plates of area $A$ separated by distance $d$ carrying charges $+Q$ and $-Q$ . Surface charge density $\sigma = Q/A$ .

E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}

V = E \cdot d = \frac{Q d}{\varepsilon_0 A}

C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}

2. Effect of Inserting Dielectric Slab (Constant K)

When a dielectric material of dielectric constant $K$ is placed between the plates, electric polarisation creates an opposing induced field $E_p$ :

Reduced Electric Field: $E_{\text{net}} = E_0 / K$
Reduced Potential Difference: $V = V_0 / K$
Increased Capacitance: $C = K C_0 = \frac{K \varepsilon_0 A}{d}$

Question 5: Moving Coil Galvanometer, Sensitivity & Conversions

5 Marks — Magnetic Effects of Current

"Explain the construction and working of a Moving Coil Galvanometer. Derive the expression for its current sensitivity. How is it converted into an ammeter and a voltmeter?"

1. Principle & Working Theory

A current-carrying coil placed in a uniform radial magnetic field experiences a deflecting torque proportional to current. The radial magnetic field created by concave pole pieces ensures that the plane of the coil remains parallel to the magnetic lines ( $\theta = 90^\circ$ ):

\tau_d = NIAB \sin 90^\circ = NIAB

\tau_r = k \phi \quad (\text{Restoring torque of spring})

\text{At equilibrium: } NIAB = k \phi \implies I = \left( \frac{k}{NAB} \right) \phi \implies I \propto \phi

2. Current Sensitivity

Current sensitivity $I_s$ is the deflection produced per unit current:

I_s = \frac{\phi}{I} = \frac{NAB}{k}

3. Conversion into Ammeter and Voltmeter

Property	Conversion to Ammeter	Conversion to Voltmeter
Connected Component	Low resistance shunt $S$ connected in parallel	High resistance $R$ connected in series
Resistance Value	Very low (ideal ammeter resistance = 0)	Very high (ideal voltmeter resistance = $\infty$ )
Connection in Circuit	Connected in series with the load	Connected in parallel across the load
Formula	$S = \frac{I_g G}{I - I_g}$	$R = \frac{V}{I_g} - G$

Frequently Asked Questions

Why is a radial magnetic field required in a Moving Coil Galvanometer?

A radial magnetic field ensures that the deflecting torque on the coil is always maximum ( $\sin \theta = 1$ ) regardless of the coil's position, resulting in a linear scale ( $I \propto \phi$ ) where deflection is directly proportional to current.

How do laminated cores reduce eddy currents in transformers?

Laminating the iron core with thin insulated varnish sheets breaks up the large closed circular loops through which eddy currents circulate. Since eddy current loss is proportional to the square of lamination thickness ( $P_e \propto t^2$ ), reducing thickness significantly minimizes heat generation.

Can I download this 12-page PDF guide for free?

Yes. The PDF is hosted on Google Drive and can be downloaded or read offline on any mobile device or desktop.

For Students

For Institutes

Company