CBSE Class 12 Physics Important Case Study Based Questions for Half Yearly 2025

CBSE Class 12 Physics Half Yearly Exam 2025: All CBSE Class 12 students in the science stream, whether they are preparing for non-medical or medical careers, must take Physics as a foundational and required subject. Practice is crucial in this subject, particularly for high-weightage subjects like optics and electromagnetics. Success on the next Half-Yearly Exam 2025 depends on students grasping the essential ideas and being proficient in the critical questions. This article is intended to assist the students in efficiently preparing by emphasizing the most essential subjects and question kinds.

The CBSE Class 12 Physics exam for 2025 will be divided into five sections, with a significant emphasis on applied knowledge. A key component of the paper, Section E, will feature two case study-based questions, each carrying 4 marks. These questions are a vital part of the exam, as they test your ability to apply theoretical knowledge to real-world scenarios. This article provides solved examples of these case study questions and a selection of other important questions to help you familiarize yourself with the exam pattern and boost your overall score.

RELATED: CBSE Physics Class 12 Syllabus 2025-26

CBSE Class 12 Physics Unit Wise Marks Distribution 2025

Why for Case Study Questions are beneficial for Class 12 Physics?

Physics is no walk in the park. It is a mind-boggling subject that’s on par with mathematics for non-medical stream students. However, physics is challenging for all students due to its conceptual and numerical-based nature. Physics requires a clear understanding of the fundamentals, memorisation of several formulas, derivations and expert calculation skills plus the ability to apply them to tricky questions. All this requires practice but not blind practice. You must know which topics are important and which chapters are frequently asked in the exam.

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CBSE 12th Physics: Important Case Study Based Questions 

Ques. 1 An ammeter and a voltmeter are connected in series to a battery with an emf  of 10V. When a certain resistance is connected in parallel with the voltmeter, the reading of the voltmeter decreases three times, whereas the reading of the  ammeter increases two times.

A: Find the voltmeter reading after the connection of the resistance.

1. 1 V
2. 2 V
3. 3 V
4. 4 V

Answer: (2) 2V

B: If the resistance of the ammeter is 2 ohm, then the resistance of the voltmeter is:-

1. 1 ohm
2. 2 ohm
3. 3 ohm
4. 4 ohm

Answer: (3) 3 ohm

C: If the resistance of ammeter is 2 ohm ,then resistance of the resistor which is added in parallel to the voltmeter is

1. ⅗ ohm
2. 2/7 ohm
3. 3/7 ohm
4. None of the above

Answer: (1) 3/5 ohm

Ques. 2 Given figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s-1 in the direction shown. Give the polarity and magnitude of the induced emf. Physics / XII (2020-21)

(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?

(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.

(d) What is the retarding force on the rod when K is closed?

(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm/ sec) when K is closed? How much power is required when K is open?

(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?

(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Answers:

(a) EMF = vBL = 0.12 0.50 x 0.15 = 9.0 mV; P positive end and Q negative end.

(b) Yes. When K is closed, the excess charge is maintained by the continuous flow of current.

(c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite signs at the ends of the rod.

(d) Retarding force = IBL

9 mV / 9 mΩ x 0.5 T x 0.15 m = 75 x 10^-3 N

9. e) Power expended by an external agent against the above retarding force to keep the rod moving uniformly at 12 cm s' = 75 x 10^-3x 12 x 10^-2= 9.0 x 10^-3 W

When K is open, no power is expended.

(f) I² R = 1x1x 9 x 10^-3 = 9.0 x 10^-3 W

The source of this power is the power provided by the external agent as calculated above.

g) Zero: motion of the rod does not cut across the field lines. [Note: length of Pg has been considered above to be equal to the spacing between the rails.]

Ques. 3 According to Ohm's law, the current flowing through a conductor is directly proportional to the potential difference across the ends of the conductor i.e I ∝ V ⇒ V/ I = R, where R is resistance of the conductor Electrical resistance of a conductor is the obstruction posed by the conductor to the flow of electric current through it. It depends upon length, area of cross-section, nature of material and temperature of the conductor.

We can write R∝l/A or R=ρl/A

Where ρ is electrical resistivity of the material of the conductor.

(i) Dimensions of electric resistance is

(a) [ML2 T−2 A−2]

(b) [ML2T−3A−2]

(c) [M−1 L−2 T−1 A]

(d) [M−1L2T2A−1]

(ii) If 1μA current flows through a conductor when potential difference of 2 volt is applied

across its ends, then the resistance of the conductor is

(a) 2×106Ω

(b) 3×105Ω

(c) 1.5×105Ω

(d) 5×107Ω

(iii) Specific resistance of a wire depends upon

(a) length

(b) cross-sectional area

(c) mass

(d) none of these

(iv) The slope of the graph between potential difference and current through a conductor is

(a) a straight line

(b) curve

(c) first curve then straight line

(d) first straight line then curve

(v) The resistivity of the material of a wire 1.0 m long, 0.4 mm in diameter and having a

resistance of 2.0 ohm is

(a) 57×10−6Ωm

(b) 5.25×10−7Ωm

(c) 7.12×10−5Ωm

(d) 2.55×10−7Ωm

Answer-

Now, ρ = RA/ l = 2×4π×10−8/ 1 = 2.55×10−7Ωm

(i) (b)

(ii) (b) As I = ε/ (R+r)

In first case, I = 0.5 A; R = 12 Ω

0.5 = ε/ (12+r) ⇒ ε = 6.0 + 0.5 r ....(i)

In second case I = 0.25 A; R=25 Ω

ε = 6.25 + 0.25 r ...(ii)

From equation (i) and (ii), r = 1 Ω

(iii) (b)

(iv) (a) Current in the circuit I= ε/ (R+r)

Power delivered to the resistance R is P = I2R = E2R/ (R+r)2

It is maximum when dP/ dR = 0

dP/ dR = E2[(r+R)2−R(r+R)]/ (r+R)4 = 0

or (r+R)2 = 2R(r+R) or R = r

(v) (b) For first case, ε/ (R+r) = 10/ R ...(i)

For second case, ε/ (5R+r) = 30/ 5R

Dividing (i) by (ii), we get r = 5R

From (i), ε/ (R+5R) = 10/ R ,ε = 60 V

Ques 4: Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms. Also, Resistivity is the electrical resistance of a conductor of unit cross-sectional area, and unit length. … A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents.

Resistivity is independent of:

1. nature of material

2. temperature

3. dimensions of material

4. none of the above

As compared to short wires, long wires have _______ resistance.

1. more
2. less
3. same
4. zero

As compared to thin wires, thick wires have _______ resistance.

1. more

2. less

3. same

4. zero

The resistance of a wire depends upon:

1. cross-sectional area

2. length of wire

3. wire’s nature

4. all of the above

A copper wire having the same size as steel wire have:

1. more resistance

2. less resistance

3. same resistance

4. none of the above

Answer:

1. Resistivity is independent of: dimensions of material.

2. As compared to short wires, long wires have _______ resistance: more.

3. As compared to thin wires, thick wires have _______ resistance: less.

4. The resistance of a wire depends upon: all of the above.

5. A copper wire having the same size as steel wire have: less resistance.

Ques 5: The Bohr model of the atom was proposed by Neil Bohr in 1915. It came into existence with the modification of Rutherford’s model of an atom. Rutherford’s model introduced the nuclear model of an atom, in which he explained that a nucleus (positively charged) is surrounded by negatively charged electrons.

Which of the following statements does not form a part of Bohr’s model of a hydrogen atom?

1. The energy of the electrons in the orbit is quantized

2. The electron in the orbit nearest the nucleus has the lowest energy

3. Electrons revolve in different orbits around the nucleus

4. The position and velocity of the electrons in the orbit cannot be determined simultaneously

What is in the center of the Rutherford model?

1. Single proton

2. Multiple electrons

3. A nucleus

4. Neutrons

When an electron remains between orbits its momentum is:

1. quantized

2. emitted

3. dequantized

4. none of the above

What were the limitations of the Rutherford model which could not explain the observed features of atomic spectra explained in Bohr’s model of a hydrogen atom?

1. It must emit a continuous spectrum

2. It loses its energy

3. Gaining its energy

4. A discrete spectrum

When an electron jumps from its orbit to another orbit, energy is:

1. emitted only

2. absorbed only

3. both (a) and (b)

4. none of these

Answer: 

1. The position and velocity of the electrons in the orbit cannot be determined simultaneously.

2. A nucleus.

3. none of the above.

4. It must emit a continuous spectrum.

5. both (a) and (b).

Also Check:

CBSE Class 12 Physics Half Yearly Sample Paper 2025

Class 12 Half Yearly Exam Physics: Important Numerical Questions