CBQ Practice›Class 12 Physics
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Class 12 Physics
CBQ Practice

Competency Based Questions · 3 chapters · 6 CBQ sets

Question types:Case StudySource BasedAssertion–Reason
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Ch 1

Electric Charges and Fields

2 sets
CBQ 1Case StudyElectric Field Due to Point Charges4 marks

Read the passage

Two point charges +4 μC and −4 μC are placed 20 cm apart in vacuum. A physics student is asked to find the electric field at various points around these charges. The student recalls that the electric field due to a point charge q at distance r is E = kq/r², directed radially outward for positive charges and radially inward for negative charges, where k = 9 × 10⁹ N m²/C². The student also learns that electric field lines start from positive charges and end at negative charges, never crossing each other. For the given dipole arrangement, the student finds the electric field at the midpoint of the line joining the two charges and at a point on the perpendicular bisector.
1

The electric field at the midpoint between charges +4 μC and −4 μC (placed 20 cm apart) is directed:

1M
(A)From −4 μC towards +4 μC
(B)From +4 μC towards −4 μC
(C)Perpendicular to the line joining charges
(D)Zero at the midpoint
2

Electric field lines never intersect each other because:

1M
(A)They carry the same charge
(B)At any point, the electric field has a unique direction, and two lines intersecting would imply two directions
(C)They travel at the speed of light
(D)They are parallel in a uniform field
3

The SI unit of electric field is:

1M
(A)N/m
(B)V/m (or N/C)
(C)C/m²
(D)J/C
4

Calculate the electric field at the midpoint of the two charges (+4 μC and −4 μC, 20 cm apart).

1M
CBQ 2Assertion–Reason1 mark
A
Assertion

The electric flux through a closed surface is zero if no net charge is enclosed within the surface.

R
Reason

According to Gauss's Law, the total electric flux through any closed surface is equal to the net charge enclosed divided by ε₀.

(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Ch 3

Current Electricity

2 sets
CBQ 1Case StudyWheatstone Bridge and Metre Bridge4 marks

Read the passage

A student in a physics lab is using a Wheatstone bridge circuit to find an unknown resistance. The Wheatstone bridge consists of four resistances P, Q, R, and S arranged in a diamond shape with a galvanometer between the middle junctions and a battery across the other two junctions. When the bridge is balanced, no current flows through the galvanometer. The balance condition is P/Q = R/S. The student sets P = 100 Ω, Q = 200 Ω, and R = 150 Ω. He adjusts S until the galvanometer shows zero deflection. The metre bridge is a practical application of the Wheatstone bridge principle where a resistive wire of 100 cm is used.
1

The balance condition for the Wheatstone bridge is:

1M
(A)P + Q = R + S
(B)P/Q = R/S
(C)PQ = RS
(D)P − Q = R − S
2

Using P = 100 Ω, Q = 200 Ω, R = 150 Ω and the balance condition, the unknown resistance S is:

1M
(A)100 Ω
(B)200 Ω
(C)300 Ω
(D)75 Ω
3

In a balanced Wheatstone bridge, the galvanometer reads:

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(A)Maximum current
(B)Zero current
(C)Half the total current
(D)Infinite resistance
4

In a metre bridge experiment, the balance point is found at 40 cm from the left end when a known resistance of 30 Ω is in the left gap. Find the unknown resistance.

1M
CBQ 2Assertion–Reason1 mark
A
Assertion

The internal resistance of an ideal voltmeter is infinity and the internal resistance of an ideal ammeter is zero.

R
Reason

A voltmeter measures potential difference and must be connected in parallel without drawing current; an ammeter measures current and must not impede the current flow.

(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Ch 9

Ray Optics and Optical Instruments

2 sets
CBQ 1Case StudyTotal Internal Reflection and Optical Fibres4 marks

Read the passage

Optical fibres are thin glass or plastic fibres used to transmit light signals over long distances. They work on the principle of total internal reflection. When light travels from a denser medium (glass, n = 1.5) to a rarer medium (air, n = 1.0), it bends away from the normal. If the angle of incidence exceeds the critical angle, the light is completely reflected back into the denser medium — this is total internal reflection. The critical angle θ_c can be found using sin(θ_c) = 1/n, where n is the refractive index of the denser medium. Optical fibres are used in telecommunications, medical endoscopes, and decorative lighting. India's national broadband network extensively uses optical fibre technology.
1

Total internal reflection occurs when light passes from:

1M
(A)Rarer to denser medium at any angle
(B)Denser to rarer medium at an angle greater than the critical angle
(C)Denser to rarer medium at an angle less than the critical angle
(D)Rarer to denser medium at an angle equal to the critical angle
2

The critical angle for glass (n = 1.5) with respect to air is approximately:

1M
(A)30°
(B)42°
(C)60°
(D)90°
3

Optical fibres are used in medical endoscopes because:

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(A)They carry electrical signals
(B)They can transmit light through bends using total internal reflection
(C)They are cheaper than wires
(D)They are transparent to X-rays
4

State the two conditions necessary for total internal reflection to occur.

1M
CBQ 2Assertion–Reason1 mark
A
Assertion

A concave mirror is used as a shaving mirror because it forms a magnified, erect, virtual image when the object is placed within the focal length.

R
Reason

When an object is placed between the pole and focus of a concave mirror, the image formed is virtual, erect, and magnified.

(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true

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