CBSE Class 12 Physics 2023 : Important Case Study Based Questions with Solution

CBSE Class 12 Physics 2023 : Important Case Study Based Questions with Solution

The annual CBSE exam is extremely important for the students, and the next paper will be Physics on 6th March. The subject also demands immense practice, there will be five sections in the Physics exam, and the last section E will have two case study-based questions of 5 marks each. These questions are very important from the exam point of view, and you can read the solved versions here.

Why are Case Study Questions Beneficial for Class 12 Physics?

It is a mind-boggling subject which is at par with Mathematics for non-medical stream students. However, Physics is challenging for all students because of its conceptual and numerical-based nature.
 
Physics requires a clear understanding of fundamentals, memorization of many formulas, derivation and expert calculation skills as well as the ability to apply them to complex problems. All this requires practice, you can read here Case Study Questions for Class 12 Physics in solved version.

CBSE Class 12 Physics - Important Questions Case Study

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

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

CBSE Class 12 Study Materials

CBSE Class 12 Syllabus 2022-23	CBSE Class 12 Previous Year Papers
NCERT Books For Class 12 Books	NCERT Class 12 Solutions
CBSE Class 12 Full Study Material	CBSE Class 12 Sample Paper 2022-2