Important Case Study Questions for Class 12 Physics: Check here the important case study based questions of Section E in the CBSE Class 12 Physics exam 2023 for last minute preparation.
CBSE Class 12 Physics Exam 2023 Important Questions: The Central Board of Secondary Education is the largest and one of the most prestigious school boards in India, and millions of students are currently enrolled in it. The annual CBSE Exams are of utmost importance for the students, and the next paper will be of Physics on 6 March. Physics is a compulsory subject in CBSE Class 12 science stream and is essential for both medical and non-medical science stream students. The subject also demands immense practice, especially the important topics for class 12 Physics, like optics and electromagnetics. There will be five sections in the 2023 CBSE Class 12 Physics exam, and the last section E will comprise two case study-based questions of 5 marks. These questions are very important from the exam point of view, and you can read the solved versions here.
Must Read: CBSE Physics Class 12 Syllabus 2023
Must Read: CBSE Topper Answer Sheet Class 12 Physics
CBSE Class 12 Physics Unit Wise Marks Distribution 2023
Name of the unit
Magnetic Effects of Current and Magnetism
Electromagnetic Induction and Alternating Currents
Dual Nature of Radiation and Matter
Atoms and Nuclei
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.
Related: CBSE Physics Previous Year Question Paper Class 12
Related: CBSE Class 12 Physics Sample Paper 2023
Important Questions For Class 12 Physics CBSE Board
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 V
- 2 V
- 3 V
- 4 V
Answer: (2) 2V
B: If the resistance of the ammeter is 2 ohm, then the resistance of the voltmeter is:-
- 1 ohm
- 2 ohm
- 3 ohm
- 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
- ⅗ ohm
- 2/7 ohm
- 3/7 ohm
- 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?
(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]
(c) [M−1 L−2 T−1 A]
(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
(iii) Specific resistance of a wire depends upon
(b) cross-sectional area
(d) none of these
(iv) The slope of the graph between potential difference and current through a conductor is
(a) a straight line
(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
Now, ρ = RA/ l = 2×4π×10−8/ 1 = 2.55×10−7Ωm
(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
(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
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