CUET · PHYSICS · PYQ PAPER 2025
A capacitor of unknown capacitance is connected with a battery of V volts. The charge stored in it is \( 200 \, \mu\text{C} \). When the potential of the capacitor is reduced by 100 volts, the charge stored in it becomes \( 150 \, \mu\text{C} \). The unknown capacitance is:
- A \( 0.5 \, \mu\text{F} \)
- B 1 pF
- C \( 4 \, \mu\text{F} \)
- D \( 1.5 \, \mu\text{F} \)
Answer & Solution
Correct Answer
(A) \( 0.5 \, \mu\text{F} \)
Step-by-step Solution
Detailed explanation
\( Q = CV \) \( \frac{Q_1}{Q_2} = \frac{V_1}{V_2} \implies \frac{200}{150} = \frac{V}{V-100} \) \( 4(V-100) = 3V \) \( 4V - 400 = 3V \implies V = 400 \, \text{V} \) \( C = \frac{Q_1}{V_1} = \frac{200 \, \mu\text{C}}{400 \, \text{V}} = 0.5 \, \mu\text{F} \)
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