CUET · PHYSICS · PYQ PAPER 2023
A conductor of length ' \(L\) ' is connected to a battery of emf ' \(E\) '. The drift velocity of electron in the conductor is ' \(v_d^{\prime}\). The drift velocity of electron in the conductor on stretching it four times, to length 4 L keeping emf ' \(E\) ' constant will be :
- A \(4 v_d\)
- B \(0.25 v_d\)
- C \(0.5 v_d\)
- D \(2 v_d\)
Answer & Solution
Correct Answer
(B) \(0.25 v_d\)
Step-by-step Solution
Detailed explanation
\(v_d = \frac{I}{neA} = \frac{E/R}{neA} = \frac{E/(\rho L/A)}{neA} = \frac{E}{ne\rho L}\) \(v_d' = \frac{E}{ne\rho L'} = \frac{E}{ne\rho (4L)} = \frac{1}{4} v_d = 0.25 v_d\)
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