JEE PYQ: Current Electricity Question 1
Question 1 - 2021 (16 Mar 2021 Shift 1)
A conducting wire of length ‘$l$’, area of cross-section A and electric resistivity $\rho$ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be:
(1) $\frac{1}{4} \frac{VA}{\rho l}$
(2) $\frac{3}{4} \frac{VA}{\rho l}$
(3) $\frac{1}{4} \frac{\rho l}{VA}$
(4) $4 \frac{VA}{\rho l}$
Show Answer
Answer: (1)
Solution
When length is doubled to $2l$ and area is halved to $A/2$, resistance $R = \frac{\rho(2l)}{(A/2)} = \frac{4\rho l}{A}$. Current $= \frac{V}{R} = \frac{VA}{4\rho l}$.
BETA
