Question

If the drift velocity of electrons in a copper wire of cross-sectional area $2\,mm^2$ carrying current $I$ is $v_1$ and that in another copper wire of cross-sectional area $1.5\,mm^2$ carrying current $2I$ is $v_2$, then the ratio $v_1 : v_2$ is

• A. $3:8$
• B. $2:4$
• C. $8:3$
• D. $4:2$
• E. $1:3$

Hint:

Drift velocity inversely proportional to area.

Answer 1

The Correct Option is A

Concept:
• Drift velocity: \[ v_d = \frac{I}{nqA} \]

Step 1: Write ratios
\[ v \propto \frac{I}{A} \]

Step 2: Compute
\[ v_1 : v_2 = \frac{I}{2} : \frac{2I}{1.5} \] \[ = \frac{1}{2} : \frac{2}{1.5} = \frac{1}{2} : \frac{4}{3} \]

Step 3: Simplify
\[ = 3 : 8 \] Final Conclusion:
Option (A)