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$(A)$ The drift velocity $v_d$ is given by the formula $v_d = \frac{eE	au}{m}$, where $E = \frac{V}{L}$.Thus, $v_d = \frac{eV	au}{mL}$.Since all rod

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$v_A = v_B > v_C$

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$(A)$ The drift velocity $v_d$ is given by the formula $v_d = \frac{eE	au}{m}$, where $E = \frac{V}{L}$.Thus, $v_d = \frac{eV	au}{mL}$.Since all rods are made of copper and are at the same temperature, the relaxation time $	au$ and the mass of electron $m$ are constant.Therefore, $v_d \propto \frac{V}{L}$.For rod $(A)$: $v_A \propto \frac{V}{L} = 1 \cdot (V/L)$.For rod $(B)$: $v_B \propto \frac{2V}{2L} = 1 \cdot (V/L)$.For rod $(C)$: $v_C \propto \frac{2V}{3L} = 0.67 \cdot (V/L)$.Comparing the values, we get $v_A = v_B > v_C$.

Rod	Length,Diameter,Potential Difference
$(A)$	$L, 3d, V$
$(B)$	$2L, d, 2V$
$(C)$	$3L, 2d, 2V$

Three copper rods are subjected to different potential differences. Compare the drift speed of electrons through them. Assume that all $3$ are at the same temperature.
Rod Length,Diameter,Potential Difference
$(A)$ $L, 3d, V$
$(B)$ $2L, d, 2V$
$(C)$ $3L, 2d, 2V$

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Three copper rods are subjected to different potential differences. Compare the drift speed of electrons through them. Assume that all $3$ are at the same temperature.RodLength,Diameter,Potential Difference$(A)$$L, 3d, V$$(B)$$2L, d, 2V$$(C)$$3L, 2d, 2V$