Question

A conductor of length \( l \) is connected across an ideal cell of emf E. Keeping the cell connected, the length of the conductor is increased to \( 2l \) by gradually stretching it. If R and \( R' \) are initial and final values of resistance and \( v_d \) and \( v_d' \) are initial and final values of drift velocity, find the relation between:
\( R' \) and \( R \)
\( R' = 4R \)

Hint:

When a conductor is stretched, its resistance increases due to the increase in length and decrease in cross-sectional area, while the drift velocity increases due to the reduced cross-sectional area.

Answer 1

(i) Relation between R′ and R:

The resistance R of a conductor is given by:

\(R = \frac {ρ l}{A}\)

When the length is increased to 2l, the cross-sectional area A decreases to A/2 (assuming the volume remains constant). Thus:

\(R′ = \frac {ρ (2l)}{(A/2) }= \frac {4ρ l}{A} = 4R\)

So, \(R^′ = 4R.\)

(ii) Relation between v′_d and v_d:

Drift velocity v_d is given by:

\(v_d =\frac{I}{(neA)}\)

When the length is doubled, the current I remains the same (since the cell is ideal), but the cross-sectional area A is halved. Thus:

\(v^′_d = 2v_d\)

So, \(v^′_d = 2v_d\).