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 = 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′ = ρ (2l) / (A/2) = 4ρ l / A = 4R

So, R′ = 4R.

(ii) Relation between v′_d and v_d:

Drift velocity v_d is given by:

v_d = 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.