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) Relationship between R′ and R:

The resistance R of a conductor is defined as:

\(R = \frac {ρ l}{A}\)

If the length is doubled to 2l, and assuming constant volume, the cross-sectional area A becomes A/2. Consequently:

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

Therefore,\(R^′ = 4R.\)

(ii) Relationship between v′_d and v_d:

Drift velocity v_d is expressed as:

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

When the length is doubled, the current I remains constant (assuming an ideal cell), while the cross-sectional area A is halved. Thus:

\(v^′_d = 2v_d\)

Hence,\(v^′_d = 2v_d\).