Question

A potential difference $V$ is applied across the ends of a cylindrical conductor of length $l$, area of cross-section $A$ and resistance $R$. It is gradually stretched till its length is tripled. How will (i) the drift speed of electrons in the conductor, and (ii) the resistance of the conductor be affected? Justify your answers.

Hint:

Volume stays constant, so the area becomes A/3. Then use \(R = \rho l/A\) and \(v_d = eE\tau/m\) with \(E = V/l\).

Answer 1

Step 1: Understanding the Concept:
The conductor is stretched, not cut or added to. So the volume of the material stays the same.
The number density of free electrons \(n\) and the relaxation time \(\tau\) also stay the same, because the material is the same.

Step 2: Find the new area:
Volume = length \(\times\) area. So \(l \cdot A = 3l \cdot A'\).
\[ A' = \frac{A}{3} \]

Step 3: Effect on resistance (ii):
Use \(R = \rho \dfrac{l}{A}\).
\[ R' = \rho \frac{3l}{A/3} = 9\,\rho \frac{l}{A} = 9R \]
The resistance becomes 9 times the original value. The length is tripled and the area falls to one third, so both changes raise the resistance.

Step 4: Effect on drift speed (i):
The potential difference \(V\) across the ends is the same. The electric field inside is \(E = V/l\).
\[ E' = \frac{V}{3l} = \frac{E}{3} \]
The drift speed is \(v_d = \dfrac{e E \tau}{m}\). Here \(e\), \(\tau\) and \(m\) do not change, so \(v_d\) is proportional to \(E\).
\[ v_d' = \frac{v_d}{3} \]

Final Answer:
(i) The drift speed becomes one third of its original value. (ii) The resistance becomes 9 times its original value. \[ \boxed{v_d' = \frac{v_d}{3}, \quad R' = 9R} \]