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, \(v^′_d = 2v_d\).

Answer 2

(i) Relation between R′ and R: 

The resistance R of a conductor is given by the formula:

\(R = \frac{\rho l}{A}\)

Here:

• \(\rho\) = resistivity of the material
• \(l\) = original length of the conductor
• \(A\) = cross-sectional area

 

Now, when the length of the conductor is increased to \(2l\), the volume remains constant. This means if length doubles, the cross-sectional area becomes half, i.e., \(A/2\).

Substituting the new values:

\(R' = \frac{\rho (2l)}{(A/2)}\)

Simplifying: \(R' = \frac{2\rho l}{A/2} = \frac{2\rho l \times 2}{A} = \frac{4\rho l}{A}\)

But \(\frac{\rho l}{A} = R\), so: \(R' = 4R\)

Therefore, \( R' = 4R \).

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(ii) Relation between v′_d and v_d:

The drift velocity of electrons is given by:

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

Here:

• \(I\) = current in the conductor
• \(n\) = number of electrons per unit volume
• \(e\) = charge of an electron
• \(A\) = cross-sectional area of the conductor

 

When the length is doubled, the resistance increases but we are told the cell is ideal, so the current \(I\) remains the same. However, the cross-sectional area is halved (\(A/2\)).

Substituting the new area into the formula:

\(v'_d = \frac{I}{ne(A/2)}\)

Simplifying: \(v'_d = \frac{I}{(neA/2)} = \frac{2I}{neA}\)

Since \( \frac{I}{neA} = v_d \), we get: \(v'_d = 2v_d\)

Therefore, \( v'_d = 2v_d \).