Question

A wire of resistance \(50 \, \Omega\) is elongated to get a resistance of \(60.5 \, \Omega\). The unelongated length of the wire is \(100 \, cm\). The wire is elongated by:

• A. 6
• B. 8
• C. 10
• D. 12

Hint:

When wire is stretched with constant volume, resistance varies as \(R \propto L^2\).

Answer 1

The Correct Option is C

Step 1: Write how resistance depends on length when volume stays fixed.
Stretching a wire keeps its volume $AL$ constant, so $A_2 = \frac{A_1 L_1}{L_2}$, and since $R = \rho \frac{L}{A}$, \[ \frac{R_2}{R_1} = \left(\frac{L_2}{L_1}\right)^2 \]
Step 2: Let the fractional elongation be $x$, so $L_2 = L_1(1+x)$.
\[ \frac{R_2}{R_1} = (1+x)^2 = 1 + 2x + x^2 \]
Step 3: Substitute the given resistances.
\[ \frac{60.5}{50} = 1.21 = 1 + 2x + x^2 \quad\Rightarrow\quad x^2 + 2x - 0.21 = 0 \]
Step 4: Solve this quadratic using the quadratic formula, instead of taking a square root of the ratio directly.
\[ x = \frac{-2 \pm \sqrt{4 + 0.84}}{2} = \frac{-2 \pm \sqrt{4.84}}{2} = \frac{-2 \pm 2.2}{2} \] Taking the positive root, \[ x = \frac{0.2}{2} = 0.1 \]
Step 5: Convert to the actual elongation.
With $L_1 = 100$ cm, the extension is $0.1 \times 100 = 10$ cm, a $10\%$ increase.

Final Answer:
\[ \boxed{10\%} \]