Question

A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ________ cm.

Hint:

$\Delta L \propto \frac{L}{d^2}$. Doubling length increases elongation $(\times 2)$, but doubling diameter decreases it $(\times 1/4)$, resulting in a net factor of $1/2$.

Answer 1

Correct Answer: 2

Step 1: Elongation formula is $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi (d/2)^2 Y} = \frac{4FL}{\pi d^2 Y}$.
Step 2: Given $\Delta L_1 = 0.04 \text{ m}$. New conditions: $L' = 2L$ and $d' = 2d$.
Step 3: $\Delta L_2 = \frac{4F(2L)}{\pi (2d)^2 Y} = \frac{8FL}{4\pi d^2 Y} = \frac{1}{2} \left( \frac{4FL}{\pi d^2 Y} \right) = \frac{1}{2} \Delta L_1$.
Step 4: $\Delta L_2 = \frac{0.04}{2} = 0.02 \text{ m} = 2 \text{ cm}$.