Question

Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has length 2l and diameter 2r. If the two wires are stretched by the same force, the elongation in A divided by elongation in B is

• A. 1/8
• B. 1/4
• C. 4
• D. 8

Answer 1

The correct option is A

1/8

Let the magnitude of constant force be F.
For wire A:

Stress=$\frac{F}{A}=\frac{F}{\left(\frac{\pi r^2}{4}\right)}$
Strain developed =$\frac{\Delta L_A}{l}$
young's modulus,$Y=\frac{4Fl}{\pi r^2\Delta L_A}$
for wire B
Stress=$\frac{F}{A}=\frac{F}{\left(\frac{\pi r^2}{16}\right)}=\frac{16F}{\pi r^2}$
Strain=$\frac{\Delta L_g}{2l}$
$Y=\frac{2\times 16Fl}{\pi r^2\Delta L_B}=\frac{32Fl}{\pi r^2\Delta L_B}$
Equating both [as both are of same material} $\Rightarrow \frac{4Fl}{\pi r^2\Delta L_A}=\frac{32Fl}{\pi r^2\Delta L_B}$
$\Rightarrow \frac{\Delta L_A}{\Delta L_B}=\frac{1}{8}$