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

#### Options

1/8

1/4

4

8

#### Solution

\[\text{ Let the Young's modulus of the wire's material be Y .} \]

\[\text{ Here: }\]

\[\text{ Force = F }\]

\[ A_1 = \pi r^2 \]

\[ L_1 = l\]

\[ A_2 = \pi \left( \frac{r}{2} \right)^2 = \frac{\pi r^2}{4}\]

\[ \text{ L }_2 = 2\text{l }\]

\[\text{ Let the elongation in A be x and that in B be y }. \]

\[\text{ Since the Young's modulus for both the wires is the same: } \]

\[Y = \frac{\frac{F}{A_1}}{\frac{x}{l}} = \frac{\frac{F}{A_2}}{\frac{y}{2l}}\]

\[ \Rightarrow \frac{x}{y} = \frac{A_2}{2 A_1}\]

\[ \Rightarrow \frac{x}{y} = \frac{1}{8}\]