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Question
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
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Solution
\[\text{ Let the Young's modulus of the wire's material be Y .} \]
\[\text{ Here: }\]
\[\text{ Force = F }\]
\[ A_1 = \pi r^2 \]
\[ L_1 = l\]
\[ A_2 = \pi \left( \frac{r}{2} \right)^2 = \frac{\pi r^2}{4}\]
\[ \text{ L }_2 = 2\text{l }\]
\[\text{ Let the elongation in A be x and that in B be y }. \]
\[\text{ Since the Young's modulus for both the wires is the same: } \]
\[Y = \frac{\frac{F}{A_1}}{\frac{x}{l}} = \frac{\frac{F}{A_2}}{\frac{y}{2l}}\]
\[ \Rightarrow \frac{x}{y} = \frac{A_2}{2 A_1}\]
\[ \Rightarrow \frac{x}{y} = \frac{1}{8}\]
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