Two wires A and B are made of same material having ratio of lengths and their diameters ratio . If both the wires are stretched using same force, what would be the ratio of their respective elongations? [2025]

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Solution

Q.
Two wires A and B are made of same material having ratio of lengths $\frac{L_{A}}{L_{B}} = \frac{1}{3}$ and their diameters ratio $\frac{d_{A}}{d_{B}} = 2$ . If both the wires are stretched using same force, what would be the ratio of their respective elongations [2025]

1 1 : 6

2 1 : 12

3 3 : 4

4 1 : 3

Ans.
(2)

$\frac{L_{A}}{L_{B}} = \frac{1}{3}$ and $\frac{d_{A}}{d_{B}} = 2$

$Strain = \frac{△ L}{L} = \frac{Stress}{Y} = \frac{F}{A Y} \Rightarrow △ L = \frac{F L}{A Y}$

$△ L_{A} = \frac{F_{A} L_{A}}{A_{A} Y_{A}} and △ L_{B} = \frac{F_{B} L_{B}}{A_{B} Y_{B}}$

Given $F_{A} = F_{B}$ and $Y_{A} = Y_{B}$

$\frac{△ L_{A}}{△ L_{B}} = \frac{\frac{F_{A} L_{A}}{A_{A} Y_{A}}}{\frac{F_{B} L_{B}}{A_{B} Y_{B}}} = (\frac{L_{A}}{L_{B}}) (\frac{A_{B}}{A_{A}})$

$\frac{△ L_{A}}{△ L_{B}} = (\frac{L_{A}}{L_{B}}) (\frac{\frac{π}{4} d_{B}^{2}}{\frac{π}{4} d_{A}^{2}}) = (\frac{L_{A}}{L_{B}}) {(\frac{d_{B}}{d_{A}})}^{2}$

$\frac{△ L_{A}}{△ L_{B}} = (\frac{1}{3}) {(\frac{1}{2})}^{2} = \frac{1}{12}$

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