Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. lf the length of the first wire is increased by $∆ l$ on applying a force F, how much force is needed to stretch the second wire by the same amount ?

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answer is B.

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Detailed Solution

We know $Y = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{\frac{∆ l}{l}}$

$\Rightarrow Y = \frac{F ∆, l}{Al} = \frac{Fl}{A ∆ l}$

For 1st wire; $Y = \frac{F_{1} l_{1}}{A . ∆ l} - - - - - - (i)$

For IInd wire; $Y = \frac{F_{2} l_{2}}{3 A . ∆ l} - - - - - (ii)$

As volume of the wire is same

Hence, ${Al}_{1} = 3 {Al}_{2}$

or $\frac{l_{1}}{l_{2}} = 3 - - - - - (iii)$

From (i) and (ii); $\frac{F_{1}}{F_{2}} = \frac{l_{2}}{3 l_{1}}$ -----------(iv)

From (iii) and (iv); $\frac{F_{1}}{F_{2}} = \frac{1}{9} or F_{2} = 9 F_{1}$

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