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Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area $A$ and wire 2 has cross-sectional area $3 A$. If the length of wire 1 increases by $\Delta x$ on applying force $F,$ how much force is needed to stretch wire 2 by the same amount?
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The correct answer is:
$9 F$
As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of crosssection $3 A$ will be $\ell / 3$ (samevolume as wire 1 ). For wire $1, \quad Y=\frac{F / A}{\Delta x / \ell}$
For wire $2, Y=\frac{F^{\prime} / 3 A}{\Delta x /(\ell / 3)}$
From (i) and (ii) $, \frac{F}{A} \times \frac{\ell}{\Delta x}=\frac{F^{\prime}}{3 A} \times \frac{\ell}{3 \Delta x}$
$\Rightarrow F^{\prime}=9 F$
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