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?

• A

  $4 F$
• B

  $6 \mathrm{~F}$
• Correct answer
  C

  $9 F$
• D

  $F$

Solution

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 $\ell / 3$ (same volume 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) $, \frac{F}{A} \times \frac{\ell}{\Delta x}=\frac{F^{\prime}}{3 A} \times \frac{\ell}{3 \Delta x}$ $\Rightarrow F^{\prime}=9 F$

Hence, the correct answer is option: 'C'.