Question

Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3 A. If the length of wire 1 increased by $\Delta x$ on applying force F, how much force is needed to stretch wire 2 by the same amount?

• A. $4F$
• B. $6F$
• C. $9F$
• D. $F$

Answer 1

Answer: (C)

$9F$

$Y=\frac{FL}{A\Delta L}$
or $F=\frac{YA\Delta L}{L}=\frac{Y{A}^2\Delta L}{AL}$

$=\frac{Y{A}^2\Delta L}{V}=\frac{Y{A}^2\Delta x}{V}$

$F^{\prime }=\frac{Y{A^{\prime }}^2\Delta L}{V}=\frac{Y{A^{\prime }}^2\Delta x}{V}$
where, $AL=V=$ Volume of wire, Young's modulus is the same as both the wires are made of same material. It is given that, both the wires have same volume and same extension in length. Let's assume $F^{\prime }$ force is needed to stretch wire 2 and its area of cross-section is A'.
$\therefore \frac{F^{\prime }}{F}=\frac{{A^{\prime }}^2}{A^2}=\frac{{\left(3A\right)}^2}{A^2}=9$
$F^{\prime }=9F$