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 $3A$. 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. $F$
• B. $4F$
• C. $6F$
• D. $9F$

Answer 1

The correct option is D $9F$

Two wires are made of the same material and have the same volume.
$\therefore A_1{l}_1=A_2{l}_2\Rightarrow l_2=\frac{A_1{l}_1}{A_2}=\frac{A\times l_1}{3A}=\frac{l_1}{3}$
$\Rightarrow \frac{l_1}{l_2}=3$

We know that, change in length $\Delta l=\frac{F}{A\gamma }\times l$
where $\gamma =$Young's modulus.
For wire $1$;
$\Delta x_1=\frac{F_1}{A\gamma }\times l_1......\left(i\right)$
For wire $2$;
$\Delta x_2=\frac{F_2}{3A\gamma }\times l_2.....\left(ii\right)$
Here $\Delta x_1=\Delta x_2$
$\frac{F_2}{3A\gamma }{l}_2=\frac{F_1}{A\gamma }{l}_1$
$F_2=3F_1\times \frac{l_1}{l_2}$
Put the value of $\frac{l_1}{l_2}$ here
$=3F_1\times 3=9F$