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 $△x$ 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)

$9F$

As shown in the figure, the wires will have the same Young's modulus (same material) an the length of the wire of area of cross-section $3A$ will be $l/3$ (same volume as wire $1$).
For wire $1,Y=\frac{F/A}{△x/l}.....\left(i\right)$
For wire $2,Y=\frac{F^{\prime }/3A}{△x/\left(l/3\right)}....\left(ii\right)$
From (i) and (ii), $\frac{F}{A}\times \frac{l}{△x}=\frac{F^{\prime }}{3A}\times \frac{l}{3△x}\Rightarrow F^{\prime }=9F$.