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. $4F$
• B. $6F$
• C. $9F$
• D. $F$

Answer 1

The correct option is C $9F$

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 cross-section 3A will be l/3(same volume as wire 1 implies Al is same, thus 3A, l/3 and A, l are the respective values)
For wire 1, $Y=\frac{FlA}{\Delta x/l}$ --- (1)
For wire 2, $Y=\frac{F/3A}{\Delta x/\left(l/3\right)}$--- (2)
Using (1) and (2) we get, $\frac{F}{A}\times \frac{l}{\Delta x}=\frac{F}{3A}\times \frac{l}{3\Delta x}\Rightarrow F^{\prime }=9F$