Question

Two wires $P$ and $Q$ are made of same material. The wire $P$ has a length $l$ and diameter $r$ while the wire $Q$ has a length $2 l$ and diameter $r / 2$. If the two wires are stretched by the same force, the elongation in $P$ divided by the elongation in $Q$ is

• A. 1/8
• B. 1/4
• C. 4
• D. 8

Answer 1

Answer: (A)

$Y=\frac{F/A}{x / L}$
So, the elongation x is given as.
$x=\frac{F.L}{A.Y}=\frac{F.L}{\pi R^{2}.Y}$
For wires P and Q, made of same material so Young’s modulus will be same and stretched by the same force,
$\frac{x_{p}}{x_{Q}}=\left(\frac{L_{P}}{R_{P}^{2}}\right)\left(\frac{R^{2}_{Q}}{L_{Q}}\right)=\left(\frac{L_{P}}{L_{Q}}\right).\left(\frac{R_{Q}}{R_{P}}\right)^{^2}$
Here $L_{P}=l, R_{P}=\frac{r}{2}, L_{Q}=21$ and $R_{Q}=\frac{r}{4}$
$\Rightarrow \frac{x_{p}}{x_{Q}}=\left(\frac{l}{2l}\right)\left(\frac{1}{2}\right)=\frac{1}{8}$