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 $2l$ 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,
$\frac{x_p}{x_Q}=\left(\frac{L_P}{R_P^2}\right)\left(\frac{R_Q^2}{L_Q}\right)=\left(\frac{L_P}{L_Q}\right).{\left(\frac{R_Q}{R_P}\right)}^{{}^2}$
Here $L_P=l,R_P=\frac{r}{2},L_Q=21$ and $R_Q=\frac{r}{4}$
$\Rightarrow \frac{x_p}{x_Q}=\left(\frac{l}{2l}\right)\left(\frac{1}{2}\right)=\frac{1}{8}$