Question

Two separate wires $A$ and $B$ are stretched by $2$ $\text{mm}$ and $4$ $\text{mm}$ respectively, when they are subjected to a force of $2$ $\text{N}$. Assume that both the wires are made up of same material and the radius of wire $B$ is $4$ times that of the radius of wire $A$. The length of the wires $A$ and $B$ are in the ratio of $a:b$; Then $\frac{a}{b}$ can be expressed as $\frac{1}{x}$ where $x$ is.

Answer 1

Given, $F_a=F_b=2\text{N}$, $r_b=4r_a$
$\Delta l_a=2$ $\text{mm}$ and $\Delta l_b=4$ $\text{mm}$

We know, $Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{\Delta l/l}$

$\Rightarrow \frac{\Delta l}{l}=\frac{F}{AY}$ $\Rightarrow l=\frac{AY\Delta l}{F}$

Now, $\frac{l_a}{l_b}=\frac{\pi r_a^2\times Y\times \Delta l_a\times F_b}{\pi r_b^2\times Y\times \Delta l_b\times F_a}$

Because both are made up of same material, their Young's modulus is same.

$\Rightarrow \frac{l_a}{l_b}=\frac{r_a^2\times \Delta l_a}{r_b^2\times \Delta l_b}=\frac{r_a^2\times 2}{(4r_a{)}^2\times 4}=\frac{r_a^2}{16r_a^2\times 2}$

$\Rightarrow \frac{l_a}{l_b}=\frac{1}{32}=\frac{1}{x}$

$\Rightarrow x=32$