Question

A uniform metallic wire is elongated by $0.04\text{m}$ when subjected to a linear force $\text{F}$. The elongation $\left(\text{in cm}\right)$, if its length and diameter is doubled and subjected to the same force, will be

Answer 1

Young modulus of metallic wire is,
$$Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$$
$\Rightarrow Y\frac{\Delta l}{l}=\frac{F}{A}$

$\Rightarrow Y\frac{0.04}{l}=\frac{F}{A}$ ....(1)

For length and diameter is doubled,

$l^{\prime }=2l\text{and}{A}^{\prime }=4A$

$\Rightarrow Y\frac{\Delta l^{\prime }}{l^{\prime }}=\frac{F^{\prime }}{A^{\prime }}$

Here, $F^{\prime }=F$

$\Rightarrow Y\frac{\Delta l^{\prime }}{2l}=\frac{F}{4A}$ ...(2)

By dividing equation(1) by equation(2),

$\Rightarrow \frac{2\times 0.04}{\Delta l^{\prime }}=4$

$\Rightarrow \frac{2\times 0.04}{4}=\Delta l^{\prime }$

$\Rightarrow \Delta l^{\prime }=0.02\text{m}$

$\Rightarrow \Delta l^{\prime }=2\text{cm}$