Question

A wire is stretched by $0.01m$ by a certain force $F$. Another wire of same material whose diameter and lengths are doubled to the original wire is stretched by the same force. Then its elongation will be:

• A. $0.005m$
• B. $0.01m$
• C. $0.02m$
• D. $0.08m$

Answer 1

The correct option is A $0.005m$

Given that,

$\Delta l_1=0.01m$

Length $l_1=l$

Length $l_2=2l$

Radius $r_1=r$

Radius $r_2=2r$

We know that,

$Y=\frac{Fl}{A\Delta l}$

$\Delta l=\frac{Fl}{\pi r^{2}Y}$

$\Delta l\propto \frac{l}{r^2}$

Now, the diameter and lengths are doubled to the original wire is stretched by the same force

So,

$\frac{\Delta l_2}{\Delta l_1}=\frac{2l}{l}\times \frac{r^2}{4r^2}$

$\frac{\Delta l_2}{\Delta l_1}=2\times \frac{1}{4}$

$\Delta l_2=\frac{0.01}{2}$

$\Delta l_2=0.005m$

Hence, the elongation is $0.005m$