Question

A mass m is suspended from a wire. Change in length of the wire is $\Delta l$. Now the same wire is stretched to double its length and the same mass is suspended from the wire. The change in length in this case will become (it is assumed that elongation in the wire is within the proportional limit)

• A. $\Delta l$
• B. $2\Delta l$
• C. $4\Delta l$
• D. $8\Delta l$

Answer 1

The correct option is C $4\Delta l$

Elongation is given by
$\Delta l=\frac{Fl}{AY}=\frac{Fl}{\frac{V}{l}Y}=\frac{F{l}^2}{VY}$
SInce the force, volume and Y are constant
$\Delta l\propto l^2$
New length is double the original length, so the elongation would be
$\frac{\Delta l_2}{\Delta l_1}={\left(\frac{l_2}{l_1}\right)}^2$
$\Delta l_2=\Delta l_1{\left(\frac{2l}{l}\right)}^2=\Delta l_1\left(4\right)$
Option C.