Question

When a metal sphere is suspended at the end of a metal wire its extension is $2.7mm$. If another metal sphere of same material with its radius one third that of previous is suspended then find its elongation.

Answer 1

Let, the original length of the metal wire is L meter. And, density of metal sphere is d. If radius of first metal sphere is R then its mass(m) = dv = $d\times \frac{4}{3}\pi R^3$

$\therefore$ Y of wire = $\frac{mg.L}{A.l}=\frac{d\times \frac{4}{3}\pi R^3\times g\times L}{A\times 2.7\times {10}^{-3}}⟶\left(I\right)$

If now radius of the sphere becomes one third of previous i.e. $\frac{K}{3}$ then,

mass (m) = $d\times \frac{4}{3}\pi {\left(\frac{R}{3}\right)}^3$

Then, Y = $\frac{mgL}{Al}$

or, Y = $\frac{d\times \frac{4}{3}\pi {\left(\frac{R}{3}\right)}^3\times g\times L}{A\times l}⟶\left(II\right)$

Equating (I) & (II) we get,

$\frac{R^3}{2.7\times {10}^{-3}}=\frac{{\left(\frac{R}{3}\right)}^3}{l}$

or, $\frac{R^3}{2.7\times {10}^{-3}}=\frac{R^3}{27\times l}$

or, $l=\frac{2.7\times {10}^{-3}}{27}=1\times {10}^{-4}m$

$\therefore$ Elongation = 0.1mm