Question

When a sphere of radius $2cm$ is suspended at the end of a wire, elongation is ${}^{\prime }e$'. When the same wire is loaded with a sphere of radius $3cm$ and made of the same material, the elongation would be :

• A. $\frac{8}{27}e$
• B. $\frac{27}{8}e$
• C. $\frac{4}{9}e$
• D. $\frac{9}{4}e$

Answer 1

Answer: (B)

$\frac{27}{8}e$

$Y=\frac{F/A}{\Delta l/l}$
$\Delta l=\frac{Fl}{YA}$
So, $\frac{\Delta l_1}{\Delta l_2}=\frac{p\frac{4}{3}\pi r{{}_1}^{3}g}{p\frac{4}{3}\pi r{{}_2}^{3}g}$
$\frac{\Delta l_1}{\Delta l_2}=\frac{(2{)}^3}{(3{)}^3}$
$\frac{\Delta l_1}{\Delta l_2}=\frac{8}{27}$
$\Delta l_2=\frac{27}{8}\Delta l_1$
$=\frac{27}{8}e$.