A circular tube of mean radius $$8\,cm$$  and thickness $$0.04\,cm$$  is melted up and recast into a solid rod of the same length. The ratio of the torsional rigidities of the circular tube and the solid rod is

Solution :
$${C_1} = \frac{{\pi \eta \left( {r_2^4 - r_1^4} \right)}}{{2\ell }},{C_2} = \frac{{\pi \eta {r^4}}}{{2\ell }}$$
Initial volume = Final volume
$$\eqalign{ & \therefore \pi \left[ {r_2^2 - r_1^2} \right]\ell \rho = \pi {r^2}\ell \rho \cr & \Rightarrow {r^2} = r_2^2 - r_1^2 \Rightarrow {r^2} = \left( {{r_2} + {r_1}} \right)\left( {{r_2} - {r_1}} \right) \cr & \Rightarrow {r^2} = \left( {8.02 + 7.98} \right)\left( {8.02 - 7.98} \right) \cr & \Rightarrow {r^2} = 16 \times 0.04 = 0.64\,cm \Rightarrow r = 0.8\,cm \cr & \therefore \frac{{{C_1}}}{{{C_2}}} = \frac{{r_2^4 - r_1^4}}{{{r^4}}} = \frac{{{{\left[ {8.02} \right]}^4} - {{\left[ {7.98} \right]}^4}}}{{{{\left[ {0.8} \right]}^4}}} \cr} $$