Question

A circular tube of mean radius $8cm$ and thickness $0.04cm$ 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

• A. $\frac{(8.02{)}^4-(7.98{)}^4}{(0.8{)}^4}$
• B. $\frac{(8.02{)}^2-(7.98{)}^2}{(0.8{)}^2}$
• C. $\frac{(0.8{)}^2}{(8.02{)}^4-(7.98{)}^4}$
• D. $\frac{(0.8{)}^2}{(8.02{)}^3-(7.98{)}^2}$

Answer 1

The correct option is A $\frac{(8.02{)}^4-(7.98{)}^4}{(0.8{)}^4}$

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