Wave Mechanics Question 533
Question: 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
Options:
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}}} $
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Answer:
Correct Answer: A
Solution:
[a] $ C _{1}=\frac{\pi \eta (r _{2}^{4}-r _{1}^{4})}{2\ell },C _{2}=\frac{\pi \eta r^{4}}{2\ell } $ Initial volume = Final volume
$ \
Therefore ,\pi [r _{2}^{2}-r _{1}^{2}]\ell \rho =\pi r^{2}\ell \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}}} $
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