A) \[\frac{16}{17}\theta \]
B) \[\frac{15}{16}\theta \]
C) \[8\theta \]
D) \[\frac{3}{2}\theta \]
Correct Answer: A
Solution :
[a] Let \[{{\theta }_{1}}\]and \[{{\theta }_{2}}\]are the angle of twist produced in cylinders A and B respectively. Given, \[{{\theta }_{1}}+{{\theta }_{2}}=\theta \] ?(i) On being in series, the torque acts at their free ends are equal. We have \[\tau =\frac{\pi \eta {{r}^{4}}\theta }{2\ell }\] \[\therefore \frac{\pi \eta {{r}^{4}}{{\theta }_{1}}}{2\ell }=\frac{\pi \eta {{(2r)}^{4}}{{\theta }_{2}}}{2\ell }\Rightarrow {{\theta }_{1}}=16{{\theta }_{2}}\] ? (ii) From (i) and (ii), we have \[{{\theta }_{1}}=\frac{16}{17}\theta \]
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