Two metallic cylinder made of same material and radii r and 2r are joined as shown in figure. If top end is fixed and lower end of cylinder B is twisted by angle ? Then angle of twist for cylinder A is (both cylinder have same length)

 

Dear student 
Let the total angle twisted by A and B is $\theta$ (not mentioned in question) and twist angle made by A and B will be ${\theta }_1 and {\theta }_2 respectively$.
hence $\theta ={\theta }_1+{\theta }_2.......1> (total angle twisted is contribution of individual angular twist)$
now the force applied produce the torque on B and a restoring torque is produce in A to counter torque of B this continuous till  both the rods does not archives the condition of equilibrium.
using the formula of torque  in both cylinder A and B 
$$\begin{gathered} torque clockwise by B = torque anticlockclockwise by A (\sin ce we assume that force on B acts clockwise) \\ {\tau }_c ={\tau }_a \\ \frac{\eta \pi (2r{)}^4}{2l}{\theta }_2 =\frac{\eta \pi (r{)}^4}{2l}{\theta }_1 \\ {\theta }_1 =16{\theta }_2 .........2> \\ now putting the values of {\theta }_2 and {\theta }_1 in equation 1 we get \\ {\theta }_2+16{\theta }_2 = \theta \\ {\theta }_2 =\frac{\theta }{17} and \\ {\theta }_1 = \frac{16}{17} \theta . \end{gathered}$$


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