Two cylinders A and B of the same material have same length, their radii being in the ratio of 1 : 2 respectively. The two are joined in series. The upper end of A is rigidly fixed. The lower end of B is twisted through an angle θ, the angle of twist of the cylinder A is

The correct option is 16θ/17.

For a cylinder of length $l$, radius $r$, and shear modulus of material $\eta$, the restoring torque $T$ required to produce an angle of twist $\phi$ is given by the relation:
$T=\frac{\pi \eta r^4\phi }{2l}=C\phi$
where $C=\frac{\pi \eta r^4}{2l}$ is the torsional rigidity of the cylinder.

Since both cylinders A and B are made of the same material and have the same length, their shear modulus $\eta$ and length $l$ are identical. Thus, the torsional rigidity is directly proportional to the fourth power of the radius:
$C\propto r^4$
Given that the ratio of their radii is $r_A:r_B=1:2$, we can write the ratio of their torsional rigidities as:
$\frac{C_A}{C_B}={\left(\frac{r_A}{r_B}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{1}{16}$
This gives:
$C_B=16C_A$

Since the two cylinders are joined in series, the same torque $T$ acts on both of them. Let ${\theta }_A$ be the angle of twist of cylinder A and ${\theta }_B$ be the angle of twist of cylinder B. Therefore:
$T=C_A{\theta }_A=C_B{\theta }_B$
Substituting $C_B=16C_A$ into the equation:
$C_A{\theta }_A=16C_A{\theta }_B\Rightarrow {\theta }_B=\frac{{\theta }_A}{16}$

The upper end of cylinder A is rigidly fixed, and the lower end of cylinder B is twisted through an angle $\theta$. The total angle of twist is the sum of the individual twists of both cylinders:
${\theta }_A+{\theta }_B=\theta$
Substitute ${\theta }_B=\frac{{\theta }_A}{16}$ into the sum:
${\theta }_A+\frac{{\theta }_A}{16}=\theta$
$\frac{17{\theta }_A}{16}=\theta$
${\theta }_A=\frac{16}{17}\theta$