Two cylinders A and B of the same materials have same length. Their radii being in the ratio of 1 : 2 respectively. The two are joined end to end as shown in the figure. 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 :

# Two cylinders A and B of the same materials have same length. Their radii being in the ratio of 1 : 2 respectively. The two are joined end to end as shown in the figure. The upper end of A is rigidly fixed. The lower end of B is twisted through an angle $\mathrm{\theta }$ , the angle of twist of the cylinder A is :

1. A

$\frac{15}{16}\mathrm{\theta }$

2. B

$\frac{16}{15}\mathrm{\theta }$

3. C

$\frac{16}{17}\mathrm{\theta }$

4. D

$\frac{17}{16}\mathrm{\theta }$

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### Solution:

We know that restoring torque,

$\mathrm{T}=\frac{{\mathrm{G\theta \pi R}}^{4}}{2\mathrm{L}}$

Since T, G, and L are constant for both cylinders,

$\mathrm{\theta }\propto \frac{1}{{\mathrm{R}}^{4}}$

$\frac{{\mathrm{\theta }}_{\mathrm{A}}}{{\mathrm{\theta }}_{\mathrm{B}}}={\left(\frac{{\mathrm{R}}_{\mathrm{B}}}{{\mathrm{R}}_{\mathrm{A}}}\right)}^{4}$

$\frac{{\mathrm{\theta }}_{\mathrm{A}}}{{\mathrm{\theta }}_{\mathrm{B}}}={2}^{4}$

${\mathrm{\theta }}_{\mathrm{A}}=16{\mathrm{\theta }}_{\mathrm{B}}$

Now as,

${\mathrm{\theta }}_{\mathrm{A}}+{\mathrm{\theta }}_{\mathrm{B}}=\mathrm{\theta }$

${\mathrm{\theta }}_{\mathrm{A}}+\frac{{\mathrm{\theta }}_{\mathrm{A}}}{16}=\mathrm{\theta }$

$\frac{17{\mathrm{\theta }}_{\mathrm{A}}}{16}=\mathrm{\theta }$

${\mathrm{\theta }}_{\mathrm{A}}=\frac{16\mathrm{\theta }}{17}$

Hence the correct answer is $\frac{16\mathrm{\theta }}{17}.$

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