Question

Two cylinders A and B of the same material have same length, their radii being in the ratio 1: 2 respectively. The two are joined end to end as shown in the figure. One end of cylinder A is rigidly clamped while free end of cylinder B is twisted through an angle $\theta$. The angle of twist of cylinder A is ________.

• A. $\frac{17}{16}\theta$
• B. $16\theta$
• C. $17\theta$
• D. $\frac{16}{17}\theta$

Hint:

In series, the thinner cylinder undergoes a much larger twist because torsional rigidity depends on the fourth power of the radius.

Answer 1

The Correct Option is D

Step 1: Concept
For cylinders in series, the torque ($\tau$) is the same for both. The torque is given by $\tau = C\phi$, where $C$ is the torsional rigidity ($C \propto r^{4}/l$) and $\phi$ is the angle of twist.
Step 2: Analysis
Let the twist in A be $\phi_{A}$ and in B be $\phi_{B}$. The total twist is $\theta = \phi_{A} + \phi_{B}$. Since material and length are same, $C \propto r^{4}$. Given $r_{A}/r_{B} = 1/2$, so $C_{B} = 16C_{A}$.
Step 3: Calculation
Equating torque: $C_{A}\phi_{A} = C_{B}\phi_{B} \Rightarrow C_{A}\phi_{A} = 16C_{A}\phi_{B} \Rightarrow \phi_{B} = \phi_{A}/16$. Substituting in total twist: $\theta = \phi_{A} + \phi_{A}/16 = \frac{17}{16}\phi_{A}$. Thus, $\phi_{A} = \frac{16}{17}\theta$.
Step 4: Conclusion
Hence, the angle of twist of cylinder A is $\frac{16}{17}\theta$.
Final Answer:(D)