Q. A rod of length $l\ $and radius $r$ is joined to a rod of length $\frac{l}{2}$ and radius $r/2$ of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of $\theta^{0}$, the twist angle at the joint will be
A) $\frac{\theta}{4}$
B) $\frac{\theta}{2}$
C) $\frac{5\theta}{6}$
D) $\frac{8\theta}{9}$
Explanation
(d) $\tau = C\theta$
$= \frac{\pi ɳr^{4}\theta}{2L} = constant$
⟹ $\frac{\pi ɳr^{4}(\theta - \theta_{0})}{2l} = \frac{\pi ɳ\left( \frac{r}{2} \right)^{4}(\theta_{0} - \theta^{''})}{2(l/2)}$
⟹ $\frac{\left( \theta - \theta_{0} \right)}{2} = \frac{\theta_{0}}{16}$
⟹ ${\ \ \ \theta}_{0} = \frac{8}{9}\theta$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.