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 A rod of length l and radius r is joined to a rod of length \frac{l}{2} and radius \frac{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^{\circ}, the twist angle at the joint will be ______

Option: 1

\theta / 4


Option: 2

\theta / 2


Option: 3

5 \theta / 6


Option: 4

8 \theta / 9


Answers (1)

best_answer

\theta_1+\theta_2=\theta~~~~~~~~~~...(i)

We know that -

\begin{aligned} \frac{\tau}{\theta} & =\frac{\pi \eta r^4}{2 l} \\ \therefore \quad \frac{\theta_1}{\theta_2} & =\frac{l_1}{l_2} \times \frac{r_2^4}{r_1^4}=\frac{l / 2 \times r^4}{l \times(r / 2)^4} \\ \frac{\theta_1}{\theta_2} & =8 ~~~~~~~~~~...(ii)\end{aligned}

After solving above equations, we get-
\begin{aligned} & \theta_1+\theta_2=\theta \\ & \theta_1-8 \theta_2=0 \\ &\text{-----------------}\\ & 9 \theta_2=\theta \\ &\Rightarrow \theta_2=\frac{\theta}{9} \end{aligned}
Put this value in equation (i).
\begin{aligned} \theta_1+\left(\frac{\theta}{9}\right)&=\theta \\ \theta_1&=\frac{8 \theta}{9} \end{aligned}

Posted by

Gautam harsolia

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