Question

A uniform rod of length $L$ pivoted at the bottom of a pond of depth $L/2$ stays in stable equilibrium as shown in the figure. Find the angle $\theta$ if the density of the material of the rod is half of the density of water.

• A. ${60}^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${37}^o$

Answer 1

The correct option is C ${45}^o$

Weight of the rod $W=ALdg$, where $d$ is the density of the rod, $A$ is the uniform cross section area of rod. $L$ is the total length of the rod.

Lets say $x$ is the length of submerged part, we have

$x\sin \theta =\frac{L}{2}\Rightarrow x=\left(\frac{L}{2}\right)cosec\theta$

Force of buoyancy $B=A\left(\frac{L}{2}cosec\theta \right)\rho g$

Since the rod is at rest i.e. it is in rotational equilibrium, we have

$W\times \frac{L}{2}\cos \theta =B\times \frac{1}{2}\left(\frac{L}{2}\right)cosec\theta \Rightarrow ALdg\times \frac{L}{2}\cos \theta =A\left(\frac{L}{2}cosec\theta \right)\rho g\times \frac{1}{2}\left(\left(\frac{L}{2}\right)cosec\theta \right)\cos \theta \Rightarrow d=\left(\frac{1}{2}cosec\theta \right)\rho \times \frac{1}{2}cosec\theta \Rightarrow \frac{\rho }{2}=\left(\frac{1}{2}cosec\theta \right)\rho \times \frac{1}{2}cosec\theta \because d=\frac{\rho }{2}\Rightarrow cose{c}^2\theta =2\Rightarrow cosec\theta =\sqrt{2}\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta ={45}^o$