Question

A wooden plank of length $1m$ and uniform cross section is hinged at one end to the bottom of a tank as shown. The tank is filled with water upto a high of $0.5m$. The specific gravity of the plank is $0.5$. The angel $\theta$ made by the plank in equilibrium position is

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

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

$\begin{array}{c}{f}_B.\left(\frac{1}{4\cos \theta }\right)\sin \theta =\left(\frac{1}{2}\sin \theta \right)mg\\ \cos \theta =\frac{f_B}{2mg}=\frac{{\rho }_{i}A\frac{1}{\cos }g}{2\rho A1g}\\ \cos \theta =\frac{1}{\sqrt{2}}\\ \theta ={45}^o\\ Hence,\\ optionBiscorrectanswer.\end{array}$