Question

A wooden plank of length $1\text{m}$ and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water upto a height $0.5\text{m}$. The specific gravity of the plank is $0.5$. Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position. (Exclude the case $\theta =0^{\circ })$

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

Answer 1

The correct option is B ${45}^{\circ }$

For equilibrium $F_{\text{net}}=0$ and ${\tau }_{\text{net}}=0$


Let $F_B$ be the buoyant force acting on the plank and $m$ be the mass of plank.
Taking moment about $O$,
$mg\times \frac{l}{2}\sin \theta =F_B\left(\frac{l-x}{2}\right)\sin \theta$
$\Rightarrow mg\times l=F_B\left(l-x\right)$ .......(1)

Also $F_B=$ wt. of fluid displaced
$\Rightarrow F_B=\left[\right(l-x\left)A\right]\times {\rho }_{w}g$
And, $m=\left(lA\right)0.5{\rho }_w$
where $A$ is the area of cross section of the rod.

Putting the value of $F_B$ and $m$ in Eq.(1),
$\left(lA\right)0.5{\rho }_{w}g\times l=\left[\right(l-x\left)A\right]{\rho }_{w}g\times \left(l-x\right)$
Here, $l=1\text{m}$
$\therefore (1-x{)}^2=0.5\Rightarrow x=0.293\text{m}$
From the diagram,
$\cos \theta =\frac{0.5}{1-x}=\frac{0.5}{0.707}\Rightarrow \theta \approx {45}^{\circ }$
Hence, option $\left(b\right)$ is the correct answer.