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. The tank is filled with water up to a height of $0.5\text{m}$. The specific gravity of the plank is $0.5$. The angle $\theta$ made by the plank in equilibrium position is

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

Answer 1

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

The force acting on the plank are shown in the figure.

Let the height of water level is $l$.

$\therefore$ The length of the plank is $2l$.

The weight of the plank acts through the centre $B$ of the plank. We have $OB=l$.

The buoyant force $F$ acts through the point $A$ which is the mid-point of the dipped part $OC$ of the plank.

We have $OA=\frac{OC}{2}=\frac{l}{2\cos \theta }$.

Let the cross-sectional area of plank$=a$

$\therefore$ Weight of plank,
$mg=0.5{\rho }_w\times 2la\times g$

The mass of the part $OC$ of the plank $=0.5{\rho }_w\left(\frac{l}{\cos \theta }\right)a$

$\therefore$ Buoyant force acting on the plank = weight of the water displaced

$\Rightarrow F=0.5{\rho }_w\left(\frac{l}{\cos \theta }\right)ag$

For equilibrium, the moment of all the forces about hinge $O$ should be zero.

$\therefore F\times OA\sin \theta -mg\times OB\sin \theta =0$

$\Rightarrow F\times OA-mg\times OB=0(\text{For}\theta \ne 0)$

$\Rightarrow 0.5{\rho }_w\left(\frac{l}{\cos \theta }\right)ag\times \frac{l}{2\cos \theta }=0.5{\rho }_w\times 2la\times g\times \frac{l}{2}$

$\Rightarrow {\cos }^2\theta =\frac{1}{2}$

$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \theta ={45}^{\circ }$

Hence, option $\left(\text{B}\right)$ is correct.