Question

A wooden rod of a uniform cross-section and of length $120\text{cm}$ is hinged at the bottom of the tank, which is filled with water to a height of $40\text{cm}$. In the equilibrium position, the rod makes an angle of ${60}^{\circ }$ with the vertical. The centre of buoyancy is located on the rod at a distance (from the hinge) of

• A. $20\text{cm}$
• B. $40\text{cm}$
• C. $60\text{cm}$
• D. $75\text{cm}$

Answer 1

The correct option is B $40\text{cm}$

$OP$ is the portion of the rod immersed in water.


From figure, $\cos {60}^{\circ }=\frac{40}{OP}$

$\Rightarrow OP=\frac{40}{\cos {60}^o}=80\text{cm}$

The centre of gravity of the displaced fluid is called the buoyant centre or the centre of buoyancy. Buoyant force acts on buoyant centre.

Therefore, the centre of buoyancy is at the centre of the immersed part of the rod.

Hence, centre of buoyancy of rod $=40\text{cm}$

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