First slide

Pressure in a fluid-mechanical properties of fluids

Question

A wooden rod of a uniform cross section and of length 120 cm is hinged at the bottom of the tank which is filled with water to a height of 40 cm. In the equilibrium position, the rod makes an angle of 60⁰ with the vertical. The centre of buoyancy is located on the rod at a distance (from the hinge) of

Moderate

A

20 cm
B

40 cm
C

60 cm
D

75 cm

Solution

OP is the portion of the rod immersed in water.
$\begin{array}{l} \cos 60^{\circ} = \frac{40}{O P} \\ or O P = \frac{40}{\cos 60^{\circ}} cm = \frac{40}{1 / 2} cm = 80 cm \end{array}$
The centre of buoyancy is at the centre of the immersed part of the rod. So, the required distance is 40 cm.

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