# Pressure in a fluid-mechanical properties of fluids

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Question

# A wooden plank of length 1 m and uniform cross section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. If the angle $\mathrm{\theta }$ by the inclination of that the plank makes with the vertical in the equilibrium position (exclude the case $\mathrm{\theta }$ = 0). Find the value of $\frac{1}{{\mathrm{cos}}^{2}\mathrm{\theta }}$.

Moderate
Solution

## Let y be the length of the plank inside water.          $\mathrm{y}=\frac{0.5}{\mathrm{cos}\mathrm{\theta }}$Let A be the cross-sectional area of the plank. Then the buoyant force on it is ${\mathrm{F}}_{\mathrm{b}}={\mathrm{V\rho }}_{\mathrm{\omega }}\mathrm{g}=\left(\mathrm{Av}\right){\mathrm{\rho }}_{\mathrm{\omega }}\mathrm{g}$Since the plank is in rotational equilibrium, so $\sum {\stackrel{\to }{\mathrm{\tau }}}_{0}=0$

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