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 fig 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 $\theta = 0^\circ$)

Step 1: Identify the given parameters.

Let the parameters of the system be:

• Length of the plank: $L = 1\,\text{m}$
• Height of the water: $h = 0.5\,\text{m}$
• Specific gravity of the plank: $s = 0.5$
• Uniform cross-sectional area of the plank: $A$
• Density of water: $\rho_w$
• Density of the plank: $\rho_p = s\rho_w$
• Immersed length of the plank: $x$

Step 2: Express the forces acting on the plank.

The weight $W$ of the plank acts downwards at its center of gravity (at a distance of $\frac{L}{2}$ from the hinge).

$$W = \rho_p A L g = s \rho_w A L g$$

The buoyant force $F_B$ acts upwards at the center of gravity of the submerged portion (at a distance of $\frac{x}{2}$ from the hinge).

$$F_B = \rho_w A x g$$

Step 3: Set up the torque balance about the hinge.

For rotational equilibrium, the torque due to the weight must balance the torque due to the buoyant force.

$$\tau_W = \tau_B$$ $$W \left(\frac{L}{2} \sin\theta\right) = F_B \left(\frac{x}{2} \sin\theta\right)$$

Since $\theta \neq 0^\circ$, we can divide both sides by $\sin\theta$.

$$W L = F_B x$$

Step 4: Solve for the immersed length $x$.

Substitute the expressions for $W$ and $F_B$ into the torque balance equation.

$$(s \rho_w A L g) L = (\rho_w A x g) x$$ $$s L^2 = x^2$$ $$x = L\sqrt{s}$$

Substitute the given values $L = 1\,\text{m}$ and $s = 0.5$.

$$x = 1 \cdot \sqrt{0.5} \approx 0.707\,\text{m}$$

Step 5: Determine the angle $\theta$ using geometry.

The vertical projection of the immersed length $x$ must equal the water height $h$.

$$x \cos\theta = h$$ $$\cos\theta = \frac{h}{x}$$

Substitute $h = 0.5\,\text{m}$ and $x = \sqrt{0.5}\,\text{m}$.

$$\cos\theta = \frac{0.5}{\sqrt{0.5}} = \sqrt{0.5} = \frac{1}{\sqrt{2}}$$ $$\theta = 45^\circ$$

Answer: $\theta = 45^\circ$