Solution
Concept: For the plank to be in equilibrium, the net torque about the point of suspension (pivot) must be zero. The forces creating torque are the Weight ($W$) and the Buoyant Force ($B$).
Let $V$ be the total volume of the plank. Since the diagonal coincides with the water surface, exactly half the volume is submerged: $$ V_{sub} = \frac{V}{2} $$
1. Forces:
- Weight ($W = mg = \rho_{wood} V g$) acts downwards at the geometric center of the rectangle ($G$).
- Buoyant Force ($F_B = \rho_{water} V_{sub} g = \rho_{water} \frac{V}{2} g$) acts upwards at the centroid of the submerged triangle.
2. Torque Equation: We take torque about the pivot point (the midpoint of the bottom edge). From the geometry of a rectangle divided by a diagonal, if the horizontal distance of the Center of Mass ($G$) from the vertical axis passing through the pivot is $x$, then the horizontal distance of the Center of Buoyancy (centroid of the triangle) from the same axis is $\frac{2x}{3}$.
For rotational equilibrium ($\tau_{net} = 0$): $$ \tau_{weight} = \tau_{buoyancy} $$ $$ W \cdot x = F_B \cdot \left( \frac{2x}{3} \right) $$
3. Substitution and Calculation: Substitute the expressions for $W$ and $F_B$: $$ (\rho_{wood} V g) \cdot x = \left( \rho_{water} \frac{V}{2} g \right) \cdot \frac{2x}{3} $$ Cancel common terms ($V, g, x$): $$ \rho_{wood} = \rho_{water} \cdot \frac{1}{2} \cdot \frac{2}{3} $$ $$ \rho_{wood} = \rho_{water} \cdot \frac{1}{3} $$
Specific Gravity ($SG$) is the ratio of the density of the wood to the density of water: $$ SG = \frac{\rho_{wood}}{\rho_{water}} = \frac{1}{3} $$
The specific gravity of the wood is 1/3.
Answer: (c)