Solution
Let $N$ be the number of discs. The stack of discs acts as a floating cylinder.
Step 1: Equilibrium of the Stack
Let $d_{sub}$ be the submerged depth of the stack. The weight of $N$ discs is balanced by the buoyant force.
$$ \text{Weight} = N \cdot (\pi r^2 t) \cdot \rho_{disc} \cdot g $$
$$ \text{Buoyancy} = (\pi r^2 d_{sub}) \cdot \rho_{water} \cdot g $$
Equating them:
$$ N t \rho_{disc} = d_{sub} \rho_{water} \implies d_{sub} = N t \left( \frac{\rho_{disc}}{\rho_{water}} \right) $$
Given $\rho_{disc} = 0.8$, $\rho_{water}=1$, and $t=1$:
$$ d_{sub} = 0.8 N \quad \text{…(i)} $$
Step 2: Conservation of Water Volume
Let $H_0 = 8 \text{ cm}$ be the initial water height. Let $H_f$ be the final water height. When the discs are added, the water level rises. The volume of water remains constant. The water occupies the cylindrical vessel minus the submerged volume of the discs.
$$ \text{Initial Volume} = \pi R^2 H_0 $$
$$ \text{Final Volume} = \pi R^2 H_f – \pi r^2 d_{sub} $$
$$ \pi R^2 H_0 = \pi R^2 H_f – \pi r^2 d_{sub} $$
Step 3: Condition for Touching the Bottom
For the lowest disc to touch the bottom, the submerged depth of the stack must equal the final height of the water level.
$$ d_{sub} = H_f $$
Substitute $H_f$ with $d_{sub}$ in the volume equation:
$$ \pi R^2 H_0 = \pi R^2 d_{sub} – \pi r^2 d_{sub} $$
$$ R^2 H_0 = d_{sub} (R^2 – r^2) $$
$$ d_{sub} = H_0 \frac{R^2}{R^2 – r^2} $$
Step 4: Solve for N
Equate the two expressions for $d_{sub}$:
$$ 0.8 N = H_0 \frac{R^2}{R^2 – r^2} $$
$$ N = \frac{H_0}{0.8} \left( \frac{R^2}{R^2 – r^2} \right) $$
Calculation
Values: $R=10$, $r=5$, $H_0=8$.
$$ N = \frac{8}{0.8} \left( \frac{100}{100 – 25} \right) $$
$$ N = 10 \left( \frac{100}{75} \right) = 10 \times \frac{4}{3} = \frac{40}{3} \approx 13.33 $$
Since we need the number of discs required for the lowest one to touch (reach) the bottom, we must exceed 13.33. Therefore, we need 14 discs.