Question

long cylindrical vessel ofradius R = 10 cm placed on a horizontal table is filled with water up to a height h = 8 cm. A disc of radius r = 5 cm, thickness t = 1 cm and made of material of density p = 0.8 g/cm' when put on the water it floats. When another identical disc is coaxially placed on it the former shifts further down and now both of them float. If we keep on putting more identical discs in this manner one above the other, eventually the lowest disc will touch the bottom of the vessel. How much minimum number of these discs is needed so that the lowest disc touches the bottom of the vessel? Density of water is p0 = 1.0 g/cm3•

Answer 1

14

Answer 2

The volume of water in the vessel is constant. When $N$ discs are placed and the lowest disc touches the bottom, the water level rises to a height $h_{water,N}$. This height is determined by the original volume of water and the space occupied by the submerged part of the discs. Assuming the lowest disc is at the bottom, the submerged part of the stack is a cylinder of radius $r$ and height $h_{water,N}$ (provided $h_{water,N} \le Nt$). The volume of water is the volume of the vessel up to $h_{water,N}$ minus the volume of the submerged discs: $V_{water} = \pi R^2 h_{water,N} - \pi r^2 h_{water,N} = \pi (R^2 - r^2) h_{water,N}$. Using the given values, $800\pi = \pi (10^2 - 5^2) h_{water,N} = 75\pi h_{water,N}$, which gives $h_{water,N} = 800/75 = 32/3$ cm.

For the stack of $N$ discs to touch the bottom, its weight must be greater than or equal to the buoyant force acting on it when the lowest disc is at the bottom. With the lowest disc at the bottom, the water level is $h_{water,N} = 32/3$. The height of the stack is $N$. The submerged height is $\min(N, h_{water,N}) = \min(N, 32/3)$. The buoyant force is $F_B = \pi r^2 \min(N, 32/3) \rho_0 g = 25\pi \min(N, 32/3) g$. The weight of $N$ discs is $W = N m_d g = 20\pi N g$. The condition for the lowest disc to touch the bottom is $W \ge F_B$, which means $20\pi N g \ge 25\pi \min(N, 32/3) g$, or $20N \ge 25 \min(N, 32/3)$, or $0.8N \ge \min(N, 32/3)$.

If $N \le 32/3$, then $\min(N, 32/3) = N$, and $0.8N \ge N$ implies $0.2N \le 0$, which is impossible for $N>0$.

If $N > 32/3$, then $\min(N, 32/3) = 32/3$, and $0.8N \ge 32/3$ implies $N \ge 32/(3 \times 0.8) = 32/2.4 = 320/24 = 40/3 = 13.33...$.

So, we need $N > 32/3 \approx 10.67$ and $N \ge 13.33...$. The minimum integer $N$ satisfying both conditions is $N=14$.