Question

A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 10³kg/m³. If outer diameter and the density of the bowl are 1 m and 2 × 10⁴kg/m³ respectively, then the inner diameter of the bowl will be

• A. 0.94 m
• B. 0.97 m
• C. 0.98 m
• D. 0.99 m

Answer 1

Answer: (C)

0.98 m

Answer 2

Weight of the bowl = mg = $V \rho g$ $= \frac { 4 } { 3 } \pi \left[ \left( \frac { D } { 2 } \right) ^ { 3 } - \left( \frac { d } { 2 } \right) ^ { 3 } \right] \rho g$

where D is the outer diameter , d is the inner diameter and $\rho$

is the density of bowl

Weight of the liquid displaced by the bowl

$= \frac { 4 } { 3 } \pi \left( \frac { D } { 2 } \right) ^ { 3 } \sigma g$

where $\sigma$ is the density of the liquid.

For the flotation $\frac { 4 } { 3 } \pi \left( \frac { D } { 2 } \right) ^ { 3 } \sigma g = \frac { 4 } { 3 } \pi \left[ \left( \frac { D } { 2 } \right) ^ { 3 } - \left( \frac { d } { 2 } \right) ^ { 3 } \right] \rho g$

⇒ $\left( \frac { 1 } { 2 } \right) ^ { 3 } \times 1.2 \times 10 ^ { 3 } = \left[ \left( \frac { 1 } { 2 } \right) ^ { 3 } - \left( \frac { d } { 2 } \right) ^ { 3 } \right] 2 \times 10 ^ { 4 }$

By solving we get d = 0.98 m.