Question

A hemispherical bowl just floats without sinking in a liquid of density $1.2\times {10}^{3}kg/m^3$. If the outer diameter and the density of the bowl are $1m$ and $2\times {10}^{4}kg/m^3$ respectively, then the inner diameter of the bowl will be :

• A. $1.91m$
• B. $0.5m$
• C. $0.98m$
• D. $1.75m$

Answer 1

Answer: (C)

$0.98m$

Total up trust force executed on the

Bowl$=\frac{2}{3}\pi \left({\frac{1m}{2}}^3\right)\times 1.2\times {10}^{3}g$

$=\frac{0.8\pi }{8}{10}^{3}g$

$=0.1\pi \times {10}^3$...................1

Now,

For bowl to just float 1 should be equal to weight of bowl,

So,

Volume of bowl=$=\frac{0.2\pi \times {10}^3\times 3}{2\times {10}^4}{m}^3=(1^3-r^3)$

$r^3=\frac{1}{8}-\frac{0.2\times 3\times {10}^3}{4\times 10}$

$r^3==0.065$

$r=\frac{0.98}{2}m$

So,

Inner Diameter$=0.98m$