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

Answer 1

Hint: A body when put into water will float only if the weight of the body and the upthrust force acting on it by the liquid is equal. The upthrust force can be calculated using the volume of the liquid displaced.

Formula used:
Upthrust force on a body by a liquid, ${F}_{up}=\rho Vg$, where $\rho$ is the density of fluid, V is the volume of water displaced and g is the acceleration due to gravity.

Complete step-by-step answer:
Let us assume the inner and outer radius of the bowl be $r$ and $R$ respectively. Density of the bowl and that of the water be ${\rho}_{b}$ and ${\rho}_{w}$ respectively.
According to the question, the bowl is floating without sinking, so according to the law of floatation, an object floats in a liquid when its weight and the upthrust by the fluid are equal.
So, we need to first find out the volume of the hemispherical bowl, $V=\dfrac{2}{3}\pi (R^{3}-r^{3})$
Therefore, mass of the ball will be, $m={\rho}_{b}\times V={\rho}_{b}\times \dfrac{2}{3}\pi (R^{3}-r^{3})$ ………. (i)
Now, upthrust force by a liquid is given by ${F}_{up}=\rho Vg$, where $\rho$ is the density of fluid, V is the volume of water displaced and g is the acceleration due to gravity.
So, ${F}_{up}={\rho}_{w}\times \dfrac{2}{3} \pi R^{3}g$ ………. (ii)
For the object to float, $mg={F}_{up}$, so from equation (i) and (ii), we can write, ${\rho}_{b}\times \dfrac{2}{3}\pi (R^{3}-r^{3}) g ={\rho}_{w}\times \dfrac{2}{3} \pi R^{3}g$
After putting the value of densities and outer radius and arranging the terms, we will get
$\implies (0.5)^3 \times 1.2 \times 10^{3} = ((0.5)^3 -r^3)\times 2 \times 10^4$
$\implies r=0.98$ m
Hence, option c is the right answer.

Note: There are chances of mistake while finding the volume of water displaced by the hemispherical bowl as one may unconsciously take its inner diameter. And it should also be noted that the volume of the object that needed to be taken while calculating the upthrust force should always be the part submerged in the liquid and not the complete volume of the object.