Question

A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The relative densities of concrete and sawdust are 2.4 and 0.3 respectively. For this sphere to float with its entire volume submerged under water, the ratio of the mass of concrete to the mass of sawdust will be

• A. 8
• B. 4
• C. 3
• D. 2

Answer 1

The correct option is B

4

Let m be the mass of concrete and $\rho$ its density and let m' be the mass of sawdust and ${\rho }^{\prime }$ its density.
Then
$m=\frac{4\pi }{3}(R^3-r^3)\rho$
and $m^{\prime }=\frac{4\pi }{3}{r}^3{\rho }^{\prime }$
$\therefore \frac{m}{m^{\prime }}=\frac{R^3-r^3}{r^3}.\frac{\rho }{{\rho }^{\prime }}$ ..... (i)
Since the entire volume V $=\frac{4\pi }{3}{R}^3$ of the sphere is submerged under water, we have, from the principle of flotation,
Weight of concrete + Weight of sawdust = weight of volume V of water displaced
$\Rightarrow mg+m^{\prime }g=V{\rho }_{0}g\Rightarrow m+m^{\prime }=V{\rho }_0$
where ${\rho }_0$ is the density of water.
Thus, $\frac{4\pi }{3}(R^3-r^3)\rho +\frac{4\pi }{3}{r}^3{\rho }^{\prime }=\frac{4\pi }{3}{R}^3{\rho }_0$
$\Rightarrow (R^3-r^3)d+r^3{d}^{\prime }=R^3$ ..... (ii)
where $d=\frac{\rho }{{\rho }_0}$ and $d^{\prime }=\frac{{\rho }^{\prime }}{{\rho }_0}$ are the relative densities of concrete and sawdust respectively.
Equation (ii), on simplification, gives $\frac{R^3}{r^3}=\frac{(d-d^{\prime })}{(d-1)}$
$\Rightarrow \frac{R^3}{r^3}-1=\frac{(d-d^{\prime })}{(d-1)}-1$
$\Rightarrow \frac{R^3-r^3}{r^3}=\frac{(1-d^{\prime })}{(d-1)}$ .... (iii)
Using (iii) in (i) and noting that $\frac{\rho }{{\rho }^{\prime }}=\frac{d}{d^{\prime }}$, we have
$\frac{m}{m^{\prime }}=\frac{(1-d^{\prime })}{(d-1)}\times \frac{d}{d^{\prime }}=\frac{(1-0.3)}{(2.4-1)}\times \frac{2.4}{0.3}=4$
Hence, the correct choice is (b).