# A concrete sphere of radius R has cavity of radius r which is packed with saw-dust. The specific gravitites of concrete and sawdust are 2.4 and 0.3 respectively. The sphere floats with its entire volume sub-merged under water. Calculate the ratio of mass of concrete and the mass of saw-dust.(a) 1(b) 2(c) 3(d) 4

The specific gravities of concrete and sawdust are ρ1 and ρ2 respectively

Given ρ1 = 2.4 and ρ2 = 0.3

According to principle of floatation

Weight of a whole sphere = upthrust on the sphere

$$\begin{array}{l}\frac{4}{3}\pi \left ( R^{3}-r^{3} \right )\rho _{1}g+\frac{4}{3}\pi r^{3}\rho _{2}g=\frac{4}{3}\pi R^{3}\times 1\times g\\\left ( R^{3}-r^{3} \right )\rho _{1}+r^{3}\rho _{2}=R^{3}\\R^{3}\rho _{1}-r^{3}\rho _{1}+r^{3}\rho _{2}=R^{3}\\R^{3}\left ( \rho _{1}-1 \right )=r^{3}\left ( \rho _{1}-\rho _{2} \right )\\\frac{R^{3}}{r^{3}}=\frac{\rho _{1}-\rho _{2}}{\rho _{1}-1}\\\frac{\left ( R^{3}-r^{3} \right )}{r^{3}}=\frac{\rho _{1}-\rho _{2}-\rho _{1}+1}{\rho _{1}-1}\\\frac{\left ( R^{3} -r^{3}\right )\rho _{1}}{r^{3}\rho _{2}}=\left ( \frac{1-\rho _{2}}{\rho _{1}-1} \right )\frac{\rho _{1}}{\rho _{2}}\end{array}$$

(mass of concrete)/(mass of saw-dust) = (1 – 0.3)/(2.4 – 1) x (2.4)/(0.3)

We get,

(mass of concrete)/(mass of saw-dust) = 4

Therefore, the ratio of mass of concrete and the mass of saw-dust is 4

So, the correct option is (d)