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

$\frac{4}{3}\Pi&space;(R^{3}-r^{3})\rho&space;_{1}g&space;+&space;\frac{4}{3}\Pi&space;r^{3}\rho&space;_{2}g=&space;\frac{4}{3}\Pi&space;R^{3}\times&space;1\times&space;g$

$(R^{3}-r^{3})\rho&space;_{1}&space;+&space;r^{3}\rho&space;_{2}=&space;R^{3}$

$R^{3}\rho&space;_{1}-r^{3}\rho&space;_{1}+r^{3}\rho&space;_{2}=&space;R^{3}$

$R^{3}(\rho&space;_{1}-1)=r^{3}(\rho&space;_{1}-\rho&space;_{2})$

$\frac{R^{3}}{r^{3}}=&space;\frac{\rho_{1}-\rho&space;_{2}}{\rho&space;_{1}-1}$

$\frac{(R^{3}-r^{3})}{r^{3}}=&space;\frac{\rho&space;_{1}-\rho&space;_{2}-\rho&space;_{1}+1}{\rho&space;_{1}-1}$

$\frac{(R^{3}-r^{3})\rho&space;_{1}}{r^{3}\rho&space;_{2}}=\frac{(1-\rho&space;_{2})}{(\rho&space;_{1}-1)}\frac{\rho&space;_{1}}{\rho&space;_{2}}$

(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)