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

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

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

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

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

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

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

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

(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)

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