A concrete sphere of radius $R$ has a cavity of radius $ r$ which is packed with sawdust. The specific gravities of concrete and sawdust are respectively $2.4$ and $0.3$ for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust will be
AIIMS 1995, Diffcult
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(b)Let specific gravities of concrete and saw dust are ${\rho _1}$ and ${\rho _2}$ respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere
$\frac{4}{3}\pi ({R^3} - {r^3}){\rho _1}g + \frac{4}{3}\pi {r^3}{\rho _2}g = \frac{4}{3}\pi {R^3} \times 1 \times g$
==> ${R^3}{\rho _1} - {r^3}{\rho _1} + {r^3}{\rho _2} = {R^3}$
==> ${R^3}({\rho _1} - 1) = {r^3}({\rho _1} - {\rho _2})$ ==> $\frac{{{R^3}}}{{{r^3}}} = \frac{{{\rho _1} - {\rho _2}}}{{{\rho _1} - 1}}$
==> $\frac{{{R^3} - {r^3}}}{{{r^3}}} = \frac{{{\rho _1} - {\rho _2} - {\rho _1} + 1}}{{{\rho _1} - 1}}$
==> $\frac{{({R^3} - {r^3}){\rho _1}}}{{{r^3}{\rho _2}}} = \left( {\frac{{1 - {\rho _2}}}{{{\rho _1} - 1}}} \right)\;\frac{{{\rho _1}}}{{{\rho _2}}}$
==> $\frac{{{\rm{Mass of concrete }}}}{{{\rm{Mass of saw dust}}}} = \left( {\frac{{1 - 0.3}}{{2.4 - 1}}} \right) \times \frac{{2.4}}{{0.3}} = 4$
According to principle of floatation weight of whole sphere = upthrust on the sphere
$\frac{4}{3}\pi ({R^3} - {r^3}){\rho _1}g + \frac{4}{3}\pi {r^3}{\rho _2}g = \frac{4}{3}\pi {R^3} \times 1 \times g$
==> ${R^3}{\rho _1} - {r^3}{\rho _1} + {r^3}{\rho _2} = {R^3}$
==> ${R^3}({\rho _1} - 1) = {r^3}({\rho _1} - {\rho _2})$ ==> $\frac{{{R^3}}}{{{r^3}}} = \frac{{{\rho _1} - {\rho _2}}}{{{\rho _1} - 1}}$
==> $\frac{{{R^3} - {r^3}}}{{{r^3}}} = \frac{{{\rho _1} - {\rho _2} - {\rho _1} + 1}}{{{\rho _1} - 1}}$
==> $\frac{{({R^3} - {r^3}){\rho _1}}}{{{r^3}{\rho _2}}} = \left( {\frac{{1 - {\rho _2}}}{{{\rho _1} - 1}}} \right)\;\frac{{{\rho _1}}}{{{\rho _2}}}$
==> $\frac{{{\rm{Mass of concrete }}}}{{{\rm{Mass of saw dust}}}} = \left( {\frac{{1 - 0.3}}{{2.4 - 1}}} \right) \times \frac{{2.4}}{{0.3}} = 4$
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