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Grade 12th passMechanics

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 flow with its entire volume submerged under water. The ratio of mass of concrete to mass of sawdust will be ?

Profile image of Dhritiman Medhi
7 Years agoGrade 12th pass
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1 Answer

Profile image of Khimraj
6 Years ago
 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 \[\frac{4}{3}\pi ({{R}^{3}}-{{r}^{3}}){{\rho }_{1}}g+\frac{4}{3}\pi {{r}^{3}}{{\rho }_{2}}g=\frac{4}{3}\pi {{R}^{3}}\times 1\times g\] Þ \[{{R}^{3}}{{\rho }_{1}}-{{r}^{3}}{{\rho }_{1}}+{{r}^{3}}{{\rho }_{2}}={{R}^{3}}\] Þ  \[{{R}^{3}}({{\rho }_{1}}-1)={{r}^{3}}({{\rho }_{1}}-{{\rho }_{2}})\] Þ \[\frac{{{R}^{3}}}{{{r}^{3}}}=\frac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}-1}\]  Þ \[\frac{{{R}^{3}}-{{r}^{3}}}{{{r}^{3}}}=\frac{{{\rho }_{1}}-{{\rho }_{2}}-{{\rho }_{1}}+1}{{{\rho }_{1}}-1}\] Þ \[\frac{({{R}^{3}}-{{r}^{3}}){{\rho }_{1}}}{{{r}^{3}}{{\rho }_{2}}}=\left( \frac{1-{{\rho }_{2}}}{{{\rho }_{1}}-1} \right)\ \frac{{{\rho }_{1}}}{{{\rho }_{2}}}\] Þ \[\frac{\text{Mass of concrete }}{\text{Mass of saw dust}}=\left( \frac{1-0.3}{2.4-1} \right)\times \frac{2.4}{0.3}=4\]

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