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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 ?

Dhritiman Medhi , 5 Years ago
Grade 12th pass
anser 1 Answers
Khimraj

Last Activity: 5 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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