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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 ?
one month ago

Khimraj
3008 Points
							 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 $\frac{\text{Mass of concrete }}{\text{Mass of saw dust}}=\left( \frac{1-0.3}{2.4-1} \right)\times \frac{2.4}{0.3}=4$

one month ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions