A) 8
B) 4
C) 3
D) Zero
Correct Answer: B
Solution :
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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