A) \[0.66\,{{m}^{3}}\]
B) \[150\,{{m}^{3}}\]
C) \[\frac{3}{1}\,{{m}^{3}}\]
D) \[\frac{3}{20}\,{{m}^{3}}\]
Correct Answer: D
Solution :
From Archimedes principle Weight of (boy\[+\]log) = weight of water displaced \[(60+V\times 0.6\times {{10}^{3}})g=V\times {{10}^{3}}g\] \[\Rightarrow \] \[0.4\times {{10}^{3}}V=60\] \[\Rightarrow \] \[V=\frac{60}{0.4\times {{10}^{3}}}=\frac{600}{400}=\frac{3}{20}{{m}^{3}}\]
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