A) 100
B) 10
C) \[10\sqrt{2}\]
D) \[10\sqrt{3}\]
Correct Answer: B
Solution :
\[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] \[{{P}_{1}}.\frac{4}{3}\pi {{r}^{3}}=10.\left[ \frac{4}{3}\pi {{(kr)}^{3}} \right]\] \[{{P}_{1}}=10{{K}^{3}}={{P}_{2}}\]+ depth of lake \[=10+{{10}^{3}}K.\] \[10{{K}^{3}}-{{10}^{3}}K-10=0\] \[{{K}^{3}}-{{10}^{2}}K-1=0\] Or \[\left( \frac{-a}{2} \right)\,\times \frac{1}{3}=1\,\Rightarrow a=6\]
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