The total internal area of the curved surface of the well and the bottom of the well is
A) \[560c{{m}^{2}}\text{ and }56c{{m}^{2}}\]
B) \[56c{{m}^{2}}\text{ and 61}6c{{m}^{2}}\]
C) \[{{r}^{2}}\]
D) \[\pi {{r}^{2}}\]
Correct Answer: A
Solution :
Required area\[h=\frac{4\times {{6}^{3}}}{3\times {{(0.2)}^{2}}}=\frac{4\times 6\times 6\times 6}{3\times 0.04}\] \[=7200cm=72cm\] to\[1c{{m}^{2}}=100m{{m}^{2}}\]
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