A) \[\text{8}0\text{85}0\text{ c}{{\text{m}}^{\text{3}}}\]
B) \[\text{8}0\text{58}0\text{ c}{{\text{m}}^{\text{3}}}\]
C) \[\text{8}0\text{5}0\text{8 c}{{\text{m}}^{\text{3}}}\]
D) \[\text{8}00\text{5}0\text{8 c}{{\text{m}}^{\text{3}}}\]
Correct Answer: A
Solution :
C.S.A. \[=2\times \frac{22}{7}\times h\times \frac{5h}{3}=\frac{220{{h}^{2}}}{21}\] \[\therefore \]\[\frac{9240}{2}=\frac{220{{h}^{2}}}{21}\] \[\therefore \]\[V=\pi {{r}^{2}}h=80850\,c{{m}^{3}}\]
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