If the radii of the circular ends of a truncated conical bucket which is 45 cm high is 28 cm and 7 cm, then the capacity of the bucket in cubic centimetres is \[\left( \text{take}\,\pi =\frac{22}{7} \right)\] [SSC (CGL) 2011] |
A) 48510
B) 45810
C) 48150
D) 48051
Correct Answer: A
Solution :
Volume of bucket \[=\frac{1}{3}\pi h({{r}^{2}}+{{R}^{2}}+rR)\] |
\[=\frac{1}{3}\times \frac{22}{7}\times 45\,[{{28}^{2}}+{{7}^{2}}+(28\times 7)]\] |
\[=\frac{1}{3}\times \frac{22}{7}\times 45\,(784+49+196)\] |
\[=\frac{1}{3}\times \frac{22}{7}\times 45\times 1029=48510\,c{{m}^{3}}\] |
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