question_answer

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 \[=\frac{1}{3}\pi h({{r}^{2}}+{{R}^{2}}+rR)\]

\[=\frac{1}{3}\times \frac{22}{7}\times 45\,[{{28}^{2}}+{{7}^{2}}+(28\times 7)]\]

\[=\frac{1}{3}\times \frac{22}{7}\times 45\,(784+49+196)\]

\[=\frac{1}{3}\times \frac{22}{7}\times 45\times 1029=48510\,c{{m}^{3}}\]