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 cu cm is) (take \[\pi =\frac{22}{7}\])                                                                  [SSC (CGL) 2011]

A) 48510

B) 45810

C) 48150              

D) 48051

Correct Answer: A

Solution :

R = 28 cm

\[r=7\,\,cm\]

Volume of bucket \[=\frac{1}{3}\pi h\,\,({{R}^{2}}+{{r}^{2}}+Rr)\]

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

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

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