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 \[=\frac{1}{3}\pi h\,\,({{R}^{2}}+{{r}^{2}}+Rr)\] |
\[=\frac{1}{3}\times \frac{22}{7}\times 45\,\,[{{28}^{2}}+{{7}^{2}}+(28\times 7)]\] |
\[=\frac{1}{3}\times \frac{22}{7}\times 45\,\,(784+49+196)\] |
\[=\frac{1}{3}\times \frac{22}{7}\times 45\times 1029=48510\,\,c{{m}^{3}}\] |
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