Question

A reservoir is in the shape of a frustum of a right circular cone It is $8m$ across at the top and $4m$ across at the bottom. If it is $6m$ deep its capacity is

• A. $176m^3$
• B. $196m^3$
• C. $200m^3$
• D. $110m^3$

Answer 1

Answer: (A)

$176m^3$

Volume of a frustum of a cone $=\frac{1}{3}\pi h(R^2+r^2+Rr)$

Where $h$ is the height and $R$ and $r$ are the radii of the upper and lower ends of a frustum of a cone.

Height $h=6m$
Since, the reservoir is $8m$ across the top, upper radius $R=4m$
Similarly, lower radius $r=2m$

So, volume of the reservoir $=\frac{1}{3}\times \frac{22}{7}\times 6(4^2+2^2+4\times 2)m^3$

$=\frac{22}{7}\times 2(16+4+8)m^3$

$=\frac{22}{7}\times 2\times 28m^3$

$=176m^3$

Hence, the volume of the reservoir is $176m^3$