Question

A semi-circular sheet of iron of diameter 28 cm is bent into an open conical cup. Find the capacity of the cup.

Answer 1

Let the radius of cone = $\pi$
$2\pi r=28$
$\pi =\frac{28}{2\pi }$
$=\frac{14}{\pi }=\frac{14}{22}\times 7$
(diametric portion is folded to make a cone of radius ${}^{\prime }{\pi }^{\prime }$
hot ocne = 14 cm
Volume =$\frac{1}{3}51{\pi }^{2}h=\frac{1}{33}\times \frac{22}{7}\times \frac{14}{22}\times 7\times \frac{14}{22}\times 7\times 14$
$=\frac{1}{3}\times \frac{\text{/}2\text{/}2}{\text{/}7}\times \frac{\text{/}1\text{/}4}{\text{/}2\text{/}2}\times 7\times \frac{\text{/}{1\text{/}4}^7}{\text{/}{2\text{/}2}_{11}}\times 7\times 14$
$=\frac{49\times 14}{33}$
$=\frac{686}{33}c{m}^3$