Question

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. 0.5 cm are dropped into the vessel one- fourth a the water flows out. Find the number of lead shots dropped in the vessel

Answer 1

Let the number of lead shots dropped $=n$
The total number of number of lead shots $=1/4$ volume of the conical vessel
Lead shots
Radius, $r=0.5cm$
Volume, $V=\frac{4}{3}\pi r^3$
$\frac{4}{3}.\frac{22}{7}.0.5\times 0.5\times 0.5=\frac{4}{3}\times \frac{22}{7}\times 0.125$
Cone
Radius, $r=5cm$
Height, $h=8cm$
Volume, $V=\frac{1}{3}\pi r^{2}h$
$=\frac{1}{3}.\frac{22}{7}\times 5\times 5\times 8=\frac{1}{3}\times \frac{22}{7}\times 200$
$\frac{1}{4}thvolume=\frac{1}{4}\times \frac{1}{3}\times \frac{22}{7}\times 200$
Total volume of number of shots
$=n\times \frac{4}{3}\times \frac{22}{7}\times 0.125$
$n\times \frac{4}{3}\times \frac{22}{7}\times 0.125=\frac{1}{4}\times \frac{1}{3}\times \frac{22}{7}\times 200$
$n=\frac{1}{4}\times \frac{1}{3}\times \frac{22}{7}\times 200\times \frac{3}{4}\times \frac{7}{22}\times \frac{1}{0.125}$
$=\frac{200}{4\times 4\times 0.125}=\frac{200}{2}=100$
$\therefore$ The number of lead shots $=100$