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. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

• A. 110
• B. 100
• C. 200
• D. 210

Answer 1

The correct option is B

100

Let number of lead shots = n
Height of cone $=h=8cm$
Radius of cone $=r_1=5cm$
Volume of water present in the cone
= Volume of cone
$=\frac{1}{3}\pi (r_1{)}^{2}h$
$=\frac{1}{3}\times \frac{22}{7}\times 5^2\times 8$
Volume of water that flowed out
$=\frac{1}{4}\times (\frac{1}{3}\times \frac{22}{7}\times 5^2\times 8)c{m}^2$

Radius of lead shot $=r=0.5cm$
Volume of each lead shot
$=\frac{4}{3}\pi r^3=(\frac{4}{3}\times \frac{22}{7}\times {0.5}^3)c{m}^3$

According to given situation, n lead shots are thrown into cone such that ¼th of water present in cone flows out.
It means, volume of n lead shots = volume of water that flowed out
$\Rightarrow n\times \frac{4}{3}\times \frac{22}{7}\times (0.5{)}^3=\frac{1}{4}\times (\frac{1}{3}\times \frac{22}{7}\times 5^2\times 8)$

$\Rightarrow$ $n=\frac{1}{4}\times \frac{1}{3}\times 5^2\times 8\times \frac{1}{\left({0.5}^3\right)}\times \frac{3}{4}$

$\Rightarrow$ $n=100$