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. The number of lead shots dropped in the vessel is ______.

• A. 100
• B. 50
• C. 20
• D. 10

Answer 1

The correct option is A

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 $=\frac{1}{3}.\pi .(r_1{)}^2.h$

$=(\frac{1}{3}\times \frac{22}{7}\times 5^2\times 8c{m}^2)$
Volume of water flown out $=(\frac{1}{4}\times \frac{1}{3}\times \frac{22}{7}\times 5^2\times 8)c{m}^3$

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 1/4th of water present in cone flows out.

It means volume of n lead shots = volume of water flown out

$\Rightarrow$ $(n\times \frac{4}{3}\times \frac{22}{7}\times (0.5{)}^3)=(\frac{1}{4}\times \frac{1}{3}\times \frac{22}{7}\times 5^2\times 8)$

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

$\Rightarrow$ $n=100$