Question

A vessel is in the form of an inverted cone of height 8cm and radius of the base as 5cm. It is filled with water up to the brim. A certain number of spherical lead shots of radius 0.5cm are dropped in the cone, and one-fourth of water flows out. Find the total number of lead shots dropped.

Answer 1

Hint: Find the volume of the cone and hence find the volume of water that overflowed. Hence find the number of the spheres that will have total volume as the volume of water overflowed. Assume that the number of spherical lead shots dropped to be x. Hence form a linear equation in x and solve for x. The value of x will be the total number of lead shots dropped. Use volume of a right circular cone of radius r and height h is given by $\dfrac{1}{3}\pi {{r}^{2}}h$ , and the volume of a sphere of radius r is given by $\dfrac{4}{3}\pi {{r}^{3}}$.

Complete step-by-step answer:
Here the radius of the cone (r) = 5cm and the height of the cone (h) = 8cm.
Hence the volume of the cone $\dfrac{1}{3}\pi {{r}^{2}}h=\dfrac{1}{3}\pi {{\left( 5 \right)}^{2}}\left( 8 \right)=\dfrac{200}{3}\pi $ cubic centimeters.
Hence the volume of the water overflowed $=\dfrac{1}{4}\left( \dfrac{200\pi }{3} \right)=\dfrac{50\pi }{3}$ cubic centimeters
Also, radius of a spherical lead shot (r) = 0.5cm
Hence the volume of spherical lead shot $=\dfrac{4}{3}\pi {{\left( 0.5 \right)}^{3}}=\dfrac{4}{3}\pi \left( \dfrac{1}{8} \right)=\dfrac{\pi }{6}$ cubic centimeters.
Let the number of the spherical lead shots dropped be x.
Hence, we have
$\dfrac{\pi x}{6}=\dfrac{50\pi }{3}$
Multiplying both sides by $\dfrac{6}{\pi }$, we get
$x=\dfrac{50\pi }{3}\times \dfrac{6}{\pi }=100$
Hence the total number of the spherical lead shots dropped in the cone is 100.

Note: If the value of x came out to be fractional, then there was no value of x for which the above condition is possible. This is because the number of lead shots is a natural number and cannot be fractional. Hence the fractional answer depicts that the situation is not possible.