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.

Answer 1

Hint: We use the formula of volume of sphere which is given by \[V'=\dfrac{4}{3}\pi {{R}^{3}}\], where R is the radius of the sphere and the formula of volume of the cone which is given by \[V=\dfrac{1}{3}\pi {{r}^{2}}h\], where r is the radius of the circular end and h is the height of the cone.

Complete step-by-step answer:

Given an inverted cone of height 8 cm, the radius of its top is 5cm.
Therefore, we have,
Height of cone h = 8cm, radius of the cone r = 5cm and the Radius of sphere R = 0.5cm.
The figure of the problem is as given below,

We know that the Volume of cone is \[V=\dfrac{1}{3}\pi {{r}^{2}}h\], where r is the radius of the circular end and h is the height of the cone.

Substituting the values of r and h we get the volume of come as,

\[\begin{align}
  & \Rightarrow V=\dfrac{1}{3}\pi {{(5)}^{2}}(8) \\
 & \Rightarrow V=\dfrac{200}{3}\pi c{{m}^{3}} \\
\end{align}\]

Now because the lead shots are spherical in shape, we use the volume of sphere which is given as,
\[V'=\dfrac{4}{3}\pi {{R}^{3}}\].
Now substituting the value of the radius R = O.5cm we get the Volume of on lead shot as,

\[\begin{align}
  & \Rightarrow V'=\dfrac{4}{3}\pi {{(0.5)}^{3}} \\
 & \Rightarrow V'=\dfrac{1}{6}\pi c{{m}^{3}} \\
\end{align}\]


Now we have to calculate the number of lead shots, which is given by,
Volume of 1 lead shot multiplied by number of lead shots is equal to the volume of the cone V.
Implying this we have,

\[\begin{align}
  & Number\text{ }of\text{ }lead\text{ }shots~=\dfrac{\text{Volume of the cone V}}{\text{volume of one lead shot V }\!\!'\!\!\text{ }} \\
 & \Rightarrow Number\text{ }of\text{ }lead\text{ }shots~=\dfrac{\dfrac{200}{3}\pi }{\dfrac{1}{6}\pi } \\
 & \Rightarrow Number\text{ }of\text{ }lead\text{ }shots~=400 \\
\end{align}\]

Now it is given that only one-fourth of the water is flown out therefore only one fourth of the lead shots will drop in the cone , Thus total number of lead shots will be dropped in the vessel is (1/4)400=100.


Note: The possibility of error in this question can be multiplying the number of lead shot to the volume of the cone which would be wrong because the volume of one lead short multiplied by the number of lead shots gives the total volume of all the lead shot which will be equal to the volume of the cone.