# A vessel is in the form of an inverted cone. Its height is 8cm and the radius of its top, which is open is 5cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5cm are dropped into the vessel, $\dfrac{1}{4}$ of the water flows out. Find the number of lead shots dropped in the vessel.

Answer

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Hint: - Volume of cone $ = \dfrac{1}{3}\pi {\left( r \right)^2}h$

Given:

Height$\left( h \right)$of conical vessel$ = 8cm$

Radius$\left( r \right)$ of conical vessel$ = 5cm$

Radius$\left( {{r_1}} \right)$of the lead shots$ = 0.5cm$

Let$x$number of lead shots were dropped in the vessel

Water spilled$ = \dfrac{1}{4}$times of the volume of cone$ = x \times $volume of spherical balls

As we know volume of cone is$ = \dfrac{1}{3}\pi {\left( r \right)^2}h$

And volume of spherical balls$ = \dfrac{4}{3}\pi r_1^3$

\[

\Rightarrow \dfrac{1}{4} \times \dfrac{1}{3} \times \pi {\left( r \right)^2}h = x \times \dfrac{4}{3}\pi r_1^3 \\

\Rightarrow \dfrac{1}{4} \times \dfrac{1}{3} \times \dfrac{{22}}{7}{\left( 5 \right)^2} \times 8 = x \times \dfrac{4}{3} \times \dfrac{{22}}{7}{\left( {0.5} \right)^3} \\

\Rightarrow \dfrac{{200}}{4} = 4x \times .125 \\

\Rightarrow x = \dfrac{{200}}{{16 \times .125}} = \dfrac{{200}}{2} = 100 \\

\]

So, the number of lead shots dropped into the vessel is equal to 100.

So, 100 lead shots is the required answer.

Note: - In such types of questions the key concept we have to remember is that always remember the formula of cone and sphere which is stated above, then simplify it according to given condition we will get the number of required lead shots which is dropped into the vessel.

Given:

Height$\left( h \right)$of conical vessel$ = 8cm$

Radius$\left( r \right)$ of conical vessel$ = 5cm$

Radius$\left( {{r_1}} \right)$of the lead shots$ = 0.5cm$

Let$x$number of lead shots were dropped in the vessel

Water spilled$ = \dfrac{1}{4}$times of the volume of cone$ = x \times $volume of spherical balls

As we know volume of cone is$ = \dfrac{1}{3}\pi {\left( r \right)^2}h$

And volume of spherical balls$ = \dfrac{4}{3}\pi r_1^3$

\[

\Rightarrow \dfrac{1}{4} \times \dfrac{1}{3} \times \pi {\left( r \right)^2}h = x \times \dfrac{4}{3}\pi r_1^3 \\

\Rightarrow \dfrac{1}{4} \times \dfrac{1}{3} \times \dfrac{{22}}{7}{\left( 5 \right)^2} \times 8 = x \times \dfrac{4}{3} \times \dfrac{{22}}{7}{\left( {0.5} \right)^3} \\

\Rightarrow \dfrac{{200}}{4} = 4x \times .125 \\

\Rightarrow x = \dfrac{{200}}{{16 \times .125}} = \dfrac{{200}}{2} = 100 \\

\]

So, the number of lead shots dropped into the vessel is equal to 100.

So, 100 lead shots is the required answer.

Note: - In such types of questions the key concept we have to remember is that always remember the formula of cone and sphere which is stated above, then simplify it according to given condition we will get the number of required lead shots which is dropped into the vessel.

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