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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.
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Radius of cone = $5$ cm, height of cone = $8$ cm
Volume of water in the cone = $$\begin{aligned}& \frac{1}{3}\pi\times (5)^2 \times 8 \\ & = \frac{200\pi}{3}\end{aligned}$$ cm$^3$
Volume of water flows out = $\frac{1}{4}$ (Volume of water in the cone)
= $\frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}$ cm$^3$
Radius of sphere (lead shot) = $0.5 = \frac{1}{2}$ cm
Volume of one lead shot = $$\begin{aligned}& \frac{4}{3}\pi\times \left(\frac{1}{2}\right)^3 \\ & = \frac{4}{3}\pi\times \frac{1}{8} = \frac{\pi}{6}\end{aligned}$$ cm$^3$
Number of lead shots = $\frac{\frac{50\pi}{3}}{\frac{\pi}{6}} = \frac{50\pi}{3} \times \frac{6}{\pi} = 100$
Volume of water in the cone = $$\begin{aligned}& \frac{1}{3}\pi\times (5)^2 \times 8 \\ & = \frac{200\pi}{3}\end{aligned}$$ cm$^3$
Volume of water flows out = $\frac{1}{4}$ (Volume of water in the cone)
= $\frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}$ cm$^3$
Radius of sphere (lead shot) = $0.5 = \frac{1}{2}$ cm
Volume of one lead shot = $$\begin{aligned}& \frac{4}{3}\pi\times \left(\frac{1}{2}\right)^3 \\ & = \frac{4}{3}\pi\times \frac{1}{8} = \frac{\pi}{6}\end{aligned}$$ cm$^3$
Number of lead shots = $\frac{\frac{50\pi}{3}}{\frac{\pi}{6}} = \frac{50\pi}{3} \times \frac{6}{\pi} = 100$