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, then the number of lead shots dropped in the vessel is
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Answer 1

Step-by-step explanation:

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Answer 2

Solution :

$\sf{}Volume \: of \: the \: water \: in \: the \: cone = ( \frac{1}{3} \pi \times (5) ^{2} \times 8) {cm}^{3} \\ \\ \sf{}Volume = \frac{200\pi}{3} {cm}^{3}$

Volume of the water that flows out of the cone :

$\sf{} = \dfrac{1}{4} \times \dfrac{200\pi}{3} {cm}^{3} \\ \\ = \sf{}\frac{50\pi}{3} {cm}^{3}$

Volume of one spherical shot of radius 0.5 cm :

$\sf{} = \frac{4}{3} \pi \times (0.5) ^{3} {cm}^{3} \\ \\ \sf{} = (\frac{4\pi}{3} \times 0.125) {cm}^{3} \\ \\ \sf{}( \frac{4\pi}{3} \times \frac{125}{1000} ) {cm}^{3} \\ \\ \sf{} = \frac{\pi}{6} {cm}^{3}$

Let the number shot put into the vessel be equal to n when 1/4 th of the water flows out of the cone .

Volume of n spherical lead shots = volume of the water that flows out

$\sf{}n \times \dfrac{\pi}{6} \\ \\ = \frac{50\pi}{3 } \\ \\ \sf{}n = 100$

Therefore, number of shots is equal s to 100.