A metallic cone with radius 5 , cm and height | vedclass

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Question

(C) The volume of the metallic cone is given by the formula $V_{cone} = \frac{1}{3} \pi r^2 h$. Substituting the given values $r = 5 \, cm$ and $h = 8

Vedclass · Accepted Answer

$3200$

Vedclass · Answer

(C) The volume of the metallic cone is given by the formula $V_{cone} = \frac{1}{3} \pi r^2 h$. Substituting the given values $r = 5 \, cm$ and $h = 8 \, cm$,we get $V_{cone} = \frac{1}{3} \pi (5)^2 (8) = \frac{200}{3} \pi \, cm^3$.The volume of each spherical ball with diameter $0.5 \, cm$ (radius $r_b = 0.25 \, cm = \frac{1}{4} \, cm$) is $V_{ball} = \frac{4}{3} \pi r_b^3 = \frac{4}{3} \pi (\frac{1}{4})^3 = \frac{4}{3} \pi (\frac{1}{64}) = \frac{1}{48} \pi \, cm^3$.The number of balls produced is $n = \frac{V_{cone}}{V_{ball}} = \frac{\frac{200}{3} \pi}{\frac{1}{48} \pi} = \frac{200}{3} 	imes 48 = 200 	imes 16 = 3200$.

$A$ metallic cone with radius $5 \, cm$ and height $8 \, cm$ is melted and recast to produce identical balls with diameter $0.5 \, cm$. Find the number of balls produced.

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