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Question

(C) Radius $(r)$ of the hemispherical part = Radius $(r)$ of the conical part = $60\, cm$.Height $(h_2)$ of the conical part of the solid = $120\, cm$

Vedclass · Accepted Answer

$1.131$

Vedclass · Answer

(C) Radius $(r)$ of the hemispherical part = Radius $(r)$ of the conical part = $60\, cm$.Height $(h_2)$ of the conical part of the solid = $120\, cm$.Height $(h_1)$ of the cylinder = $180\, cm$.Radius $(r)$ of the cylinder = $60\, cm$.The volume of water left in the cylinder = Volume of the cylinder - Volume of the solid.Volume of the solid = Volume of the cone + Volume of the hemisphere = $\frac{1}{3}\pi r^2 h_2 + \frac{2}{3}\pi r^3$.Volume of water left = $\pi r^2 h_1 - (\frac{1}{3}\pi r^2 h_2 + \frac{2}{3}\pi r^3)$.Substituting the values: $\pi(60)^2(180) - (\frac{1}{3}\pi(60)^2 	imes 120 + \frac{2}{3}\pi(60)^3)$.$= \pi(60)^2 [180 - (40 + 40)] = \pi(3600)(100) = 360,000\pi\, cm^3$.Using $\pi = \frac{22}{7}$,Volume $\approx 360,000 	imes 3.14159 = 1,130,973.36\, cm^3$.Since $1\, m^3 = 1,000,000\, cm^3$,the volume in $m^3$ is $\approx 1.131\, m^3$.

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