A cylindrical container whose diameter is 12, | vedclass

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(A) Volume of the cylindrical container $= \pi r^2 h = \pi 	imes (6)^2 	imes 15 = 540\pi \, cm^3$.This ice cream is distributed equally among $10$ c

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Volume of the cylindrical container $= \pi r^2 h = \pi 	imes (6)^2 	imes 15 = 540\pi \, cm^3$.This ice cream is distributed equally among $10$ children,so the volume of one ice cream unit $= \frac{540\pi}{10} = 54\pi \, cm^3$.Let the diameter of the base of the cone be $D$. Then the radius $r = \frac{D}{2}$.The height of the conical portion is given as $h = 2D$.The volume of one ice cream unit (cone + hemisphere) $= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$.Substituting $r = \frac{D}{2}$ and $h = 2D$:Volume $= \frac{1}{3}\pi (\frac{D}{2})^2 (2D) + \frac{2}{3}\pi (\frac{D}{2})^3 = \frac{1}{3}\pi (\frac{D^2}{4}) (2D) + \frac{2}{3}\pi (\frac{D^3}{8}) = \frac{\pi D^3}{6} + \frac{\pi D^3}{12} = \frac{2\pi D^3 + \pi D^3}{12} = \frac{3\pi D^3}{12} = \frac{\pi D^3}{4}$.Equating the volumes: $\frac{\pi D^3}{4} = 54\pi$.$D^3 = 54 	imes 4 = 216$.$D = \sqrt[3]{216} = 6 \, cm$.

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