Q) An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled in it so that lower part of the cone, which is \frac{1}{6} th of volume of the cone, is unfilled but hemisphere is formed on the top. Find volume of the of ice-cream.

An empty cone


Volume of the cone = \frac{1}{3}\pi r2h = \frac{1}{3} \pi x (3)2 x 12 = 36 \pi

When Ice cream is filled in this cone, its  \frac{1}{6}th portion is unfilled and  \frac{5}{6}th gets filled.

Hence, volume of this  \frac{5}{6}th cone = \frac{5}{6} x 36 \pi  = 30 \pi ………. (i)

Volume of Icecream’s hemispeherical shape on top of the cone

=        \frac{2}{3} \pi r3     =        \frac{2}{3} \pi x (3)= 18 \pi ..……… (ii)

From equations (i) and (ii), we get,

Total Volume of Icecream = 30 \pi + 18 \pi = 48 \pi  = 48 x \frac{22}{7}

=          150.72 cm3

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