Question

Question 9
An ice cream cone full of ice cream having radius 5 cm and height 10 cm as shown in figure. Calculate the volume of ice- cream, provided that its $\frac{1}{6}$part is left unfilled with ice – cream.

Answer 1

Given, ice cream cone is the combination of a hemisphere and a cone.
Also, radius of hemisphere = 5 cm
$\therefore$ Volume of hemisphere
$=\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times (5{)}^3=\frac{5500}{21}=261.90c{m}^3$

Now, radius of the cone = 5 cm
And height of the cone = 10 – 5 = 5 cm
$Volumeofthecone=\frac{1}{3}\times \frac{22}{7}\times (r{)}^2\times h)\therefore Volumeofthecone=\frac{1}{3}\times \frac{22}{7}\times (5{)}^2\times 5=\frac{2750}{21}=130.95c{m}^3$

Now, total volume of ice cream cone $=Volumeofthecone+Volumeofhemisphere=261.90+130.95=392.85c{m}^3$

Since $\frac{1}{6}$ part is left unfilled with ice – cream.
$\therefore$ Required volume of ice cream
$=392.85c{m}^3–(392.85\times \frac{1}{6})c{m}^3=392.85c{m}^3-65.475c{m}^3=327.4c{m}^3$