Question

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water.A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub.(Take $\pi =22/7$)

Answer 1

The tub is half of a cylinder
∴ Volume of water = $\frac{1}{2}$ × $\frac{22}{7}$ × $5^2$ × 9.8 = 385 $c{m}^3$
Volume of the solid = Volume of cone + Volume of hemisphere
= $\frac{1}{3}$ x $\frac{22}{7}$ × ${3.5}^2$ × 5 + $\frac{1}{3}$ × $\frac{22}{7}$ × ${3.5}^3$

= $\frac{385}{6}$ + $\frac{539}{6}$
= 154 $c{m}^3$
Remaining volume = 385 - 154 = 231 $c{m}^3$