A 90 degree inverted cone contains water as shown The volume

A 90 degree inverted cone contains water as shown. The volume of the water in the cone is given by = (pi/3)h^3. The original depth of the water is 10 cm. A block with a volume of 200 cm^3 and a specific gravity of 0.60 is floated in the water. What will be the change in cm in water surface height in the cone?

Solution

Volume of water in the cone before placing block V1 = (/3) x h³    --(1)

Where,

h is the depth of water = 10 cm

Now block of 200 cm³ is placed inside the cone

As specific gravity is 0.6

Specific gravity = mass density of block / mass density of water

0.6 = mass density of block/ 1 (g/cm³)

mass density of block = 0.6 g/cm³

Hence total mass of block = 200 x 0.6 = 120 g

1 g object will sink until it displaces 1 g of water. This is independent of its size or shape. Since water has a density of 1 g/cm3, a 1 g object will displace 1 cm3 of water

Hence 120 g will displace 120 cm³ of volume which can say as V2

New volume V2 = (/3) x y³   

Y = 4.858 cm