# ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.5 in Chapter 17 - Mensuration

Question 17 Mensuration Exercise 17.5

A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows.

What is the volume of each of the solid cone submerged?

Solution:

Assume that a cone has a height of "h" and a circular base with radius "r." This cone will have a volume that is one-third of the product of the base's area and its height.

Given height of the cone, h = 20 cm

Diameter of the cone = 16.8 cm

Radius of the cone, r = 16.8/2 = 8.4 cm

Volume of water in the vessel = (1/3)π r2h

= (1/3)π x8.42 x20

= (1/3)x(22/7)x8.4 x8.4 x20

= 1478.4 cm3

One third of volume of water in the vessel = (1/3)× 1478.4

= 492.8 cm3

One third of volume of water in the vessel = Volume of water over flown

Volume of water over flown = volume of two equal solid cones dropped into the vessel.

volume of two equal solid cones dropped into the vessel = 492.8 cm3

Volume of one solid cone dropped into the vessel = 492.8/2 = 246.4 cm3

Hence the volume of each of the solid cone submerged is 246.4 cm3.

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