ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.4 in Chapter 17 - Mensuration

Question 21 Mensuration Exercise 17.4

A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid. (Take π = 3.14)

Solution:

Since the given solid is a combination of a cone, cylinder, and hemisphere, we must sum the volumes of the cone, cylinder, and hemisphere to determine its volume.

Given height of the cylinder, H = 10 cm

Height of the cone, h = 6 cm

Common diameter = 3.5 cm

Common radius, r = 3.5/2 = 1.75 cm

Volume of the solid = Volume of the cone + Volume of the cylinder + Volume of the hemisphere

= (1/3)π r2h +π r2H + (2/3) π r3

= π r2((h/3) + H + (2r/3))

= 3.14×1.752×((6/3)+10+(2×1.75)/3)

= 3.14×3.0625×(2+10+1.167)

= 3.14×3.0625×13.167

= 9.61625×13.167

= 126.617 cm3

= 126.62 cm3

Hence the volume of the solid is 126.62 cm3.

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