Class 9 Volume of a Cylinder

Topic Covered

• Volume of a Cylinder

Volume of a Cylinder

Just as a cuboid is built up with rectangles of the same size, we have seen that a right circular cylinder can be built up using circles of the same size.

So, using the same argument as for a cuboid, we can see that the volume of a cylinder can be obtained
as : base area × height

`color{green}("= area of circular base × height" = πr^2h)`


`color{red}"Volume of a Cylinder" = πr^2h)`

where r is the base radius and h is the height of the cylinder.
Q 3220178911

The pillars of a temple are cylindrically shaped (see Fig. 13.26). If each pillar has a circular base of radius 20 cm and height 10 m, how much concrete mixture would be required to build 14 such pillars?
Class 9 Chapter 13 Example 13

Since the concrete mixture that is to be used to build up the pillars is going to occupy the entire space of the pillar, what we need to find here is the volume of the cylinders.

Radius of base of a cylinder `= 20 cm`
Height of the cylindrical pillar `= 10 m = 1000 cm`
So, volume of each cylinder `= πr^2h`
`= (22)/7 xx20xx20 xx1000 cm^3`
`= (8800000)/7 cm^2`
`(8.8)/7 m^3 ("Since" 1000000 cm^3 = 1m^3)`
Therefore, volume of 14 pillars = volume of each cylinder × 14
`=17.6 m^3`
So, 14 pillars would need `17.6 m^3` of concrete mixture.
Q 3240178913

At a Ramzan Mela, a stall keeper in one of the food stalls has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses (see Fig. 13.27) of radius 3 cm up to a height of 8 cm, and sold for Rs. 3 each. How much money does the stall keeper receive by selling the juice completely?
Class 9 Chapter 13 Example 14

The volume of juice in the vessel
= volume of the cylindrical vessel
`= πR^2H`
(where R and H are taken as the radius and height respectively of the vessel)
`= π × 15 × 15 × 32 cm^3`
Similarly, the volume of juice each glass can hold `= πr^2h`
(where r and h are taken as the radius and height respectively of each glass)
`= π × 3 × 3 × 8 cm^3`
So, number of glasses of juice that are sold
`text(volume of the vessel)/text(volume of each glass)`
`= (pixx15xx15xx32)/(pixx3xx3xx8)`
`= 100`
Therefore, amount received by the stall keeper `= Rs. 3 × 100`
`= Rs. 300`