## Problems on Volume

The page  "Problems on Volume". On this page we are going to see a practice problem on volume for some particular solids.

To understand this topic much better you can go through the following questions.

Problem 1

Mr.David has a toy  in the form of hemisphere with a cone surmounted on it and  is having the radius  3.5cm.The total height of the toy is 15.5cm.He would like to find the volume of the toy.Please help him to find the volume of the toy.

Solution: Problems on Volume The radius of the cone = Radius of the hemisphere = 3.5cm
Total height of the toy=15.5cm
Total height of the toy = radius of the hemisphere +

height of the cone.

15.5cm = 3.5cm + height of the cone
Height of the cone = 15.5cm - 3.5cm
Height of the cone = 12cm
Volume of the toy = volume of the cone + volume of the hemisphere

= (1/3) π r² h + (2/3) π r³
Here r = 3.5 =7/2     h=12
= (1/3)π(7/2)(7/2)(12) + (2/3)π(7/2)(7/2)(7/2)
= 49π + 343π/12
= 49π+28.58π
= (77.58)π
Hence, volume of the toy = (77.58)π cm³

Problem 2

A vessel is in the form of hollow cylinder which has been surmounted on a hemispherical bowl.The radius of the hemisphere is 7cm and the total height of the vessel is 13cm.Determine the total capacity of the vessel.

Solution: Radius if the hemisphere =  7 cm
Radius if the cylinder = radius of the hemisphere = 7 cm
Total height of the vessel = 14 cm
Total height of the vessel = height of the cylinder +

13 = height of the cylinder + 7
Height of the cylinder = 13 - 7 = 6 cm
Capacity of the vessel = vol. of the cylinder + vol. of the hemisphere
= π r² h + (2/3) π r³

Here "r" = 7 and "h" = 6
= π (7)² 6 + (2/3) π (7)³
= π(7)(7)6 + (2/3)π(7)(7)(7)
= π(7) (7) (6) + (2/3) π (7) (7) (7)
= 294π + 228.67π
= 522.67π
Hence, the capacity of the vessel is = (522.67)π  cm³

Student who are practicing problems on volume can go through the steps of the above problems to have better understanding. 1. Click on the HTML link code below.

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