Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones
attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its
length is 12 cm. If each cone has a height of 2 cm.
Find the volume of air contained in the model that Rachel made. (Assume the outer and inner
dimensions of the model to be nearly the same).
Which mathematical concept is used in the above problem?
Answers
Figure:-
Given:-
- The diameter of the model is 3 cm and its length is 12 cm.
- The each cone has a height of 2 cm.
To find:-
- Find the volume of air contained in the model that Rachel made...?
Solutions:-
- Height (h1) of each conical part = 2cm.
- Height (h2) of cylindrical part = 12 - 2 × height of conical part.
=> 12 - 2 × 2
=> 12 - 4
=> 8cm
Radius (r) of cylindrical part = Radius of conical part = 3/2 cm.
Volume of air present in the model = Volume cylinder +
2 × Volume of cone.
=> πr²h2 + 2 × 1/3 πr²h
=> π (3/2)² + (8) + 2 × (1/3)² (2)
=> π × 9/4 × 8 + 2/3 × π × 9/4 × 2
=> π × 9 × 2 + 3π
=> 18π + 3π
=> 21π
=> 21 × 22/7
=> 3 × 22
=> 66cm³
Hence, Volume of air present in the model is 66cm³.
Volume of Air Contained in Model
= Volume of Model
= Volume of Cylinder + Volume of 1st cone + Volume of 2nd Cone
= πr²h + 1/3πr²h + 1/3πr²h
= (π × r² × h) + (1/3 × π × r² × h) + (1/3 × π × r² × h)
= (22/7 × (3/2)² × h) + (1/3 × 22/7 × (3/2)² × h) + (1/3 × 22/7 × (3/2)² × h)
= (22/7 × 9/4 × h) + (1/3 × 22/7 × 9/4 × 2) + (1/3 × 22/7 × 9/4 × h)
= (198/28 × h) + (22/217 × 9/4 × h) + (22/21 × 9/4 × h)
= (198/28 × (12 - 4)) + (22/21 × 9/4 × 2) + (22/21 × 9/4 × 2)
= (198/28 × 8) + (22/21 × 9/4 × 2) + (22/21 × 9/4 × 2)
= (198/28 × 8) + (22/21 × 9/2 × 1) + (22/21 × 9/2 × 1)
= (198/28 × 8) + (22/21 × 9/2) + (22/21 × 9/2)
= (1584/28) + (198/42) + (198/42)
= (56.57) + (4.71) + (4.71)
= 65.99 cm³
≈ 66 cm³