Question

Rachel an engineering student was asked to make a model shaped like a cylinder with two cones attached at it's two ends by using a thin aluminum sheet.the diameter of model is 3cm and it's length is 12cm.if each come has a height of 2cm.find the volume of air contained in the model that Rachel made.​

Answer 1

$\huge{\underline{\underline{\red{\mathbf{Answer :}}}}}$

★ Case 1 :

Here, radius of 2 cones and cylinder = 3/2

= 1.5 cm

Height of cone (H) = 2 cm

Height of cylinderical portion(h) = 12 - 2 - 2

= 8 cm

Volume of air = Volume of cylinderlical part + 2 Volume of cone

Volume = πr²h - 2(1 * π * r²h)/3

$\sf{volume = \pi {(1.5)}^{2} \times 8 \: + \: 2( \frac{1}{3} \pi {r}^{2} h )} \\ \\ \sf{ \implies\frac{22}{7}(2.25) \times 8 \: + \: 2( \frac{1}{3} \times \frac{22}{7} \times 2.25 \times 2 ) } \\ \\ \sf{\implies \frac{22}{7} \times 18 \: + \:2( \frac{99}{21}) } \\ \\ \sf{\implies \frac{396}{7} + \frac{198}{21} } \\ \\ \sf{ \implies\frac{1188 + 198}{21} } \\ \\ \sf{\implies \frac{1384}{21} } \\ \\ \sf{ \implies 65.90 \: {cm}^{3} }$

$\large{\boxed{\green{\sf{Volume = 65.90 \: cm^3}}}}$

__________________________

★ Case 2 :

The methamatical concept in the above problem used is Mensuration.

$\rule{200}{2}$

#answerwithquality

#BAL

Answer 2

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