Question

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 3cm and its Length is 12cm. If each cone has a height of 2cm.
OR
Find the volume of air contained in the model that he made.
Scm
2cm
2cm
cm
3cm
12cm​

Answer 1

21 π cm³

Step-by-step explanation:

Given:

Diameter of Model = 3 cm

Length of Model = 12 cm

Height(h) of each cone = 2 cm

To Find:

Volume of Air Contained by model. i.e Volume of Model

Solution:

In Case Cone:

Diameter(d) = 3 cm

So, Radius (r) = 3/2 cm

Height (h) of cone = 2 cm

$\textsf{Volume}(V_1) \: \textsf{of 2 cone} = 2. \dfrac{1}{3} \pi \: {r}^{2} \: h \\ \\ = \frac{2}{3} \pi {( \frac{3}{2})}^{2}.2 \\ \\ = \frac{4}{3} \times \frac{9}{4} \times \pi \\ \\ = \frac{\cancel{4}}{\cancel{3}} \times \frac{ \cancel{9}}{\cancel{4}} \times \pi \\ \\ = 3\pi \: \: \: { \text{cm}}^{3}$

In Case of Cylinder:

Diameter (d) = 3 cm

So, Radius (r) = 3/2 cm

Height (h) = length of Model - Height of 2 cones

= (12 - 2 × 2) cm

= (12 - 4) cm = 8 cm

$\textsf{Volume} (V_2) \: \textsf{of Cylinder} = \pi {r}^{2} h \\ \\ = \pi {( \frac{3}{2})}^{2}.8 \\ \\ = 8 \times \frac{9}{4} \times \pi \\ \\ = \cancel{8} \times \frac{9}{\cancel{4}} \times \pi \\ \\ = 2 \times 9 \times \pi\\ \\ = 18\pi \: \: \: { \text{cm}}^{3}$

Therefore, Volume of Air Contained by model = V1 + V2 = (3π + 18π) cm³ = 21π cm³

Answer 2

◇ Question ◇

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 3cm and its Length is 12cm. If each cone has a height of 2cm.

OR

Find the volume of air contained in the model that he made.

◇ Given ◇

★ Height if the cone (h1) = 2cm

★ Diameter of cone = 3cm so,

★ Radius will be = Radius of the cylinder = 3/2 or 1.5cm

★ Height of the cylinder(h2) = 12-4 = 8cm

$\rule{300}2$

★ Solution ★

Volume= Volume of the cylinder + 2×volume of Cone

$V= πr^2h+ 2(\frac{1}{3}πr^2h)\\ V= π(3/2)^2×8+2×1/3π×(3/2)^2×2\\ V= π×9/4×8+2×1/3π×9/4×2\\V= 18π+ 3π \\ V= 21π \\ V=21×22/7 \\ V= 66cm^3$

$\rule{150}2$

Hence,

The required Volume is = 21π or 66cm cubic unit.