Question

Mayank, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at both ends with thin film-sheet. The radius of the model is 4cm and the total height is 13 cm. If each cone has height 3cm, find the volume of the air contained in the model.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• A model shaped like a cylinder with two cones attached at both ends with thin film-sheet.

• The radius of the model is 4cm and the total height is 13 cm.

• Each cone has height 3cm.

Let assume that radius of cylinder be represented by r, height of cylinder be h. Further assume that height of cone is represented by H.

So, we have

• Radius of cone and cylinder, r = 4 cm

• Height of cone, H = 3 cm

As, it is given that

• Total height = 13 cm

So, it means 2H + h = 13

2 × 3 + h = 13

6 + h = 13

$\rm\implies \:h = 7 \: cm$

Now,

$\rm \: Volume_{(Air\:contained)} \:$

$\rm \: = \: Volume_{(Cylinder)} + 2 \times Volume_{(Cone)}$

$\rm \: = \: \pi {r}^{2}h + 2 \times \dfrac{1}{3} \pi {r}^{2}H$

$\rm \: = \: \pi {r}^{2}\bigg(h + \dfrac{2}{3} H \bigg)$

$\rm \: = \: \frac{22}{7} \times {4}^{2} \times \bigg(7 + \dfrac{2}{3} \times 3 \bigg)$

$\rm \: = \: \frac{22}{7} \times 16 \times \bigg(7 +2 \bigg)$

$\rm \: = \: \frac{22}{7} \times 16 \times 9$

$\rm \: = \: \frac{3168}{7} \: {cm}^{3}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:Volume_{(Air\:contained)} = \: \frac{3168}{7} \: {cm}^{3} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

Brainliest please

452.57cm³

Step-by-step explanation:

Question:-

Mayank, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at both ends with thin film-sheet. The radius of the model is 4cm and the total height is 13 cm. If each cone has height 3cm, find the volume of the air contained in the model.

To Find:-

Volume of air contained in project.

Given that:-

• A figure consisting of a cylinder and it's height.

• Radius of the cylinder and cone is 4cm.

• Height of cone is 3cm.

• Height of the cylinder is 7cm.

Formula used

$\large\underline{\sf{Volume_{cylinder} + 2(Volume_{cone}) }}$

$\large\underline{\sf{π {r}^{2} h+2( \frac{\pi {r}^{2}h }{3} )}}$

$\large\boxed{\sf{\pi {r}^{2}(h + \frac{2}{3} h) }}$

Solution:-

$\large{\sf{ \frac{22}{7} \times {4}^{2}(7 + \frac{2}{3} \times 3)}}$

$\large{\sf{ \frac{22}{7} \times 16(7 + 2)}}$

$\large{\sf{ \frac{22}{7} \times 16 (9)}}$

$\large{\sf{ \frac{22}{7} \times 16 \times 9 }}$

$\large\boxed{\sf{ \red{\implies\frac{3168}{7}} }}$

$\large\boxed{\sf{ \red{ \implies452.57 {cm}^{3}} }}$

Hence, the volume of aur which can be stored is 452.57cm³.