Question

A friction clutch is in the form of a frustum of a cone, the diameter of the ends being 32cm and 20cm and height 8cm. Find its bearing surface area and volume.​

Answer 1

Step-by-step explanation:

Diagram:- Refer the attachment

Given:-

• The diameter of the ends of the frustum are 32cm and 20cm.

• The height of the frustum = 8cm

To Find:-

• The bearing surface

• Thr volume of the frustum.

Solution:-

Diameter of the lower base = 32cm

Radius of the lower base (R) = 16cm

Diameter of the upper base = 20cm

Radius of the lower base (r) = 10cm

So we have:-

• R = 16cm , r = 10cm and h = 8cm

We need to find the slant height (l)

$\rm l = \sqrt{ {h}^{2} + (R - r )^{2} }$

$\rm l = \sqrt{ {8}^{2} + (16 - 10)^{2} }$

$\rm l = \sqrt{ 64 + (6)^{2} }$

$\rm l = \sqrt{ 64 + 36 }$

$\rm l = \sqrt{ 100} = 10cm$

$\rm\therefore \underline{l = 10cm}$

Bearing surface = Lateral surface area of the frustum (S)

$\rm\dashrightarrow S = \pi(R + r)l$

$\rm\dashrightarrow S = \dfrac{22}{7}\times (16 + 10) \times 10$

$\rm\dashrightarrow S = 817.14$

$\underline{\boxed{\pink{\rm \therefore The \: bearing \: surface = 817.14cm^{2}}}}$

Now, let us find the Volume of the frustum (V)

$\rm V= \dfrac{1}{3} \pi h( {R}^{2} + Rr + {r}^{2})$

$\rm\dashrightarrow V= \dfrac{1}{3}\times \dfrac{22}{7} \times 8 times ( {16}^{2} + 16 \times 10 + {10}^{2})$

$\rm\dashrightarrow V= \dfrac{1}{3}\times \dfrac{22}{7} \times 8 times ( 256 + 160 + 100)$

$\rm\dashrightarrow V= \dfrac{176}{21}\times 516$

$\rm\dashrightarrow V= 4324.57$

$\underline{\boxed{\purple{\rm \therefore Volume \: of \: the \: frustum = 4324.57cm^{3}}}}$