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# An oil funnel made of tin sheet consists of a $ 10 \mathrm{~cm} $ long cylindrical portion attached to a frustum of a cone. If the total height is $ 22 \mathrm{~cm} $, diameter of the cylindrical portion is $ 8 \mathrm{~cm} $ and the diameter of the top of the funnel is $ 18 \mathrm{~cm} $, find the area of the tin sheet required to make the funnel (see Fig. 13.25).

"

Given:

An oil funnel made of tin sheet consists of a \( 10 \mathrm{~cm} \) long cylindrical portion attached to a frustum of a cone.

The total height is \( 22 \mathrm{~cm} \), diameter of the cylindrical portion is \( 8 \mathrm{~cm} \) and the diameter of the top of the funnel is \( 18 \mathrm{~cm} \).

To do:

We have to find the area of the tin sheet required to make the funnel.

Solution:

Diameter of the upper circular end of frustum $= 18\ cm$

This implies,

Radius of the upper circular end of frustum $(r_1) = 9\ cm$

The radius of the lower circular end of frustum $(r_2)=$ Radius of the circular end of the cylinder

$r_2 = \frac{8}{2}$

$= 4\ cm$

Height of the frustum $h_1= 22 - 10$

$= 12\ cm$

Height of the cylindrical part $h_2= 10\ cm$

Slant height $l=\sqrt{(r_1-r_2)^2+h_1^2}$

$=\sqrt{(9-4)^2+12^2}$

$=\sqrt{25+144}$

$=\sqrt{169}$

$=13\ cm$

Area of the tin sheet required $=$ Curved surface area of frustum part $+$ Curved surface area of cylindrical part

$= \pi (r_1+r_2)l+2\pi r_2h_2$

$=\frac{22}{7}(9+4)\times13+2\times\frac{22}{7}\times4\times10$

$=\frac{22}{7}(169+80)$

$=\frac{22}{7}\times249$

$=782\frac{4}{7}\ cm^2$

**The area of the tin sheet required is $782\frac{4}{7}\ cm^2$.**

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