Question

An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion be 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.

$\left[3\right]$

Answer 1

Radius (R) of upper circular end of frustum part = 18/2​ = 9 cm
Radius (r) of lower circular end of frustum part = Radius of circular end of cylindrical part =8/2 ​= 4 cm

Height ($h_1$​) of frustum part =22−10=12 cm
Height ($h_2$​) of cylindrical part =10 cm

Slant height (l) of frustum part =$\sqrt{{(R-r)}^2+{h_1}^2}=\sqrt{{(9-4)}^2+{12}^2}=\sqrt{169}=13\mathrm{cm}$$\left[1.5\right]$
Area of tin sheet required = CSA of frustum part + CSA of cylindrical part
$=\pi (R+r)l+2{\mathrm{\pi rh}}_2=\frac{22}{7}\times (9+4)\times 13+2\times \frac{22}{7}\times 4\times 10=\frac{22}{7}\times (169+80)=782.57{\mathrm{cm}}^2$

$\left[1.5\right]$