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 be 22 cm, the diameter of the cylindrical portion 8 cm and the diameter of the top of the funnel 18 cm, find the area of the tin required.
(Use π = 22/7).

Answer 1

Let r₁ and r₂ (r₁ > r₂) be the radii of the ends of the frustum of a cone. Suppose l and h be the slant height and height of frustum of cone.

Let h₁ be the height of the cylindrical portion.
Now, h₁ = 10 cm
Total height of the funnel = 22 cm
height of frustum of cone, h = 22−10 = 12 cm

$$\begin{gathered} l=\sqrt{{\left(r_1-r_2\right)}^2+h^2} \\ \Rightarrow l=\sqrt{{\left(9-4\right)}^2+{\left(12\right)}^2} \\ \Rightarrow l=\sqrt{25+144} \\ \Rightarrow l=\sqrt{169} \\ \Rightarrow l=13\mathrm{cm} \end{gathered}$$

Curved surface area of funnel = CSA of frustum of cone + CSA of cylinder

Curved surface area of funnel = $\mathrm{\pi }\left(r_1+r_2\right)l+2\mathrm{\pi }{r}_2{h}_1$

Therefore,

Area of the tin required

$$\begin{gathered} =\mathrm{\pi }\left(r_1+r_2\right)l+2\mathrm{\pi }{r}_2{h}_1 \\ =\mathrm{\pi }\left[\left(r_{\mathit{1}}+r_2\right)l+2r_2{h}_1\right] \\ =\mathrm{\pi }\left[\left(9+4\right)\times 13+2\times 4\times 10\right] \\ =\mathrm{\pi }\left[169+80\right] \\ =\mathrm{\pi }\left[249\right] \\ =249\mathrm{\pi }{\mathrm{cm}}^2 \end{gathered}$$