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


Answer :

Given,

Diameter of upper circular end of frustum part = 18 cm

So, radius (r1) = 9 cm

Now, the radius of the lower circular end of frustum (r2) will be equal to the radius of the circular end of the cylinder

So, r2 = \(\frac{8}{2} \) = 4 cm

Now, height (h1) of the frustum section = 22 - 10 = 12 cm

And,

Height (h2) of cylindrical section = 10 cm (given)

Now, the slant height will be, \( l \ = \ \sqrt{(r_1 \ - \ r_2)^2 \ + \ (h_1)^2} \)

Or, l = 13 cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

\( = \ (r_1 \ + \ r_2)l \ + \ 2 \pi r_2 h_2 \)

Solving this we get,

Area of tin sheet required = 782 × \( \frac{4}{7} \) cm2

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