An oil funnel of tin sheet consists of a cylindrical portion of length 1o cm long attached to a frustum - Maths - Probability

Question

An oil funnel of tin sheet consists of a cylindrical portion of
length 1o cm long attached to a frustum of cone if the total height
of the funnel is 22cm and the diametre of the cylindical portion be
8 cm and the diamter of the top of the funnel be 18cm find the area
of tin required to make the funnel

Gursheen kaur. · Accepted Answer

Let $r_{1}$ and $r_{2}$ ( r1 > r2) be the radii of ends of the frustum of a cone Also, $L e t l, h_{1} & h_{2}$ be the slant height of the frustum, height of the frustum and height of the cylinder Radius of upper circular end of frustum part, $r_{1} = \frac{18}{2} = 9 c m$ Radius of lower circular end of frustum part=radius of circular end of cylindrical part, $r_{2} = \frac{8}{2} = 4 c m$ Now, Height of frustum part, $h_{1} = 22 - 10 = 12 c m$ Height of cylindrical part=10 cm Slant height of frustum part , $l = \sqrt{{(r_{1} - r_{2})}^{2} + {(h)}^{2}}$ $l = \sqrt{{(9 - 4)}^{2} + {(12)}^{2}} l = \sqrt{{(5)}^{2} + {(12)}^{2}} l = \sqrt{25 + 144} l = \sqrt{169} l = 13 c m$ Area of tin sheet required = Curved Surface Area of frustum part+Curved Surface Area of cylindrical part $= π (r_{1} + r_{2}) l + 2 π r_{2} h_{2} = \frac{22}{7} \times (9 + 4) \times 13 + 2 \times \frac{22}{7} \times 4 \times 10 = \frac{22}{7} \times 13 \times 13 + \frac{22}{7} \times 80 = \frac{22}{7} \times 169 + \frac{22}{7} \times 80 = \frac{22}{7} \times (169 + 80) = \frac{22}{7} \times 249 = 782 \frac{4}{7} {c m}^{2} = 782 57 {c m}^{2}$