An oil funnel of tin sheet consist of a cylendrical portion 10 cm long attached to a frustum of cone. If the total height be 22cm , diameter of the cylendrical portion be 8cm and the diameter of the top of the funnel be 18cm find the area of the tin required to make the funnel?
Answers
Let r 1 and r 2 ( r 1 > r 2) be the radii of ends of the frustum of a cone.
Suppose l, h and H be the slant height of the frustum, height of the frustum and height of the cylinder respectively.
Diameter of top of the frustum, 2r 1 = 18 cm
∴ r 1 = 9 cm.
Diameter of bottom of the frustum, 2r 2 = 8 cm
∴ r 2 = 4 cm
Height of the frustum = Total height of funnel – Height of the cylinder = 22 cm – 10 cm = 12 cm
Slant height of the frustum, l
= h 2 + ( r 1 - r 2 ) 2
= (12 cm) 2 + (9 cm - 4 cm) 2
= 144 cm 2 + 25 cm 2
= 169 cm 2
= 13 cm
Curved surface area of the oil funnel = π( r 1 + r 2)l + 2πr 2H
∴ Area of tin sheet required to make the funnel
= π( r 1 + r 2)l + 2πr 2 H
= 22 7 (9 cm + 4 cm) × 13 cm + 2 × 22 7 × 4 cm × 10 cm
= 22 7 (13 × 13 cm 2 + 80 cm 2 )
= 22 7 × ( 169 cm 2 + 80 cm 2 )
= 22 7 × 249 cm 2
= 782.6 m 2
Thus, the area of tin sheet required to make the oil funnel is 782.6 cm2.
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