Q1 of 48 Page 270

A metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The total vertical height of the bucket is 40 cm and that of cylindrical base is 10 cm, radii of two circular ends are 60 cm and 20 cm. Find the area of the metallic sheet used. Also find the volume of water the bucket can hold. (π = 3.14)

Given.


Height of bucket is 40 cm


Height of cylinder is 10 cm


Radii of both circular ends of frustum are 60 cm, 20 cm


Formula used/Theory.


CSA of cylinder = 2πrh


CSA of frustum = π(r + R) × l


Volume of cylinder = πr2h


Volume of frustum = h[R2 + r2 + Rr]


As the bucket is always open from mouth


Metallic sheet require will be sum of CSA of Frustum and CSA of cylinder and Area of base circle


In frustum


L2 = height2 +


Height of frustum = height of bucket – height of cylinder


= 40 – 10 = 30 cm


L2 = 302 + [60 – 20]2


L2 = 900 cm2 + 1600 cm2 = 2500 cm2


L = √(2500 cm2) = 50cm


⇒ CSA of Frustum = π(60 + 20)50 cm


= 3.14 × 80cm × 50cm


= 12560cm2


⇒ CSA of cylinder = 2πrh


= 2 × 3.14 × 20cm × 10cm


= 1256 cm2


⇒ Area of base = πr2


= 3.14 × 20 × 20


= 1256 cm2


Area of metallic sheet = 12560 cm2 + 1256 cm2 + 1256 cm2


= 15072 cm2


⇒ Volume of Frustum = h[R2 + r2 + Rr]


= × 30cm × [(60 cm)2 + (20 cm)2 + 60 cm × 20 cm]


= 31.4 cm × [3600cm2 + 400cm2 + 1200cm2]


= 31.4 cm × 5200 cm2


= 163280 cm3


1 cm3 = litres


163280 cm3 = litres


= 163.280 litres


∴ Area of metallic sheet used in making bucket is 15072 cm2


Volume of bucket is 163.280 litres


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