Answer :

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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