Q. 74.6( 13 Votes )

An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet (see Fig.) The diameters of the two circular ends of the bucket are 45 cm and 25cm, the total vertical height of the bucket is 40cm and that of the cylindrical base is 6cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold. [Take π=22/7]


Answer :


Diameter of bigger circular end = 45cm


So,


Diameter of smaller circular end = 25cm


So,


Now,


Height of frustum = Total height of bucket – Height of cylinder


= 40 – 6


= 34cm


Also,


Slant height of the frustum, l = √{h2 + (R – r)2}




= √1156 + 100


= √ 1256


= 35.44cm


Curved Surface Area of frustum = πl(R + r)




= 22 × 35.44 × 5


= 3898.4 cm2


Curved Surface area of cylinder = 2πrh



= 471.428 cm2


Area of circular base = πr2



= 491.07 cm2


Now, Area of metallic sheet used


= CSA of frustum + Area of circular base + CSA of cylinder


= 3898.4 + 471.428 + 491.07


= 4860.89


= 4860.9 cm2






= 33615.47 cm3


We know that 1 cm3 = 0.001 litre


Volume of water that bucket can hold = 33.62 litres (approx.)


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