An open metallic bucket is in the shape of a frustum of a cone mounted on hollow cylindrical base made of metallic sheet. If the diameters of the two circular ends of the bucket are 45cm and 25cm, the total vertical height of the bucket is 30cm and that of the cylindrical portion is 6cm, find the area of the metallic sheet used to make the bucket. Also, find the volume of the water it can hold. (Take π=227)
We have,
Height of the frustum of the coneh=(30−6)=24cm
Radii of the circular ends are r1=225 and r2=12.5cm
l=Slant height of the frustum
l=√h2+(r1−r2)2
l=√242+(22.5−12.5)2=26cm
Area of metallic sheet used=Curved surface area of the frustum of cone+area of circular base+curved surface area of cylinder
=π(r1−r2)l+πr22+2πr2h2
=π[(22.5−12.5)×26+12.52+2×12.5×6]
=227×1216.25=3822.5cm2
Volume of water that the bucket can hold
=13π(r21+r22+r1r2)h
=13×227×(22.52+12.52+22.5×12.5)×24
=23719.02cm3
=23.719liters