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. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket i

Explanation

Here, `R=45/2 cm, r-22/2 cm = 12.5 cm.`
Height of the frustum of the cone, h = (30 - 6) cm = 24 cm.
`:." h = 24 cm"`.
Slant height of the frustum of the cone,
`l=sqrt(h^(2)+(R-r)^(2))"units="sqrt((24)^(2)+(22.5-12.5)^(2))cm`
`=sqrt((24)^(2)+(10)^(2))cm=sqrt(576+100)cm`
`=sqrt(676)cm = 26 cm.`
Area of the metallic sheet used
= (curved surface area of the frustum of the cone) + (area of the base) + ( curved surface area of the cylinder)
`=pil(R+r)+pir^(2)+2pirH," where H = 6 cm"`
`=pi{l(R+r)+r^(2)+2rH}" sq units"`
`=22/7{26(22.5+12.5)+(12.5)^(2)+2xx12.5xx6}cm^(2)`
`=22/7{(26xx35)+(12.5xx12.5)+150}cm^(2)`
`=22/7{910+156.25+150}cm^(2)=(22/7xx1216.25)cm^(2)`
`=(22xx173.75)cm^(2)=3822.5 cm^(2)`.
Volume of water which the bucket can hold
`=1/3pih{R^(2)+r^(2)+Rr}cm^(3)`
`=1/3xx22/7xx24xx[(45/2)^(2)+(25/2)^(2)+(45/2xx25/2)]cm^(3)`
`=176/7xx(2025/4+625/4+1125/4)cm^(3)=(176/7xx3775/4)cm^(3)`
`=166100/7cm^(3)=23728.57cm^(3)=23.73" litres".`