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प्रश्न
A test tube consists of a hemisphere and a cylinder of the same radius. The volume of the water required to fill the whole tube is `5159/6 cm^3` and `4235/6 cm^3` of water is required to fill the tube to a level which is 4 cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.
उत्तर
Volume of water filled in the test tube =`5159/6 cm^3`
Volume of water filled up to 4 cm =`4235/6 cm^3`
Let r be the radius and h be the height of test tube
∴ `2/3 pir^3 + pir^2h = 5159/6`
`=> pir^2(2/3r + h) = 5159/6`
`=> (pir^2)/3(2r + 3h) = 5159/6`
`=> pir^2(2r + 3h) = 5159/2` ...(i)
And
`2/3pir^3 + pir^2(h - 4) = 4235/6`
`=> pir^2(2/3r + h - 4) = 4235/6`
`=> (pir^2)/3(2r + 3h - 12) = 4235/6`
`=> pir^2(2r + 3h - 12) = 4235/2` ...(ii)
Dividing (i) by (ii)
`(2r + 3h)/(2r + 3h - 12) = 5259/4235` ...(iii)
Subtracting (ii) from (i)
`pir^2(12) = 5159/2 - 4235/2 = 924/2`
`=> 12 xx 22/7 xx r^2 = 924/2`
`=> r^2 = (924 xx 7)/(2 xx 12 xx 22) = (7 xx 7)/(2 xx 2)`
`=> r^2 = 49/4`
`=> r = 7/2 = 3.5 cm`
Subtracting the value of r in (iii)
`(2 xx 7/2 + 3h)/(2 xx 7/2 + 3h - 12) = 5159/4235`
`=> (7 + 3h)/(7 + 3h - 12) = 5159/4235`
`=> (7 + 3h)/(7 + 3h - 12) = 469/385`
`=>` 2695 + 1155h = 1407h – 2345
`=>` 252h = 5040
`=>` h = 20
Hence, height = 20 cm and radius = 3.5 cm
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