Question

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. 

Answer 1

 
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