A `16cm^(3)` volume of water flows per second through a capillary tube of radius r cm and length l cm , when connected to a pressure head of h cm of water. If a tube of same length and radius `r//2` is connected to the same pressure head, find the mass of water flowing per minutes through the tube.

Explanation

Here, `V_(1) = 16 cm^(3)//sec , p_(1) = h rho g , r,r_(1) =r, l_(1)=l`
`V_(2)=?, l_(2) =l , r_(2)=r//2 , p_(2) = h rho g , so, p_(1)=p_(2)`
Now , `V_(1) = (pi p_(1)r_(1)^(4))/(8 pi l_(1)) and V_(2) = (pi p_(2)r_(2)^(4))/(8 pi l_(2))`
`:. (V_2)/(V_1) = (p_(2))/(P-1) xx (r_(2)^(4))/(r_(1)^(4)) xx (l_1)/(l_2) = ((r//2)^(4))/(r^4) xx l/l = (1/2)^(4) = 1/16`
`V_(2) = (V_1)/(16) = 16/16 = 1 cm^(3)//s`,
Volume of water flowing per minute = ` 1xx 60 = 60cm^(3)//mi n`
`:.` Mass of water flowing per minute = `60 xx 1 = 60 gram`.