A capillary tube of radius r is immersed in water and water rises in to a height h. The mass of water in the capillary tube is 5g. Another capillary tube of radius 2 r is immersed in water. The mass of water that will rise in this tube is

Options

A. `5.0` g
B. 10 g
C. 20 g
D. `2.5` g

Answer

B

Explanation

` = (rhpg)/(2 cos theta) " For water " theta = 0^(@) and cos theta = 1 `
` :. h = (2T)/(rrhog)`
The mass of water that is lifted in the capillary tube is
` m = pir^(2) h rho = pi r^(2) rho ((2T)/(rrhog)) = ((2piT)/g)r `
Thus ` m prop r `
` (m')/m = (r')/r = 2 ". m' = 2m = 10 g `
Second method :
For capillary rise , `h_(1)r_(1) = h_(2)r_(2)`
` :. (h_(2))/(h_(1)) = (r_(1))/(r_(2)) = (r_(1))/(2r_(1)) = 1/2 " " ` ...(1)
The mass of water lifted ` M = pir^(2) h rho p`
` (M_(2))/(M_(1)) = (pir_(2)^(2)h_(2)rho)/(pir_(1)^(2)h_(1)rho) = ((r_(2))/(r_(1)))^(2) xx ((h_(2))/(h_(1)))`
` = (2/1)^(2) xx(1/2)=2 `
` :. M _(2) = 2M_(1) = 2 xx 5 = 10 g`