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

Question

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 5 g. Another capillary tube of radius 2 r is immersed in water. The mass of water that will rise in this tube is (A) 2.5 g (B) 5.0 g (C) 10 g (D) 20 g

Tardigrade Admin · Accepted Answer

h=(2S cos θ/rρ g) Mass of water in the first tube, m=π r2hρ=π r2×((2S cos θ/rρ g))×ρ =(2π rS cos θ/g) ∴ m propto r. Hence, (m'/m)=(2r/r)=2 or m'=2m=2× 5 g=10 g.

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

2.5 g

5.0 g

10 g

20 g

$h = \frac{2 S cos θ}{r ρ g}$
Mass of water in the first tube,
$m = π r^{2} h ρ = π r^{2} \times (\frac{2 S cos θ}{r ρ g}) \times ρ$
$= \frac{2 π r S cos θ}{g}$
$∴ m \propto r .$ Hence, $\frac{m ^{'}}{m} = \frac{2 r}{r} = 2$
or $m^{'} = 2 m = 2 \times 5 g = 10 g .$

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