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

Explanation:

Given

Radius, r1 = r 

 r2 = 2r

height, h1 = h

and, mass, M1 = 5 g

TO FIND:

The Mass of the water will rise in the second capillary tube.

As we have, r h = constant

Therefore, r₂h₂ = r₁h₁

Putting the given values we have;

(2r)h₂ = r h

⇒ h₂ = \(\frac{h}{2}\) ---(1)

Now, the mass of the water that rises in the first capillary tube we have;

\(Density = \frac{mass}{volume}\)

⇒ \(M_1 =\rho(\pi r^2 h)\)

⇒ \(5 =\rho(\pi r^2 h)\)

Here, \(\rho\) is the density of the water in the capillary tube.

and, the mass of the second capillary tube we have,

\(M_2 =\rho(\pi r^2_2 h_2)\) ---(2)

Putting equation (1) in equation (2) we have;

\(M_2 =\rho(\pi (2r)^2 \frac{h}{2})\)

⇒ \(M_2=2(\rho\pi r^2 \frac {h}{2})\)

⇒ M2 = 2 M1

⇒ M₂ = 2 × 5

⇒ M₂ = 10 g

Hence, option 1) is the correct answer.