A capillary tube of radius \(\mathrm{R}\) is immersed in water and water rises in it to a height \(\mathrm{H}\). Mass of water in the capillary tube is M. If the radius of the tube is doubled. Mass of water that will rise in the capillary tube will now be (A) \(\mathrm{M}\) (B) \(2 \mathrm{M}\) (C) \((\mathrm{M} / 2)\) (D) \(4 \mathrm{M}\)

We know that mass is the product of volume and density. The volume of the water in the capillary tube can be written as the product of the cross-sectional area of the tube, which is given by \(πR^2\), and the height to which the water rises, which is given by H. Therefore, the volume of the water in the capillary tube can be written as: Volume = \(πR^2H\) Since mass = volume × density, the mass of the water in the tube can be written as: M = ρ (density of water) × \(πR^2H\)

Let's assume the new radius is \(2R\). In the problem statement, it's not explicitly mentioned how the height will change when the radius is doubled, but since the same amount of water will be in the new tube, we can relate the heights of the two tubes by saying: \(Volume_1 = Volume_2\) or \(πR^2H = (π(2R)^2H_2)\) Solving this equation for \(H_2\), we get: \(H_2 = \frac{H}{4}\)

Now that we have the new height, we can find the new mass of the water in the capillary tube: M' (New mass) = ρ (density of water) × \(π(2R)^2H_2\) Substitute the value of \(H_2\) in the equation: M' = ρ (density of water) × \(π(2R)^2\left(\frac{H}{4}\right)\) M' = ρ (density of water) × \(\frac{π(4R^2)H}{4}\) Now, we can rewrite this equation as: M' = ρ (density of water) × \(πR^2H\) Since we already know that: M = ρ (density of water) × \(πR^2H\) Therefore, after canceling out the terms, we can see that: M' = M So, the answer is (A) \(\mathrm{M}\). When the radius of the capillary tube is doubled, the mass of the water that will rise in the capillary tube will remain the same as the initial mass M.