A capillary tube of a certain radius is dipped vertically into a liquid. The mass of the liquid that rises in the tube is measured. If the radius of the tube decreases by 1 % , the surface tension of the liquid increases by 2 % , and the density of the liquid increases by 1.5 % (assuming the angle of contact remains constant), the percentage change in the mass of the liquid raised in the capillary tube will be:

Step-by-step solution

The height h of the liquid raised in a capillary tube is given by: h = 2S g r The mass M of the liquid in the capillary tube is: M = Volume density = ( r^2 h) Substituting the expression for h : M = r^2 ( 2S g r ) = 2 r S g From this relation, the mass M is directly proportional to the radius r and the surface tension S , and is completely independent of the density . For small percentage changes, we can write the relative error equation as: M M 100 = ( r r + S S ) 100 Given that the radius decreases by 1 % ( r r =