Let 2 .4×10−4 J of work is done to increase the area of a film of soap bubble from 50 cm2 to 100 cm2. Calculate the value of surface tension of soap solution.
A soap bubble has two free surfaces, therefore increase in surface area ∆A = A2−A1 = 2(100-50) × 10-4m2 = 100 × 10-4m2.
Since, work done W = T ×ΔA ⇒T =
If excess pressure is balanced by a column of oil (with specific gravity 0.8) 4 mm high, where R = 2.0 cm, find the surface tension of the soap bubble.
The excess of pressure inside the soap bubble is
T = 15.68 ×10− 2 N m−1
Water rises in a capillary tube to a height of 2.0cm. How much will the water rise through another capillary tube whose radius is one-third of the first tube?
From equation (7.34), we have
h ∝ 1/r ⇒hr =constant
Consider two capillary tubes with radius r1 and r2 which on placing in a liquid, capillary rises to height h1 and h2, respectively. Then,
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 2 mm, made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside?. Surface tension of mercury T=0.456 N m-1; Density of mercury ρ = 13.6 × 103 kg m-3
where, negative sign indicates that there is fall of mercury (mercury is depressed) in glass tube.