Physics : Properties of Matter - Solved Example Problems for Surface tension

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.

*Solution:*

A soap bubble has two free surfaces, therefore increase in surface area ∆*A* = *A*2−*A*1* = *2(100-50) × 10-4*m*2* *= 100 × 10-4*m*2.

Since, work done *W* = *T* ×Δ*A* ⇒*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

*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 *r*1 and *r*2 which on placing in a liquid, capillary rises to height *h*1 and *h*2, 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

Capillary descent,

where, negative sign indicates that there is fall of mercury (mercury is depressed) in glass tube.

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11th Physics : UNIT 7 : Properties of Matter : Solved Example Problems for Surface tension |

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