Advertisements
Advertisements
प्रश्न
Figure (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.
Advertisements
उत्तर १
Take case (a):
The length of the liquid film supported by the weight, l = 40 cm = 0.4 cm
The weight supported by the film, W = 4.5 × 10–2 N
A liquid film has two free surfaces.
∴Surface tension = `W/(2l)`
`= (4.5 xx 10^(-2))/(2xx0.4) = 5.625 xx 10^(-2) Mn^(-1)`
In all the three figures, the liquid is the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a), i.e., 5.625 × 10–2 N m–1.
Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10–2 N.
उत्तर २
(a) Here, length of the film supporting the weight = 40 cm = 0.4 m. Total weight supported (or force) = 4.5 x 10-2 N.
Film has two free surfaces, Surface tension, S =4.5 x 10-2/2 x 0.4 =5.625 x 10-2 Nm-1
Since the liquid is same for all the cases (a), (b) and (c), and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same = 5.625 x 10-2. In Fig. 7(b), 38(b) and (c), the length of the film supporting the weight is also the saihe as that of (a), hence the total weight supported in each case is 4.5 x 10-2 N.
संबंधित प्रश्न
Show that the surface tension of a liquid is numerically equal to the surface energy per unit
area.
When a glass capillary tube is dipped at one end in water, water rises in the tube. The gravitational potential energy is thus increased. Is it a violation of conservation of energy?
If water in one flask and castor oil in other are violently shaken and kept on a table, which will come to rest earlier?
If two soap bubbles of different radii are connected by a tube,
Consider a small surface area of 1 mm2 at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = 1.0 × 105 Pa and surface tension of mercury = 0.465 N m−1. Neglect the effect of gravity. Assume all numbers to be exact.
A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J m−2.
Solve the previous problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping into water.
A solid sphere of radius 5 cm floats in water. If a maximum load of 0.1 kg can be put on it without wetting the load, find the specific gravity of the material of the sphere.
How much amount of work is done in forming a soap bubble of radius r?
Explain the phenomena of surface tension on the basis of molecular theory.
Describe an experiment to prove that friction depends on the nature of a surface.
The wettability of a surface by a liquid depends primarily on
A square frame of each side L is dipped in a soap solution and taken out. The force acting on the film formed is _____.
(T = surface tension of soap solution).
What is surface tension? Explain the applications of surface tension.
Two mercury droplets of radii 0.1 cm. and 0.2 cm. collapse into one single drop. What amount of energy is released? The surface tension of mercury T = 435.5 × 10–3 Nm–1.
Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water Lv = 540 k cal kg–1, the mechanical equivalent of heat J = 4.2 J cal–1, density of water ρw = 103 kg l–1, Avagadro’s No NA = 6.0 × 1026 k mole–1 and the molecular weight of water MA = 18 kg for 1 k mole.
- Estimate the energy required for one molecule of water to evaporate.
- Show that the inter–molecular distance for water is `d = [M_A/N_A xx 1/ρ_w]^(1/3)` and find its value.
- 1 g of water in the vapor state at 1 atm occupies 1601 cm3. Estimate the intermolecular distance at boiling point, in the vapour state.
- During vaporisation a molecule overcomes a force F, assumed constant, to go from an inter-molecular distance d to d ′. Estimate the value of F.
- Calculate F/d, which is a measure of the surface tension.
A liquid flows out drop by drop from a vessel through a vertical tube with an internal diameter of 2 mm, then the total number of drops that flows out during 10 grams of the liquid flow out ______. [Assume that the diameter of the neck of a drop at the moment it breaks away is equal to the internal diameter of tube and surface tension is 0.02 N/m].
A drop of water and a soap bubble have the same radii. Surface tension of soap solution is half of that of water. The ratio of excess pressure inside the drop and bubble is ______.
Calculate (i) the pressure due to the weight of the water at a depth of 2.5 m and (ii) the depth below the surface of water at which the pressure due to the weight of the water equals 1.0 atm.
