Advertisements
Advertisements
Question
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.
Solution
Given:
Surface area of mercury drop, A = 1 mm2 = 10−6 m2
Radius of mercury drop, r = 4 mm = 4 × 10−3 m
Atmospheric pressure, P0 = 1.0 × 105 Pa
Surface tension of mercury, T = 0.465 N/m
(a) Force exerted by air on the surface area:
F = P0A
⇒ F= 1.0 × 105 × 10−6 = 0.1 N
(b) Force exerted by mercury below the surface area :
\[\text{ Pressure P' = P}_0 + \frac{2T}{\text{r}}\]
\[F = P'A = \left( P_0 + \frac{2T}{r} \right)A\]
\[ = \left( 0 . 1 + \frac{2 \times 0 . 465}{4 \times {10}^{- 3}} \right) \times {10}^{- 6} \]
\[ = 0 . 1 + 0 . 00023 = 0 . 10023 \text{N}\]
(c) Force exerted by mercury surface in contact with it:
\[P = \frac{2T}{r}\]
\[F = PA = \frac{2T}{r}A\]
\[ = \frac{2 \times 0 . 465}{4 \times {10}^{- 3}} \times {10}^{- 6} = 0 . 00023 \text{ N}\]
APPEARS IN
RELATED QUESTIONS
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.
A big drop of radius R is formed from 1000 droplets of water. The radius of a droplet will be _______
A) 10 R
B) R/10
C) R/100
D) R/1000
It is said that a liquid rises or is depressed in capillary due to the surface tension. If a liquid neither rises nor depresses in a capillary, can we conclude that the surface tension of the liquid is zero?
The contact angle between water and glass is 0°. When water is poured in a glass to the maximum of its capacity, the water surface is convex upward. The angle of contact in such a situation is more than 90°. Explain.
Frictional force between solids operates even when they do not move with respect to each other. Do we have viscous force acting between two layers even if there is no relative motion?
A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of water raised in the capillary. (b) If the length of the capillary tube is half the answer of part , find the angle θ made by the water surface in the capillary with the wall.
Derive an expression for capillary rise for a liquid having a concave meniscus.
Calculate the rise of water inside a clean glass capillary tube of radius 0.1 mm, when immersed in water of surface tension 7 × 10-2 N/m. The angle of contact between water and glass is zero, the density of water = 1000 kg/m3, g = 9.8 m/s2.
The property of _______ of a liquid surface enables the water droplets to move upward in plants.
A capillary of diameter d mm is dipped in water such that the water rises to a height of 30 mm. If the radius of the capillary is made `(2/3)` of its previous value, then compute the height up to which water will rise in the new capillary?
Two spherical rain drops reach the surface of the earth with terminal velocities having ratio 16 : 9. The ratio of their surface area is ______.
A molecule of water on the surface experiences a net ______.
Why is raindrop spherical in nature?
The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius r = 2.5 × 10–5 m. The surface tension of sap is T = 7.28 × 10–2 Nm–1 and the angle of contact is 0°. Does surface tension alone account for the supply of water to the top of all trees?
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 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 ______.
When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes `(5r)/4`. Taking the atmospheric pressure to be equal to the 10 m height of the water column, the depth of the lake would approximately be ______.
(ignore the surface tension and the effect of temperature)
Work done to blow a bubble of volume V is W. The work done in blowing a bubble of volume 2V will be ______.
A drop of water of radius 8 mm breaks into number of droplets each of radius 1 mm. How many droplets will be formed?
