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A certain number of spherical drops of a liquid of radius R coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

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प्रश्न

A certain number of spherical drops of a liquid of radius R coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

पर्याय

  • energy = `4 "VT" (1/"r" - 1/"R")` is released

  • energy = `3 "VT" (1/"r" + 1/"R")` is absorbed

  • energy = `3 "VT" (1/"r" - 1/"R")` is released

  • energy is neither released nor absorbed

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उत्तर

energy = `3 "VT" (1/"r" - 1/"R")` is released

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पाठ 7: Properties of Matter - Evaluation [पृष्ठ ९१]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 7 Properties of Matter
Evaluation | Q I. 12. | पृष्ठ ९१

संबंधित प्रश्‍न

Derive Laplace’s law for spherical membrane of bubble due to surface tension.


Define surface tension and surface energy.


The force of surface tension acts tangentially to the surface whereas the force due to air pressure acts perpendicularly on the surface. How is then the force due to excess pressure inside a bubble balanced by the force due to the surface tension?


The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly ______.


Numerical Problem.

A stone weighs 500 N. Calculate the pressure exerted by it if it makes contact with a surface of area 25 cm2.


How is surface tension related to surface energy?


A drop of oil placed on the surface of water spreads out. But a drop of water place on oil contracts to a spherical shape. Why?


Soap solution is used for cleaning dirty clothes because ______.


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.

  1. Estimate the energy required for one molecule of water to evaporate.
  2. Show that the inter–molecular distance for water is `d = [M_A/N_A xx 1/ρ_w]^(1/3)` and find its value.
  3. 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.
  4. 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.
  5. Calculate F/d, which is a measure of the surface tension.

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