A thin,circular wire ring (radius = 0.04 m) and total mass 0.0007 kg
is gradually drawn apart from the water surface in a vertical direction by means of a sensity spring with spring constant k=0.75 N.m-1. When
the spring is stretched 0.034 m from its equilibrium extension in air,
the ring is on the verge of being pulled free from the water surface.
Find the coefficient of surface tension of water.
Neglect the mass of the lifted water.
|
Strategy - the rule "Must have" of Mr.Zhang ®
|
| Coefficient of surface tension(γ) | = |
Tension Force(Ft)
Circumference(L)
|
|
→ |
SI unit N.m-1
|
Ftension → (Forced required to stop movable side from sliding)
|
| L(circumference ) =2(2πr) ; because of the ring shape. |
| Equilibrium ;
|
|
↑ Fspring
|
= |
↓
mg + γl
↓
|
→(1) |
|
Solution |
|
Compute the coefficient of surface tension (γ) of water by solving the equation (1) |
| kx
|
=
|
mg + γl
|
| kx
|
=
|
mg + γ2(2πr)
|
| γ
|
=
|
|
| γ
|
=
|
(0.75 N.m-1)(3.4x10-2m) - (0.7x10-3kg)(9.81 m.s-2)
4π(2x10-2m)
|
|
| γ
|
=
|
0.026 N - 0.0069 N
0.25 m
|
|
| γ
|
=
|
0.076 N.m-1
|
|
Ans.
The coefficient of surface tension of water is 0.076 N.m-1
|
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