Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion

Discussion Forum : Hydraulics and Fluid Mechanics - Section 3 (Q.No. 10)
10.
With an increase in size of tube, the rise or depression of liquid in the tube due to surface tension will
decrease
increase
remain unchanged
depend upon the characteristics of liquid
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Shashikant Rai said:   6 years ago
@Rahu.

Due to capillary action the rise or fall of liquid in a tube is calculated by the formula.

h = ( 2*t*cosY)/(ρ*g*r) or = ( 4*t*cosY)/(w*d).

Therefore h is inversely proportional to d.
(1)

Velmurugan R said:   6 years ago
P = (4*Σ)/d.

P is inversely proportional to the diameter

Kataiah sajja said:   6 years ago
Units of surface tension is N/m, so surface tension is inversaly proportional to the diameter of tube. The diameter mentioned in m, mm, cm etc.

Rahu said:   6 years ago
How? Please explain in clear.

Jai said:   7 years ago
h=4 * ρ 8cosq/wd.
ρ(surface tension) is directly proportional to d.
So, the surface tension increases.

Kishan Ahir said:   7 years ago
D is inverlsly propositional to H.

Rajkumar Nayak said:   10 years ago
h = 4*t*cos (x)/wd, where, keeping all term constant.

We can say that rise or fall is inversely proportional to dia of tube.

Rana Dhaval said:   10 years ago
h = 4q cosQ/ 2gd. So h is inversely proportional to d.

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