Question

If surface tension of water is 0.07 N/m the weight of water supported in a capillary tube of radius 0.1 mm is?

Answer 1

Surface tension S
radius = R
density of water = d
weight of water column of height h = W
W = (2 π R) S
So W = 2 * π * 0.1 * 10^-3 m * 0.07 N/m = 4.4 * 10^-4 Newtons approx

Answer 2

Answer:

The weight of water supported in a capillary tube is 4.396×10⁻⁵N.

Explanation:

Let's see the following formula,

$h=\frac{2T}{\rho gr}$       (1)

Where,

h=height of capillarity

T=surface tension of water

ρ=density of water

g=acceleration due to gravity=10m/s²

r=radius of capillarity

We can also write equation (1) as,

$(\rho gh)\times r=2T$     (2)

Also,

$P=\rho gh$     (3)

P=pressure acting

By putting equation (3) in equation (2) we get;

$P\times r=2T$     (4)

Pressure can also be calculated as,

$P=\frac{W}{\pi r^2}$     (5)

By putting equation (5) in equation (4) we get;

$\frac{W}{\pi r^2}\times r=2T$

$\frac{W}{\pi r}=2T$

$W=2T\times \pi \times r$      (6)

W=weight of water supported

From the question we have,

The surface tension of water(T)=0.07N/m

The radius of the capillary tube=0.1mm=0.1×10⁻³m

By placing the value of "T" and "r" in equation (6) we get;

$W=2\times 0.07\times \pi \times 0.1\times 10^-^3$

$W=43.96\times 10^-^6\ N$

$W=4.396\times 10^-^5\ N$

Hence, the weight of water supported in a capillary tube is 4.396×10⁻⁵N.

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