Question

A capillary tube of internal radius 2 x 10-3m immersed vertically in a beaker containing a liquid. If the weight of the liquid rising in
the capillary tube is 9 x 10-5kg, the surface tension of liquid is​

Answer 1

Surface tension of liquid = 7 x 10 ^-2 N/m

Explanation:

Given: A capillary tube of internal radius 2 x 10-3m immersed vertically in a beaker containing a liquid. Weight of the liquid rising in the capillary tube is 9 x 10-5kg.

Find: Surface tension of liquid.

Solution:

r = 2 x 10^-3 m

We know that buoyant force F = v ρ g  = weight of liquid rising in the capillary tube

 = 9 x 10^-5 kg

Surface tension can be calculated using the formula T = F / 2πr

 = (9 x 10^-5) / (2 x 3.14 x 2 x 10^-3)

 = 0.00716 kg / m

We know that 1 kg/m = 9.806 N/m

So converting 0.00716 kg/m to N/m = 0.00716 x 9.806

 = 0.0702 N/m

 = 7 x 10 ^-2 N/m

Answer 2

Explanation:

Given A capillary tube of internal radius 2 x 10-3 m immersed vertically in a beaker containing a liquid. If the weight of the liquid rising inthe capillary tube is 9 x 10-5 kg, the surface tension of liquid is

• Given radius of capillary tube = r = 2 x 10^-3 m and weight = 9 x 10^-5 kg. We need to find the surface tension of the liquid.
• So r = 2 x 10^-3 m
• F = w = v ρ g = 9 x 10^- 5 kg
• So surface tension T = F / 2 π r
•                                 T = 9 x 10^-5 / 2 x 3.14 x 2 x 10^-3
•                              T = 9 x 10^-2 / 12.56  
•                             T = 0.716 x 10^-2 kg / m
•              Or T = 0.716 x 9.806 x 10^-2
•              Or T = 7.021 x 10^-2 N / m

Reference link will be

https://brainly.in/question/2802908