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
Answers
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
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