Question

If the surface tension of liquid is $ 5\times {{10}^{-2}}N/m $ and weight of liquid column is $ 6.28\times {{10}^{-4}}N, $ then radius of capillary is

• A. $ 2\times {{10}^{-3}}m $
• B. $ 2.5\times {{10}^{-3}}m $
• C. $ 2\times {{10}^{-4}} $
• D. $ 4\times {{10}^{-3}}m $

Answer 1

Answer: (A)

$ F=2\pi rT $
$ \Rightarrow $ $ r=\frac{F}{2\pi T} $ $ =\frac{6.28\times {{10}^{-4}}}{2\times 3.14\times 5\times {{10}^{-2}}}=2\times {{10}^{-3}}m $