Question

Radius of a capillary is $2\times 10^{-3}m.\,A$ liquid of weight $6.28\times 10^{-4}N$ may remain in the capillary, then the surface tension of liquid will be

• A. $ 5\times 10^{-3}N/m $
• B. $ 5\times 10^{-2}N/m $
• C. $5\, N/m$
• D. $50\, N/m$

Answer 1

Answer: (B)

Given, $r=2 \times 10^{-3} m.$
$F=W=V \rho g=6.28 \times 10^{-4} m$
Surface tension $(T)=\frac{F}{2 \pi r}=\frac{6.28 \times 10^{-4}}{2 \times 3.14 \times 2 \times 10^{-3}}$
$=0.05\, N / m =5 \times 10^{-2} N / m$