Question

Radius of a capillary is $2 \times 10^{-3} m$. A liquid of weight $6.2 \times 10^{-4} N$ 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)

$2 \pi R T \cos \theta=mg \left(\theta=0^{\circ}\right)$
$T=\frac{m g}{2 \pi R}$
$=\frac{6.2 \times 10^{-4}}{2 \times 3.14 \times 2 \times 10^{-3}}$
$T=5 \times 10^{-2} N / m$