Question

The surface tension of water is $7 \times 10^{-2} N / m$. The weight of water supported by surface tension in a capillary tube of $0.1 \,mm$ radius, will be

• A. $44 \times 10^{-6} N$
• B. $22 \times 10^{-6} N$
• C. $33 \times 10^{-6} N$
• D. None of these

Answer 1

Answer: (A)

Surface tension $(T)$ is defined as work done $(W)$ in increasing the surface area $(\Delta A)$ of a film by unity at constant temperature.
$\therefore T=\frac{W}{\Delta A}$
Also, $W=$ force $(F) \times$ displacement $(d)$
$\therefore T=\frac{F L}{2 \pi r L}$
$\Rightarrow F=2 \pi r T$
Given, $r=0.1\, mm =0.1 \times 10^{-3} m$,
$T=7 \times 10^{-2} N / m$
$\therefore F=2 \times \frac{22}{7} \times 10^{-4} \times 7 \times 10^{-2}$
$\Rightarrow F=44 \times 10^{-6} N$
Alternative The height $(h)$ of water rise in a capillary tube is given by
$h=\frac{2 T}{r \rho g}$ ...(i)
Also, Weight $(w)$ =volume $(V) \times$ density $(\rho)\times g$
$w=$ Area $(A) \times$ height $(h) \times \rho \times g$ ...(ii)
From Eqs. (i) and (ii), we get
Weight $=A h \rho g=\frac{2 T}{r} A=\frac{2 T}{r} \times \pi r^{2}=2 T \pi^{r}$
Given, $T=7 \times 10^{-2} N / m,$
$r =0.1 mm =0.1 \times 10^{-3} m$
$w =2 \times 7 \times 10^{-2} \times 3.14 \times 10^{-3} \times 0.1$
$\Rightarrow w =44 \times 10^{-6} N$