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 $0.1mm$ 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 $(d)$
$\therefore T=\frac{FL}{2\pi rL}$
$\Rightarrow F=2\pi rT$
Given, $r=0.1mm=0.1\times 10^{-3}m$,
$T=7\times 10^{-2}N/m$
$\therefore F=2\times \frac{22}{7}\times 10^{-4}\times 7\times 10^{-2}$
$\Rightarrow F=44\times 10^{-6}N$
Alternative The height $(h)$ of water rise in a capillary tube is given by
$h=\frac{2T}{r\rho g}$ ...(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 $=Ah\rho g=\frac{2T}{r}A=\frac{2T}{r}\times \pi r^2=2T{\pi }^r$
Given, $T=7\times 10^{-2}N/m,$
$r=0.1mm=0.1\times 10^{-3}m$
$w=2\times 7\times 10^{-2}\times 3.14\times 10^{-3}\times 0.1$
$\Rightarrow w=44\times 10^{-6}N$