Question

A capiIIary tube of radius ' $r$ ' is immersed in water and water rises to a height of ' $h$ '. Mass of water in the capillary tube is $5 \times 10^{-3}\, kg$. The same capillary tube is now immersed in a liquid whose surface tension is $\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $45^{\circ} .$ The mass of liquid which rises into the capillary tube now is, (in $kg$ )

• A. $5 \times 10^{-3}$
• B. $2.5 \times 10^{-3}$
• C. $5 \sqrt{2} \times 10^{-3}$
• D. $3.5 \times 10^{-3}$

Answer 1

Answer: (A)

We knows height of water rise in a capillary tube
$h=\frac{2 T \cos \theta}{r d g}$
$h_{1}=\frac{2 T_{1} \cos \theta_{1}}{r d g}, h_{2}=\frac{2 T_{2} \cos \theta_{2}}{r d g}$
Given, $h_{1}=h, T_{1}-T, \theta_{1}=0$
$\therefore h=\frac{2 T}{r d g}$ ...(i)
Given, $T_{2}=\sqrt{2} T, \theta=45^{\circ}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$\therefore h_{2}=\frac{2 \sqrt{2} T \times \frac{1}{\sqrt{2}}}{r d g}$ ...(ii)
From Eqs. (i) and (ii), we observe
$h_{2}=h .$
Hence, same mass of liquid rises into the capillary as before $5 \times 10^{-3}\, kg$.