Search any question & find its solution
Question:
Answered & Verified by Expert
A capillary tube of radius ' $r$ ' is immersed in water and water rises to a height of ' $h$ '. Mass of water in the capillary tube is $5 \times 10^{-3} \mathrm{~kg}$. 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 $\mathrm{kg}$ )
Options:
Solution:
2498 Upvotes
Verified Answer
The correct answer is:
$5 \times 10^{-3}$
We knows height of water rise in a capillary tube
$$
\begin{gathered}
h=\frac{2 T \cos \theta}{r d g} \\
h_1=\frac{2 T_1 \cos \theta_1}{r d g}, \quad h_2=\frac{2 T_2 \cos \theta_2}{r d g}
\end{gathered}
$$
Given, $h_1=h, T_1=T, \theta_1=0$
$$
\therefore \quad h=\frac{2 T}{r d g}
$$
Given, $T_2=\sqrt{2} T, \theta=45^{\circ}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$$
\therefore \quad h_2=\frac{2 \sqrt{2} T \times \frac{1}{\sqrt{2}}}{r d g}
$$
From Eqs. (i) and (ii), we observe
$$
h_2=h \text {. }
$$
Hence, same mass of liquid rises into the capillary as before $5 \times 10^{-3} \mathrm{~kg}$.
$$
\begin{gathered}
h=\frac{2 T \cos \theta}{r d g} \\
h_1=\frac{2 T_1 \cos \theta_1}{r d g}, \quad h_2=\frac{2 T_2 \cos \theta_2}{r d g}
\end{gathered}
$$
Given, $h_1=h, T_1=T, \theta_1=0$
$$
\therefore \quad h=\frac{2 T}{r d g}
$$
Given, $T_2=\sqrt{2} T, \theta=45^{\circ}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$
$$
\therefore \quad h_2=\frac{2 \sqrt{2} T \times \frac{1}{\sqrt{2}}}{r d g}
$$
From Eqs. (i) and (ii), we observe
$$
h_2=h \text {. }
$$
Hence, same mass of liquid rises into the capillary as before $5 \times 10^{-3} \mathrm{~kg}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.