TS EAMCET · Physics · Mechanical Properties of Fluids
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 \(\mathrm{kg}\) )
- 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 & Solution
Correct Answer
(A) \(5 \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
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, \(h_1=h, T_1=T, \theta_1=0\)…
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