KCET · Physics · Mechanical Properties of Fluids
Two capillary tubes \(P\) and \(Q\) are dipped vertically in water. The height of water level in capillary tube \(P\) is \(\frac{2}{3}\) of the height in capillary tube \(Q\). The ratio of their diameter is
- A \(2: 3\)
- B \(3: 2\)
- C \(3: 4\)
- D \(4: 3\)
Answer & Solution
Correct Answer
(B) \(3: 2\)
Step-by-step Solution
Detailed explanation
As we know, height of capillary rise or fall of a liquid in a capillary tube is given as
\(h=\frac{2 T \cos \theta}{r \rho g}\)
\(\Rightarrow h \propto \frac{1}{r}...(i)\)
where, \(r\) is the radius of the capillary tube.
Given, height of water level in capillary tube \(P=\frac{2}{3}\) height of water in capillary tube \(Q\).
\(\Rightarrow h_{P}=\frac{2}{3} h_{Q}\)
\(\text {or }\frac{h_{P}}{h_{Q}}=\frac{2}{3}\)
From Eq. (i), we can write
\(\frac{h_{P}}{h_{Q}}=\frac{r_{Q}}{r_{P}}\) radius.
So,
\(\frac{(\text {diameter})_{P}}{(\text {diameter})_{Q}}=\frac{3}{2}\)
\(h=\frac{2 T \cos \theta}{r \rho g}\)
\(\Rightarrow h \propto \frac{1}{r}...(i)\)
where, \(r\) is the radius of the capillary tube.
Given, height of water level in capillary tube \(P=\frac{2}{3}\) height of water in capillary tube \(Q\).
\(\Rightarrow h_{P}=\frac{2}{3} h_{Q}\)
\(\text {or }\frac{h_{P}}{h_{Q}}=\frac{2}{3}\)
From Eq. (i), we can write
\(\frac{h_{P}}{h_{Q}}=\frac{r_{Q}}{r_{P}}\) radius.
So,
\(\frac{(\text {diameter})_{P}}{(\text {diameter})_{Q}}=\frac{3}{2}\)
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