Q156 / 180 · KCET 2021 · Physics
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$2: 3$
- Correct answerB$3: 2$
- C$3: 4$
- D$4: 3$
Solution
As we know, height of capillary rise or fall of a liquid in a capillary tube is given as
$\begin{array}{ll}
& h=\frac{2 T \cos \theta}{r \rho g} \\
\Rightarrow \quad & h \propto \frac{1}{r}...(i)
\end{array}$
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 $Q$.
$\begin{array}{ll}\Rightarrow & h_{P}=\frac{2}{3} h_{Q} \\ \text { or } & \frac{h_{P}}{h_{Q}}=\frac{2}{3}\end{array}$
From Eq. (i), we can write
$\quad \frac{h_{P}}{h_{Q}}=\frac{r_{Q}}{r_{P}}$ radius.
So, $\quad \frac{(\text { diameter })_{P}}{(\text { diameter })_{Q}}=\frac{3}{2}$
Hence, the correct answer is option: 'B'.