Search any question & find its solution
Question:
Answered & Verified by Expert
In capillary tube having area of cross-section ' $\mathrm{A}$ ', water rises to a height ' $h$ '. If cross sectional area is reduced to $\frac{A}{9}$, the rise of water in the capillary tube is
Options:
Solution:
2148 Upvotes
Verified Answer
The correct answer is:
$3 \mathrm{~h}$
$\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r \rho g}} \\ & \Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{r}}\end{aligned}$
And $\mathrm{A}=\pi \mathrm{r}^2 \Rightarrow \mathrm{r} \propto \sqrt{\mathrm{A}}$
$\Rightarrow \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{A}}}$
Now, $\frac{h_2}{h_1}=\sqrt{\frac{A_1}{A_2}}$
$\begin{aligned} & \Rightarrow \frac{h_2}{h}=\sqrt{\frac{A}{A / 9}}=3 \\ & \Rightarrow h_2=3 h\end{aligned}$
And $\mathrm{A}=\pi \mathrm{r}^2 \Rightarrow \mathrm{r} \propto \sqrt{\mathrm{A}}$
$\Rightarrow \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{A}}}$
Now, $\frac{h_2}{h_1}=\sqrt{\frac{A_1}{A_2}}$
$\begin{aligned} & \Rightarrow \frac{h_2}{h}=\sqrt{\frac{A}{A / 9}}=3 \\ & \Rightarrow h_2=3 h\end{aligned}$
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.