Search any question & find its solution
Question:
Answered & Verified by Expert
A liquid kept in a cylindrical vessel is rotated about vertical axis through the centre
of circular base. The difference in the heights of the liquid at the centre of vessel and
its edge is $(\mathrm{R}=$ radius of vessel, $\omega=$ angular velocity of rotation, $\mathrm{g}=$ acceleration
due to gravity)
Options:
of circular base. The difference in the heights of the liquid at the centre of vessel and
its edge is $(\mathrm{R}=$ radius of vessel, $\omega=$ angular velocity of rotation, $\mathrm{g}=$ acceleration
due to gravity)
Solution:
2016 Upvotes
Verified Answer
The correct answer is:
$\frac{R^{2} \omega^{2}}{2 g}$
When the cylindrical vessel is rotated at angular speed $\omega$ about its axis, the velocity of the liquid at the sides is maximum, given by $\mathrm{v}_{\mathrm{s}}=\mathrm{r} \omega$
Applying Bernoulli's theorem at the sides and at the center of the vessel, we have
$\mathrm{P}+\frac{1}{2} \rho \mathrm{v}^{2}=$ constant $\mathrm{P}_{\mathrm{s}}+\frac{1}{2} \rho \mathrm{v}_{\mathrm{s}}^{2}=\mathrm{P}_{\mathrm{c}}+\frac{1}{2} \rho \mathrm{v}_{\mathrm{c}}{ }^{2}$
where
$\mathrm{P}_{\mathrm{s}}=$ pressure at the sides
$\mathrm{v}_{\mathrm{s}}=$ velocity of the liquid at the sides
$\mathrm{P}_{\mathrm{c}}=$ pressure at the center
$\mathrm{v}_{\mathrm{c}}=$ velocity of the liquid at the center
$\mathrm{P}_{\mathrm{c}}-\mathrm{P}_{\mathrm{s}}=\frac{1}{2} \rho \mathrm{v}_{\mathrm{s}}^{2}=\frac{1}{2} \rho \mathrm{r}^{2} \omega^{2} \ldots(\mathrm{I}) \quad \because \mathrm{v}_{\mathrm{c}}=0$
Since $\mathrm{P}_{\mathrm{c}}$ is greater than $\mathrm{P}_{\mathrm{s}}$, the liquid rises at the sides of the vessel. Let $\mathrm{h}$ be the difference in the levels of the liquids at the sides and at the center, so we have
$\mathrm{P}_{\mathrm{c}}-\mathrm{P}_{\mathrm{s}}=\rho g h \ldots(\mathrm{II})$
from (I) and (II) we have
$\rho g h=\frac{1}{2} \rho r^{2} \omega^{2}$
$\Rightarrow h=\frac{r^{2} \omega^{2}}{2 g}$
Applying Bernoulli's theorem at the sides and at the center of the vessel, we have
$\mathrm{P}+\frac{1}{2} \rho \mathrm{v}^{2}=$ constant $\mathrm{P}_{\mathrm{s}}+\frac{1}{2} \rho \mathrm{v}_{\mathrm{s}}^{2}=\mathrm{P}_{\mathrm{c}}+\frac{1}{2} \rho \mathrm{v}_{\mathrm{c}}{ }^{2}$
where
$\mathrm{P}_{\mathrm{s}}=$ pressure at the sides
$\mathrm{v}_{\mathrm{s}}=$ velocity of the liquid at the sides
$\mathrm{P}_{\mathrm{c}}=$ pressure at the center
$\mathrm{v}_{\mathrm{c}}=$ velocity of the liquid at the center
$\mathrm{P}_{\mathrm{c}}-\mathrm{P}_{\mathrm{s}}=\frac{1}{2} \rho \mathrm{v}_{\mathrm{s}}^{2}=\frac{1}{2} \rho \mathrm{r}^{2} \omega^{2} \ldots(\mathrm{I}) \quad \because \mathrm{v}_{\mathrm{c}}=0$
Since $\mathrm{P}_{\mathrm{c}}$ is greater than $\mathrm{P}_{\mathrm{s}}$, the liquid rises at the sides of the vessel. Let $\mathrm{h}$ be the difference in the levels of the liquids at the sides and at the center, so we have
$\mathrm{P}_{\mathrm{c}}-\mathrm{P}_{\mathrm{s}}=\rho g h \ldots(\mathrm{II})$
from (I) and (II) we have
$\rho g h=\frac{1}{2} \rho r^{2} \omega^{2}$
$\Rightarrow h=\frac{r^{2} \omega^{2}}{2 g}$
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.