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  • A liquid is kept in a cylindrical vessel. When the vessel is rotated about its axis, the liquid rises at its sides. If the radius of the vessel is <img alt="0.05 \mathrm{~m}" src="https://entrancec

A liquid is kept in a cylindrical vessel. When the vessel is rotated about its axis, the liquid rises at its sides. If the radius of the vessel is 0.05 \mathrm{~m} and the frequency of rotation is 2 revolutions per second, the difference in the heights of the liquid at the centre and at the sides of the vessel will be \mathrm{(take \, \, g=10 \mathrm{~ms}^{-2}\, \, and \, \, \pi^2=10 )}

Option: 1

2 cm


Option: 2

4 cm


Option: 3

1 cm


Option: 4

8 cm


Answers (1)

best_answer

Using Bernoulli's theorem, we have

                       \mathrm{ \begin{aligned} \frac{1}{2} \rho v^2 & =p=\rho g h \\ h & =\frac{v^2}{2 g} \end{aligned} }
Now \mathrm{v=r \omega=r(2 \pi v)}.

Using this in (1), we get

                     \mathrm{ h=\frac{2 \pi^2 v^2 r^2}{g} }
Given, \mathrm{v=2} rev. per second, \mathrm{r=0.05 \mathrm{~m}, g=10 \mathrm{~ms}^{-2}} and \mathrm{\pi^2=10}.

Using these values, we get

\mathrm{h=0.02 \mathrm{~m}=2 \mathrm{~cm}},

which is choice (a).

Posted by

Rishi

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