JEE PYQ: Mechanical Properties Of Fluids Question 9
Question 9 - 2020 (02 Sep 2020 Shift 1)
A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is 5 cm and the angular speed of rotation is $\omega$ rad s$^{-1}$. The difference in the height, $h$ (in cm) of liquid at the centre of vessel and at the side will be:
(1) $\frac{2\omega^2}{25g}$
(2) $\frac{5\omega^2}{2g}$
(3) $\frac{25\omega^2}{2g}$
(4) $\frac{2\omega^2}{5g}$
Show Answer
Answer: (3) $\frac{25\omega^2}{2g}$
Solution
Here, $\rho d r \omega^2 r = \rho g , dh$ $\Rightarrow \omega^2 \int_0^R r , dr = g \int dh$ $\Rightarrow \frac{\omega^2 R^2}{2} = gh$ (Given $R = 5$ cm) $\therefore h = \frac{\omega^2 R^2}{2g} = \frac{25\omega^2}{2g}$
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