Question 453: Properties of Matter

Question

A U-tube is rotated about one of its limbs with an angular velocity $$\omega$$. Find the difference in height $$\mathrm{H}$$ of the liquid (density $$\rho$$ ) level, where the diameter of the tube is $$d < <\mathrm{L}$$.

Accepted Answer

Weight of liquid column of height $$\mathrm{H}, \mathrm{W}=ho \mathrm{V} g$$
$$=\frac{ho\left(\pi d^{2}ight)}{4} \mathrm{Hg}$$ ....... (i)
Let us consider a small element $$d x$$ of mass $$d m$$ at a distance $$x$$ (as shown in figure).
Centripetal force required, in rotating this mass $$=d m \omega^{2}$$
$$	herefore$$ Total force required to rotate mass of length $$\mathrm{L}$$,
$$\begin{aligned}
\mathrm{F} & =\int d m \omega^{2} \
& =\int_{0}^{\mathrm{L}} \frac{ho\left(\pi d^{2}ight)}{4} x d x \omega^{2} \
& =\frac{ho\left(\pi d^{2}ight)}{4} \frac{\mathrm{L}^{2}}{2} \omega^{2}
~~........ (\mathrm{ii})\end{aligned}$$
This centripetal force is provided by weight of 
liquid column (of height H), Here on comparing 
equation (i) and (ii), we get
$$\begin{aligned}
W & =\mathrm{F} \
\frac{ho\left(\pi d^{2}ight)}{4} \mathrm{Hg} & =\frac{ho\left(\pi d^{2}ight)}{4} \frac{\mathrm{L}^{2}}{2} \omega^{2} \
\mathrm{H} & =\frac{\mathrm{L}^{2} \omega^{2}}{2 g}
\end{aligned}$$

Answer

$$\mathrm{H=\omega^2L^2/2g}$$

Answer

$$\mathrm{H=\omega^2L^2/3g}$$

Answer

$$\mathrm{H=5\omega^2L^2/2g}$$

Answer

$$\mathrm{H=\omega^2L^2/g}$$

Properties of Matter

A U-tube is rotated about one of its limbs with an angular velocity $\omega. Find the difference in height \mathrm{H} of the liquid (density \rho ) level, where the diameter of the tube is d < <\mathrm{L}$.

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