A beaker of radius r is filled with water (re | vedclass

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(B) In $	riangle OAB$,where $O$ is the center of curvature,$A$ is a point on the circumference,and $B$ is the center of the water surface:$R^2 = (R-h

Vedclass · Accepted Answer

$A, C$

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(B) In $	riangle OAB$,where $O$ is the center of curvature,$A$ is a point on the circumference,and $B$ is the center of the water surface:$R^2 = (R-h)^2 + r^2$$R^2 = R^2 - 2Rh + h^2 + r^2$$2Rh = h^2 + r^2$$R = \frac{h^2 + r^2}{2h}$. Since $h \ll r$,$R \approx \frac{r^2}{2h}$. Thus,$(A)$ and $(B)$ are both technically correct approximations,but $(A)$ is the exact geometric relation.For the rotating fluid,the surface profile is $y = y_0 + \frac{\omega^2 r^2}{2g}$. The height difference is $h = \frac{\omega^2 r^2}{2g}$.Using the refraction formula at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here $\mu_1 = \frac{4}{3}$ (water),$\mu_2 = 1$ (air),$u = -H$ (approx),and the surface is concave,so $R$ is negative: $-R$.$\frac{1}{v} - \frac{4/3}{-H} = \frac{1 - 4/3}{-R} \Rightarrow \frac{1}{v} + \frac{4}{3H} = \frac{1}{3R}$.$\frac{1}{v} = \frac{1}{3R} - \frac{4}{3H} = \frac{2h}{3r^2} - \frac{4}{3H}$.Substituting $h = \frac{\omega^2 r^2}{2g}$:$\frac{1}{v} = \frac{2(\omega^2 r^2 / 2g)}{3r^2} - \frac{4}{3H} = \frac{\omega^2}{3g} - \frac{4}{3H} = -\frac{4}{3H} (1 - \frac{\omega^2 H}{4g})$.$v = -\frac{3H}{4} (1 - \frac{\omega^2 H}{4g})^{-1} \approx -\frac{3H}{4} (1 + \frac{\omega^2 H}{4g})^{-1}$.Thus,$(C)$ is correct.

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