Question

A beaker of radius $r$ is filled with water $($ refractive index $\frac{4}{3})$ up to a height $H$ as shown in the figure on the left. The beaker is kept on a horizontal table rotating with angular speed $\omega$. This makes the water surface curved so that the difference in the height of water level at the center and at the circumference of the beaker is $h(h<<H,h<<r)$, as shown in the figure on the right. Take this surface to be approximately spherical with a radius of curvature $R$. Which of the following is/are correct? ($g$ is the acceleration due to gravity)

• A. $R=\frac{h^2+r^2}{2h}$
• B. $R=\frac{3r^2}{2h}$
• C. Apparent depth of the bottom of the beaker is close to $\frac{3H}{2}{\left(1+\frac{{\omega }^{2}H}{2g}\right)}^{-1}$
• D. Apparent depth of the bottom of the beaker is close to $\frac{3H}{4}{\left(1+\frac{{\omega }^{2}H}{4g}\right)}^{-1}$

Answer 1

Answer: (A), (D)

$\tan \theta =\frac{dy}{dx}=\frac{{\omega }^{2}x}{g}$
$\Rightarrow y=\frac{{\omega }^2{x}^2}{2g}$
$\Rightarrow h=\frac{{\omega }^2{r}^2}{2g}$
Pythagoras $\Rightarrow r^2+(R-h{)}^2=R^2$
$\Rightarrow R=\frac{r^2+h^2}{2h}$
$\Rightarrow$ A option is correct
For apparent depth applying refraction formula
$\frac{1}{v}-\frac{4}{3(H-h)}=\frac{1-\frac{4}{3}}{R}$
$\Rightarrow \left|\frac{1}{v}\right|=\frac{1}{3R}+\frac{4}{3H}($ as $H>>h)$
$\Rightarrow |v|=\frac{3H}{4}\left[1+\frac{{\omega }^{2}H}{4g}\right]$
$\Rightarrow$ D option is correct.