Question

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

Answer 1

Answer: (A), (D)

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