A beaker of radius $r$ is filled with water$\left(\text{refractive index} \frac{4}{3}\right)$ 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 centre and at the circumference of the beaker is $h\left(h\ll H\text{,} h\ll r\right)$, 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)

• Correct answer
  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 $\frac{3H}{2}{\left(1+\frac{{\omega }^{2}H}{2g}\right)}^{-1}$
• Correct answer
  D

  Apparent depth of the bottom of the beaker is close to $\frac{3H}{4}{\left(1+\frac{{\omega }^{2}H}{4g}\right)}^{-1}$

Hence, the correct answer is options: 'A', 'D'.