A beaker of radius r is filled with water (refractive index(4)/(3))...

$h = \dfrac{{\omega}^{2}{r}^{2}}{2g}$

${r}^{2} + ( R - h{)}^{2} = {R}^{2}$

${r}^{2} = {R}^{2} - ( R - h{)}^{2} = ( 2R - h ) h$

${r}^{2} = 2Rh - {h}^{2}$

$∴ R = \dfrac{{r}^{2} + {h}^{2}}{2h}$       ......option (A)

For apparent depth, now

$\text{As}r ≪ h$,

$R = \dfrac{{r}^{2}}{2h} = \dfrac{g}{{\omega}^{2}}$

Refraction formula,

$\dfrac{1}{v} - \dfrac{4}{3 ( H - h )} = \dfrac{1 - 4 / 3}{R}$

\left|\dfrac{1}{v}\right| = \dfrac{1}{3R} + \dfrac{4}{3 ( H - h )}$</p><p>Since$H ≫ h$,</p><p>\left|\dfrac{1}{v}\right| = \dfrac{{\omega}^{2}}{3g} + \dfrac{4}{3H} = \dfrac{4}{3H}\left[1 + \dfrac{3H}{4} \times \dfrac{{\omega}^{2}}{3g}\right]$

$⇒ | v | = \dfrac{3H}{4}{\left[1 + \dfrac{{\omega}^{2}H}{4g}\right]}^{- 1}$        .......option (D)