Q: A beaker of radius r is filled with water (Refractive index = 4/3) up to height H as shown in the figure on the left . The beaker is kept on a horizontal table rotating with angular speed ω . This makes water surface curved so that the difference in 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) $\displaystyle R = \frac{h^2 + r^2}{2h}$

(b) $\displaystyle R = \frac{3 r^2}{2h}$

(c) Apparent depth of the bottom of beaker is close to $\displaystyle v = [\frac{3H}{2}(1 + \frac{\omega^2 H}{2g})^{-1}] $

(d) Apparent depth of the bottom of beaker is close to $\displaystyle v = [\frac{3H}{4}(1 + \frac{\omega^2 H}{4g})^{-1}] $

Ans: (a,d)

Solution:

$\displaystyle (R-h)^2 + r^2 = R^2 $

$\displaystyle R^2 + h^2 – 2 R h + r^2 = R^2 $

$\displaystyle h^2 + r^2 = 2Rh $

$\displaystyle R = \frac{h^2 + r^2}{2h} $

Since h << r

$\displaystyle R = \frac{r^2}{2h} $

$\displaystyle h = \frac{\omega^2 r^2}{2 g} $

$\displaystyle R = \frac{r^2 2 g}{2\omega^2 r^2} $

$\displaystyle R = \frac{g}{\omega^2} $

$\displaystyle \frac{\mu_1}{v} – \frac{\mu_2}{u} = \frac{\mu_1 – \mu_2}{R}$

$\displaystyle \frac{1}{v} – \frac{4}{3u} = \frac{1 – 4/3}{R}$

$\displaystyle \frac{1}{v} = – (\frac{1}{3R} + \frac{4}{3(H-h)}) $

$\displaystyle \frac{1}{v} = – (\frac{1}{3R} + \frac{4}{3H}) $ (Since h << H)

Putting the value of R we get ,

$\displaystyle v = – [\frac{3H}{4}(1 + \frac{\omega^2 H}{4 g})^{-1}] $