A beaker of radius r is filled with water refractive index 4 3 up to a height H as shown in the figure on…

A beaker of radius r is filled with waterrefractive index 43 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 ω. 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 hh≪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)

  1. R=h2+r22h
  2. R=3r22h
  3. Apparent depth of the bottom of the beaker is close to 3H21+ω2H2g-1
  4. Apparent depth of the bottom of the beaker is close to 3H41+ω2H4g-1

Solution

$h = \frac{\omega^2 r^2}{2g}$ $r^2 + (R - h)^2 = R^2$ $r^2 = R^2 - (R - h)^2 = (2R - h)h$ $r^2 = 2Rh - h^2$ $\therefore R = \frac{r^2 + h^2}{2h}$ ......option (A) For apparent depth, now As $r \ll h$, $R = \frac{r^2}{2h} = \frac{g}{\omega^2}$ Refraction formula, $\frac{1}{v} - \frac{4}{3}(H - h) = \frac{1 - \frac{4}{3}}{R}$ $\left| \frac{1}{v} \right| = \frac{1}{3R} + \frac{4}{3}(H - h)$ Since $H \gg h$, $\left| \frac{1}{v} \right| $= $\frac{\omega^2}{3g}$ + $\frac{4}{3H}$ = $\frac{4}{3H} \left[ 1 + \frac{3H}{4} \times \frac{\omega^2}{3g} \right]$ $\Rightarrow |v|$ = $\frac{3H}{4} \left[ 1 + \frac{\omega^2 H}{4g} \right]^{-1}$ ......option (D)

Asked in: JEE Advanced 2020 (Paper 2)

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