Question

A beaker of radius r is filled with water (refractive index $$\dfrac{4}{3}$$) 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 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)

Answer 1

Correct option is D. Apparent depth of the bottom of the beaker is close to $$\dfrac{3H}{4} \left(1 + \dfrac{\omega^2 H}{4g} \right)^{-1}$$
$$9th \ \Delta ABC$$

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

$$R^2 + h^2 - 2Rh + r^2 = R^2$$

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

$$\dfrac{h^2 + r^2}{2h} = R$$ ___(i)

$$r >>> h$$

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

$$\because h = \dfrac{w^2 r^2}{2g}$$

$$R = \dfrac{r^2 2g}{2w^2 r^2}$$

$$R = \dfrac{g}{w^2}$$ ___(ii)

$$\dfrac{\mu_1}{V} - \dfrac{\mu_2}{u} = \dfrac{\mu_1 - \mu_2}{R}$$

$$\dfrac{1}{V} + \dfrac{4}{3u} = \dfrac{1 - \dfrac{4}{3}}{R}$$

$$\dfrac{1}{V} = -\left(\dfrac{1}{3R} + \dfrac{4}{3(H - h)}\right)$$

$$\dfrac{1}{V} =-\left[\dfrac{1}{3R} + \dfrac{4}{3H}\right]$$

Put the value of equation (ii) $$h <<H$$

$$\dfrac{1}{V} = -\left[\dfrac{4}{3H} \left[1 + \dfrac{3H}{4} \dfrac{w^2}{39} \right]\right]$$

$$V = -\left[\dfrac{3H}{4} \left(1 + \dfrac{w^2 H}{4}\right)^{-1}\right]$$