Apparent Depth and Curvature Radius of Rotating Liquid Surface

To determine which of the statements are correct, we analyze the geometry of the curved liquid surface and the optical refraction through it step-by-step.

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1. Determination of Radius of Curvature ($R$)

The question states that the water surface is approximately spherical. Consider the cross-section of a spherical cap:

• Let the lowest point (center) of the water surface be the origin/pole.
• The height difference between the rim (radius $r$) and the center is $h$.
• The center of curvature lies at a distance $R$ vertically above the lowest point.

Using the Pythagorean theorem for the right-angled triangle formed by the center of curvature, the center of the circular rim, and a point on the rim: $(R - h)^2 + r^2 = R^2$

Expanding the equation: $R^2 - 2Rh + h^2 + r^2 = R^2$

Rearranging terms to solve for $R$: $2Rh = r^2 + h^2 \implies R = \frac{h^2 + r^2}{2h}$

Thus, Option (A) is correct and Option (B) is incorrect.

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2. Radius of Curvature at the Center of the Rotating Liquid

The profile of a liquid surface rotating with angular velocity $\omega$ forms a paraboloid of revolution given by: $y(x) = \frac{\omega^2 x^2}{2g}$

The radius of curvature $R$ at the lowest point ($x = 0$) is calculated as: $R = \left. \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\left|\frac{d^2y}{dx^2}\right|} \right|_{x=0}$

Differentiating $y(x)$: $\frac{dy}{dx} = \frac{\omega^2 x}{g} \quad \implies \quad \left.\frac{dy}{dx}\right|_{x=0} = 0$ $\frac{d^2y}{dx^2} = \frac{\omega^2}{g}$

Substituting these into the expression for $R$: $R = \frac{g}{\omega^2}$

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3. Determination of the Apparent Depth

Since $h \ll H$, the depth of water at the center can be approximated as $z_0 \approx H$.

Now, consider the refraction of light rays from an object at the bottom of the beaker viewed normally from above through the curved surface:

• Refractive index of medium 1 (water): $\mu_1 = \frac{4}{3}$
• Refractive index of medium 2 (air): $\mu_2 = 1$
• Object distance: $u = -H$
• Radius of curvature: Since the center of curvature lies in air (above the surface), $R_{surface} = +R = +\frac{g}{\omega^2}$

Applying the spherical surface refraction formula: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_{surface}}$

Substituting the known values: $\frac{1}{v} - \frac{4/3}{-H} = \frac{1 - 4/3}{R}$

$\frac{1}{v} + \frac{4}{3H} = -\frac{1}{3R}$

$\frac{1}{v} = -\frac{4}{3H} - \frac{1}{3R} = -\frac{4}{3H} \left(1 + \frac{H}{4R}\right)$

Taking the magnitude to find the apparent depth $d_{\text{app}} = -v$: $d_{\text{app}} = \frac{3H}{4} \left(1 + \frac{H}{4R}\right)^{-1}$

Substituting $R = \frac{g}{\omega^2}$: $d_{\text{app}} = \frac{3H}{4} \left(1 + \frac{\omega^2 H}{4g}\right)^{-1}$

Thus, Option (D) is correct and Option (C) is incorrect.

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Conclusion

energy The correct statements are A and D.