Part 1: Analysis of the water surface and radius of curvature (Options A and B)

1. Equation of the Surface: When the beaker rotates with angular speed $\omega$, a fluid particle of mass $m$ on the surface at a radial distance $x$ from the center experiences a downward gravitational force $mg$ and a horizontal centrifugal force $m\omega^2x$. The surface of the water aligns itself perpendicular to the effective force. The slope of the surface is given by: $\tan \theta = {{dy} \over {dx}} = {{m{\omega ^2}x} \over {mg}} = {{{{\omega ^2}x} \over g}}$ Integrating this equation, we get the shape of the surface: $\int {dy} = \int {{{{\omega ^2}x} \over g}dx} \Rightarrow y = {{{\omega ^2}{x^2}} \over {2g}} + C$ Taking the lowest point of the surface at the center ($x=0$) as the origin for the vertical displacement ($y=0$), we get $C=0$. So, the surface is a parabola described by $y = {{{{\omega ^2}{x^2}} \over {2g}}}$.

2. Relating h and r: The problem states that the height difference between the center ($x=0$) and the circumference ($x=r$) is $h$. $h = y(r) - y(0) = {{{{\omega ^2}{r^2}} \over {2g}}} \Rightarrow {{{\omega ^2}} \over g} = {{2h} \over {{r^2}}}$

3. Radius of Curvature (R): The radius of curvature of a curve $y=f(x)$ is given by the formula R = {{{{\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]}^{3/2}}} \over {\left| {{{{d^2}y} \over {d{x^2}}}} \right|}}}. We have $y' = {{dy} \over {dx}} = {{{{\omega ^2}x} \over g}}$ and $y'' = {{{{d^2}y} \over {d{x^2}}}} = {{{{\omega ^2}} \over g}}$. For optical calculations near the center (paraxial approximation), we need the radius of curvature at the vertex ($x=0$). $R = {{{{\left[ {1 + 0} \right]}^{3/2}}} \over {{{{\omega ^2}} \over g}}} = {g \over {{\omega ^2}}}$ Substituting the expression for ${{{\omega ^2}} \over g}$ from step 2: $R = {1 \over {{{2h} \over {{r^2}}}}} = {{{r^2}} \over {2h}}$

4. Evaluating Options A and B:

   • Option B is $R = {{3{r^2}} \over {2h}}$, which is incorrect.
   • Option A is R = {{{h^2} + {r^2}} \over {2h}}}. This formula is the exact radius of a spherical cap with base radius $r$ and height (sagitta) $h$. In many optics problems, parabolic surfaces are approximated as spherical. If we assume $h \ll r$, then $h^2$ is negligible, and Option A simplifies to $R \approx {{{r^2}} \over {2h}}$, which matches our result for the vertex curvature. Given that this is a common approximation in geometrical optics, we can consider Option A to be the intended correct model for the radius of curvature.

Part 2: Apparent Depth (Options C and D)

1. Setup: We use the formula for refraction at a single spherical surface: ${{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{{\mu _2} - {\mu _1}} \over R}}$

   • Light travels from water (medium 1) to air (medium 2). So, ${\mu _1} = \mu = {4 \over 3}$ and ${\mu _2} = 1$.
   • The object is at the bottom of the beaker. We need to find its distance ($u$) from the vertex of the curved surface. The initial water height (flat surface) is $H$. By conservation of volume, the initial volume $V = \pi r^2 H$ must equal the final volume under rotation. The volume of a paraboloid of revolution is ${1 \over 2}(\text{base area}) \times (\text{height})$. $V = \pi r^2 H_{min} + {1 \over 2}\pi r^2 h$, where $H_{min}$ is the height at the center. $\pi r^2 H = \pi r^2 (H_{min} + h/2)$, which gives $H_{min} = H - h/2$.
   • The object distance from the vertex is $u = - H_{min} = -(H - h/2)$. The negative sign is per standard sign convention (origin at vertex, light traveling in +ve direction).
   • For light rays coming from the bottom, the water surface is convex. Therefore, its radius of curvature is positive: $R = + {{{r^2}} \over {2h}} = + {g \over {{\omega ^2}}}$.

2. Calculation: ${1 \over v} - {{{4/3}} \over { - (H - h/2)}} = {{1 - 4/3} \over { + {r^2}/(2h)}}$ ${1 \over v} + {4 \over {3(H - h/2)}} = {{ - 1/3} \over {{r^2}/(2h)}} = - {{2h} \over {3{r^2}}}$ Using ${{{{\omega ^2}} \over g} = {{2h} \over {{r^2}}}}$: ${1 \over v} = - {{{\omega ^2}} \over {3g}} - {4 \over {3(H - h/2)}}$ The image position $v$ is negative, indicating a virtual image, as expected. The apparent depth $d_{app}$ is $|v|$. $d_{app} = |v| = {1 \over {{{{\omega ^2}} \over {3g}} + {4 \over {3(H - h/2)}}}} = {3 \over {{{{\omega ^2}} \over g} + {4 \over {H - h/2}}}}$ $d_{app} = {{3(H - h/2)} \over {{{{\omega ^2}} \over g}(H - h/2) + 4}}$

3. Approximation and Evaluation of Options C and D: The options are given as