A beaker of radius r is filled with water (refractive index `(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 makes the water surface curved so that the difference in the height of water level at the centre and the circumference of the beaker is h `(h lt lt H, h lt lt r)` as shown in the figure on the right . Take this surface to be approximately spherical with a radius of curvalture R . Which of the following is/are correct ? (g is acceleration due to gravity)

Options

A. `R = (h^(2) + r^(2))/(2h)`
B. `R = (3r^(2))/(2h)`
C. Apparent of depth of the bottom of the beaker is close to `(3H)/(2) (1 + (omega^(2)H)/(2g))^(-1)`
D. Aperent depth of the bottom of the beaker is close to `(3H)/(4) (1+ (omega^(2)H)/(4g))^(-1)`

Answer

AD

Explanation

`r^(2) + (R - h)^(2) = R^(2)`
`r^(2) = R^(2) - (R - h)^(2) = (2R - h)h`
`r^(2) = 2Rh - h^(2)`
`therefore " " R = (r^(2) + h^(2))/(2h)` ……….option (A) .
For apparent depth, now .
As `r lt lt h`
`R = (r^(2))/(2h) = (g)/(omega^(2))`
` (1)/(V) - (4)/(3(H-h)) = (1-4//3)/(R)`
`|(1)/(v)| = (1)/(3R) + (4)/(3(H - h))`
`H gt gt h` br> `|(1)/(v)| = ( omega^(2))/(3g) + (4)/(3H) = (4)/(3H) [ 1+(3H)/(4) xx (omega^(2))/(3g)]`
`|V| = (3H)/(4) [1+(omega^(2)H)/(4g)]^(-1) ` .............option D