A beaker of radius r is filled with waterrefractive index 43 up to...

A beaker of radius $r$ is filled with water $(refractive index \frac{4}{3})$ 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 $ω$ . 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 $h (h ≪ H, h ≪ r)$ , 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)

R = \frac{h^{2} + r^{2}}{2 h}

R = \frac{3 r^{2}}{2 h}

Apparent depth of the bottom of the beaker is close to

\frac{3 H}{2} {(1 + \frac{ω^{2} H}{2 g})}^{- 1}

Apparent depth of the bottom of the beaker is close to

\frac{3 H}{4} {(1 + \frac{ω^{2} H}{4 g})}^{- 1}

Solution:

$h = \frac{ω^{2} r^{2}}{2 g}$

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

$r^{2} = R^{2} - (R - h)^{2} = (2 R - h) h$

$r^{2} = 2 R h - h^{2}$

$∴ R = \frac{r^{2} + h^{2}}{2 h}$ ......option (A)

For apparent depth, now

$As r ≪ h$ ,

$R = \frac{r^{2}}{2 h} = \frac{g}{ω^{2}}$

Refraction formula,

$\frac{1}{v} - \frac{4}{3 (H - h)} = \frac{1 - 4 / 3}{R}$

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

Since $H ≫ h$ ,

$|\frac{1}{v}| = \frac{ω^{2}}{3 g} + \frac{4}{3 H} = \frac{4}{3 H} [1 + \frac{3 H}{4} \times \frac{ω^{2}}{3 g}]$

$\Rightarrow | v | = \frac{3 H}{4} {[1 + \frac{ω^{2} H}{4 g}]}^{- 1}$ .......option (D)

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A beaker of radius r is filled with waterrefractive index 43 up to...

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