A beaker of radius is filled with water (refractive index ) up to a height as shown in the figure on the left. The beaker is kept on a horizontal table rotating with angular speed . [2020]

Question

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 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) [2020]

Smart Achievers · Accepted Answer

(1, 4)

From the figure, in $∆ O A B$ ,

$h (h ≪ H, h ≪ r)$

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

$\Rightarrow 2 h R = h^{2} + r^{2}$

$∴$ $R = \frac{h^{2} + r^{2}}{2 h}$

Hence, option (1) is correct.

Now, from the equation of the surface,

$y = y_{0} + \frac{ω^{2} r^{2}}{2 g}$

$h = \frac{ω^{2} r^{2}}{2 g}$

Now using the formula for refraction at a curved surface,

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{1}{v} - \frac{\frac{4}{3}}{- (H - h)} = \frac{1 - \frac{4}{3}}{R}$ $\Rightarrow \frac{1}{v} - \frac{1}{3 R} - \frac{4}{3 H}$ $[∵ h ≪ H]$

$\Rightarrow \frac{1}{v} = \frac{2 h}{3 r^{2}} - \frac{4}{3 H}$ $\Rightarrow \frac{1}{v} = - \frac{4}{3 H} [1 - \frac{ω^{2} H}{4 g}]$

$∴$ $v = \frac{3 H}{4} {[1 + \frac{ω^{2} H}{4 g}]}^{- 1}$ $(Apparent depth)$

Hence, option (4) is correct.

Solution

1 $R = \frac{h^{2} + r^{2}}{2 h}$

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

3 Apparent depth of the bottom of the beaker is close to $\frac{3 H}{2} {(1 + \frac{ω^{2} H}{2 g})}^{- 1}$

4 Apparent depth of the bottom of the beaker is close to $\frac{3 H}{4} {(1 + \frac{ω^{2} H}{4 g})}^{- 1}$

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Solution

1 R=h2+r22h

2 R=3r22h

3 Apparent depth of the bottom of the beaker is close to 3H2(1+ω2H2g)-1

4 Apparent depth of the bottom of the beaker is close to 3H4(1+ω2H4g)-1

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1 $R = \frac{h^{2} + r^{2}}{2 h}$

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

3 Apparent depth of the bottom of the beaker is close to $\frac{3 H}{2} {(1 + \frac{ω^{2} H}{2 g})}^{- 1}$

4 Apparent depth of the bottom of the beaker is close to $\frac{3 H}{4} {(1 + \frac{ω^{2} H}{4 g})}^{- 1}$