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JEE Advanced Physics Ray Optics 2020 JEE Advanced 2020 (Paper 2)

JEE Advanced Physics Question (2020) — Solution

Question

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 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)

Options

  1. A. R = h 2 + r 2 2 h
  2. B. R = 3 r 2 2 h
  3. C. Apparent depth of the bottom of the beaker is close to 3 H 2 1 + ω 2 H 2 g - 1
  4. D. Apparent depth of the bottom of the beaker is close to 3 H 4 1 + ω 2 H 4 g - 1

Answer

D. Apparent depth of the bottom of the beaker is close to 3 H 4 1 + ω 2 H 4 g - 1

Step-by-step solution

h = 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 = r 2 + h 2 2 h ......option (A) For apparent depth, now As r ≪ h , R = r 2 2 h = g ω 2 Refraction formula, 1 v - 4 3 ( H - h ) = 1 - 4 / 3 R 1 v = 1 3 R + 4 3 ( H - h ) Since H ≫ h , 1 v = ω 2 3 g + 4 3 H = 4 3 H 1 + 3 H 4 × ω 2 3 g ⇒ | v | = 3 H 4 1 + ω 2 H 4 g - 1 .......option (D)

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