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JEE Main202125 Feb 2021Morning ShiftPhysicsGravitationActual

A solid sphere of radius R gravitationally attracts a particle placed at 3 R from its centre with a force F 1 . Now a spherical cavity of radius R 2 is made in the sphere (as shown in figure) and the force becomes F 2 . The value of F 1 : F 2 is:

Options

  1. A41 : 50
  2. B50 : 41
  3. C25 : 36
  4. D36 : 25

Correct answer

B. 50 : 41

Step-by-step solution

Let the initial mass of the sphere is m ' . Hence, mass of a removed portion will be m ' / 8 , F 1 = m . E . = m . G m ' 9 R 2 F 2 = m G · m ' 3 R 2 - G · m ' / 8 5 R / 2 2 = G m ' 9 R 2 - G m ' × 4 8 × 25 = 1 9 - 1 50 G m ' R 2 F 2 = 41 50 × 9 · G m ' R 2 ⇒ F 1 F 2 = 1 9 × 50 × 9 41 = 50 41

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