A solid sphere of radius $ R $ gravitationally attracts a particle placed at $ 3 \mathrm{R} $ from its centre with a force $ F_{1} $. Now a spherical cavity of radius $ \left(\frac{R}{2}\right) $ is made in the sphere (as shown in figure) and the force becomes $ F_{2} $. The value of $ F_{1}: F_{2} $ is:

Question

A solid sphere of radius $ R $ gravitationally attracts a particle placed at $ 3 \mathrm{R} $ from its centre with a force $ F_{1} $. Now a spherical cavity of radius $ \left(\frac{R}{2}\right) $ is made in the sphere (as shown in figure) and the force becomes $ F_{2} $. The value of $ F_{1}: F_{2} $ is:, $ 41: 50 $, $ 50: 41 $, $ 25: 36 $, $ 36: 25 $

MARKS App · Accepted Answer

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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