NTA Abhyas JEE Main2020PhysicsGravitationPractice
From a solid sphere of mass M and radius R , a spherical portion of radius R 2 is removed, as shown in the figure. Taking gravitational potential V = 0 at r = ∞ , the potential at the centre of the cavity thus formed is: G =Universal g r a v i t a t i o n a l c o n s t a n t
Options
- A- 2 G M R
- B- G M 2 R
- C- G M R
- D- 2 G M 3 R
Correct answer
C. - G M R
Step-by-step solution
By superposition principle, ( v ₁= - GM 2 R ³ [3 R ²- ( R 2 )² ] ) (=- 11 GM 8 R ³ ) Also, ( v ₂=- 3 2 G ( M / 8) ( R / 2) = -3 GM 8 R ) The required potential is, ( v = v ₁- v ₂ ) ( array l =- 11 GM 8 R - (- 3 GM 8 R ) V =- GM R array )