AP EAMCET20227 Jul 2022Evening ShiftPhysicsGravitationActual
As shown in the figure, two spherical cavities are made in the uniform solid sphere of radius R . The boundaries of the cavities touch at the centre of the sphere. The centers of the cavities and the sphere lie on the X -axis. The mass of the solid sphere before the cavities were created was M . The gravitational force on a point mass m at a distance d away from the centre of the solid sphere is
Options
- AG M m d^2 [1- 1 8 1 (1+ R 2 d )^2 - 1 8 1 (1- R 2 d )^2 ]
- BG M m d^2 [1- 1 8 1 (1+ R d )^2 - 1 8 1 (1- R d )^2 ]
- CG M m d^2 [1- 1 8 1 (1+ d R )^2 - 1 8 1 (1- d R )^2 ]
- DG M m d^2 [1- 1 8 1 (1+ d R )^2 + 1 8 1 (1- d R )^2 ]
Correct answer
A. G M m d^2 [1- 1 8 1 (1+ R 2 d )^2 - 1 8 1 (1- R 2 d )^2 ]
Step-by-step solution
Given situation is as shown Radius of sphere given =R Mass of sphere =M Density of sphere, d= M 4 3 R^3 Radius of each of cavity =R / 2 Mass of each of portion removed to create a cavity = density volume = M 4 3 R^3 4 3 ( R 2 )^3= M 8 Now, force of gravity on m , F = force due to complete sphere of mass M - force of mass of cavity of sphere centre at A - force of mass of cavity of sphere centre at B . aligned & = G M m d^2 = G M^ m (d+ R 2 )^2 - G M^ m (d- R 2 )^2 & F= G M m d^2 - G M m 8 (d+ R 2 )^2 - G M m 8 (d-