A solid sphere of radius R gravitationally attracts a particle placed at 3R form 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:

Inside a uniform sphere of mass M and radius R, a cavity of radius R//3 , is made in the sphere as shown :

A solid sphere of uniform density and radius R applies a gravitational force attraction equal to F_(1) on a particle placed at P , distance 2R from the centre O of the sphere. A spherical cavity of radius R//2 is now made in the sphere as shows in figure. The particle with cavity now applies a gravitational force F_(2) on same particle placed at P . The radio F_(2)//F_(1) will be

The gravitational force acting on a particle, due to a solid sphere of uniform density and radius R , at a distance of 3R from the centre of the sphere is F_1. A spherical hole of radius (R//2) is now made in the sphere as shown in diagram. The sphere with hole now exerts a force F_2 on the same particle. ratio of F_1 to F_2 is

A solid of radius 'R' is uniformly charged with charge density rho in its volume. A spherical cavity of radius R/2 is made in the sphere as shown in the figure. Find the electric potential at the centre of the sphere.

A hemi-spherical cavity is created in solid sphere (of radius 2R ) as shown in the figure. Then

A solid sphere ofradius 'R' is uniformly charged with charge density rho in its volume. A spherical cavity of radius (R)/(2) is made in the sphere as shown in the figure. It is given that (rhoR^(2))/(in_(0)) = 48 V . Find the electric potential at the centre C of the sphere:

A solid sphere having uniform charge density rho and radius R is shown in figure. A spherical cavity ofradius (R)/(2) is made in it. What is the potential at point O?

A uniform solid sphere of radius R has a cavity of radius 1m cut from it if centre of mass of the system lies at the periphery of the cavity then