Sat March 24, 2012 By:

A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of the coordinates. Two spheres of equal radii 1 unit with their centres at A(-2, 0, 0) and B(2, 0 ,0) respectively, are taken out of the solid leaving behind cavities as shown in the figure (a) The gravitational force due to this object at the origin is zero (b) The gravitational force at point B(2,0,0) is zero (c) The gravitational potential is same at all points of the circle y² + z² = 36 (d) The gravitational potential is same at all points of the circle y² + z² = 4 (One or more options may be correct)

Expert Reply
Mon March 26, 2012
Let us assume that given sphere is made of three spheres, sphere A of radius 1 unit, Sphere B of radius 1 unit, and the remaining sphere C.
a) YZ is the axis of symmetry. It follows that from the symmetry of arguments that the force at center O of sphere is zero. 
Also, FA + FB + FC = 0
But, FB = -FA
Hence, FC =0
so, a is true.
Force due to whole of sphere is at a distance r(<R) from the center of sphere =
(GMm/R^3 )*r
So, force due to whole of sphere at B = GMm/32
B is at the center of sphere B
Hence, FB= 0
Again mass of sphere A is M/64
This is also the mass of sphere B
Now, FA = GMm/64 X 1/4^2
FC = GMm/32 - GMm/(64X16)
FC = 31/16 X GMm/64
Which is nopt equl to 0
So b is not true.
The circle y² + z² = 6^2
Has a radius 6 units. All points on this circle is at a distance of 6 units from the center of sphere. So, gravitational potential is same at all points of the circle.
So, c is true.
Same argument as in option C
So, d is true.

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