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A solid sphere of uniform density and radius 4 units is located with centre at origin.Two spheres of equal radii 1 unit, with centres at A(-2,0,0) and B (2,0,0) are taken out of the solidleaving behind spherical cavities.Then: 1. the gravitational force due to this at origin is zero. 2. the gravitational force at B is zero 3. the gravitational potential is same at all points of the circle y2+ z2= 36 4. the gravitational potential at all points on the circle y2 + z2= 4

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Native English Speaker with 32 years experience in teaching English and communication skills.

The gravitational potential at all points on the circle y2 + z2= 4
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Physicist in outlook and thought

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Science allrounder

first option is correct as centre of mass does not changes on removal of the two like sphere from two symmetric position. you find the new centre of mass by using formula for centre of mass. that comes out to be centre of mass is again origin.
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Trainer

the gravitational potential at all points on the circle y2 + z2= 4
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Tutor

See, you must understand the similarity and difference between First Case (or initial case) and the Final Case. Also, note that Gravitational Force acts from Centre of Mass, which in this case of Solid Sphere will be at the origin (Why Origin? Density is uniform.) This means at origin, that is at...
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See, you must understand the similarity and difference between First Case (or initial case) and the Final Case. Also, note that Gravitational Force acts from Centre of Mass, which in this case of Solid Sphere will be at the origin (Why Origin? Density is uniform.) This means at origin, that is at Centre of Mass, there will be no Graviational Pull (since the distance from this point itself is ZERO). This condition has not changed after removing the two spheres also. read less
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Physics and Maths made easy

Answer: Choice numbered (1) ie ' the gravitational force due to this at origin is zero' is the right answer. To arrive at the answer, use (1) Principle of superposition of gravitational fields and hence gravitational forces too (note that here superposition means those of vectors since gravitational...
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Answer: Choice numbered (1) ie ' the gravitational force due to this at origin is zero' is the right answer. To arrive at the answer, use (1) Principle of superposition of gravitational fields and hence gravitational forces too (note that here superposition means those of vectors since gravitational field and gravitational force are vectors). (2) Gravitational field (and gravitational force for a particle of known, negligible mass) at the centre of a solid sphere is zero. (3) making use of statement (1), [Total gravitational field (and gravitational force for a particle of known, negligible mass) at the centre of the sphere with the two solid spheres of equal radii 1 unit with centres at (-2,00) and (2,00) cut out] = [ (Gravitational field (and gravitational force for a particle of known, negligible mass) at the origin (centre of the solid sphere of radius 4 units)] (minus) [vector sum of Gravitational field(and gravitational forces for a particle of known, negligible mass)s at the origin due to the solid spheres of equal radii (1 unit) at (2,0,0) and (-2,0,0)] = 0 - [ [vector sum of [gravitational field (or gravitational forces for a particle of known, negligible mass) due to the solid sphere of unit radius with its centre at (2,0,0)] and [that due to an identical solid sphere with its centre at (-2,0,0)]. (4) We also make use of the fact that the gravitational field due to a solid sphere (of uniform density) at a point outside it is numerically equal to -[G*(Mass of the sphere)/(distance to the point from the centre of the sphere)]*(unit vector pointing from the centre of the sphere to the point in question) and hence the gravitational forces for a particle of known, negligible mass placed at the point to be the product of the mentioned field (as a vector) with the mass (as scalar in Newtonian, non relativistic limits) of the particle. (5) Therefore it follows the sum (vectorial) of gravitational field(and hence the gravitational forces for a particle of known, negligible mass placed at the origin)s point due to the spheres (cut out) at (2,0,0) and (-2,0,0) turn out to be zero. (6) Therefore, due to statements from (1) to (5), the total gravitational field (and hence the total gravitational force for a particle of known, negligible mass placed at the origin) at the origin is zero. Note: This is answer to the query Edifycoaching Classes By Iitian posted a Question in Class XI-XII Tuition (PUC): A solid sphere of uniform density and radius 4 units is located with centre at origin.Two spheres of equal radii 1 unit, with centres at A(-2,0,0) and B (2,0,0) are taken out of the solidleaving behind spherical cavities.Then: 1. the gravitational force due to this at origin is zero. 2. the gravitational force at B is zero 3. the gravitational potential is same at all points of the circle y2+ z2= 36 4. the gravitational potential at all points on the circle y2 + z2= 4 read less
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Option 1
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Option 1 (Justified by the Principle of superposition)
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Practical Physics Tutor

OPtion 1. As even after removal of two small spheres the centre of gravity is still at the geometric origin of the sphere, hence gravitational force at origin is still zero. PLus in option 3 is also correct. gravitational potential for circle y2 + z2 =36 means circle has radius of 6 units.All the points...
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OPtion 1. As even after removal of two small spheres the centre of gravity is still at the geometric origin of the sphere, hence gravitational force at origin is still zero. PLus in option 3 is also correct. gravitational potential for circle y2 + z2 =36 means circle has radius of 6 units.All the points are symmetrically loacated hence potential is same. Similarly Option 4 is also correct on same arguement read less
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4th option i guess.
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