JEE Main20204 Sep 2020Morning ShiftPhysicsGravitationActual
On the x -axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by A x x 2 + a 2 3 / 2 in the x -direction. The magnitude of the gravitational potential on the x -axis at a distance x , taking its value to be zero at infinity is:
Options
- AA x 2 + a 2 1 / 2
- BA x 2 + a 2 3 / 2
- CA x 2 + a 2 1 / 2
- DA x 2 + a 2 3 / 2
Correct answer
A. A x 2 + a 2 1 / 2
Step-by-step solution
V x = − ∫ ∞ x A x A 2 + x 2 3 / 2 − d x V x = - A A 2 + x 2