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JEE Main201912 Jan 2019Morning ShiftPhysicsGravitationActual

A straight rod of length L extends from x = a to x = L + a . The gravitational force it exerts on a point mass ' m ' at x = 0 , if the mass per unit length of the rod is A + B x 2 , is given by:

Options

  1. AG m A 1 a + L - 1 a + B L
  2. BG m A 1 a + L - 1 a - B L
  3. CG m A 1 a - 1 a + L - B L
  4. DG m A 1 a - 1 a + L + B L

Correct answer

D. G m A 1 a - 1 a + L + B L

Step-by-step solution

F = ∫ a a + L G m d M x 2 = Gm ∫ a a + L​ ( A + B x 2 ) d x x 2 = G m ∫ a a + L A x 2 d x + ∫ a a + L B d x = G m [ A [ - 1 x ] a a + L + B L ] = G m A 1 a - 1 a + L + B L

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