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JEE Main 2019 (12 Jan, Shift 1) · Newton's Law of Gravitation

Newton's Law of Gravitation question from JEE Main 2019 (12 Jan, Shift 1)

Q16JEE Main 2019 · 12 JAN · SHIFT 1Difficult

Newton's Law of Gravitation

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 + Bx2, is given by:

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Answer: (b)

  1. Element dxdx at xx has mass (A+Bx2)dx(A + Bx^2)dx and pulls mm with dF=Gm(A+Bx2)dxx2dF = \frac{Gm(A + Bx^2)dx}{x^2}.
  2. F=Gm∫aa+L(Ax2+B)dxF = Gm\int_a^{a+L}\left(\frac{A}{x^2} + B\right)dx
  3. F=Gm[A(1a−1a+L)+BL]F = Gm\left[A\left(\frac{1}{a} - \frac{1}{a+L}\right) + BL\right]
  4. So the answer is (b).
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