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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2019 · 12 January · Morning Shift · Physics · Gravitation

A straight rod of length L extends from x = a to x = L + a . The...

A straight rod of length LL extends from x=ax = a to x=L+a.x = L + a . The gravitational force it exerts on a point mass 'm' at x=0,x = 0, if the mass per unit length of the rod is A+Bx2,A + B{x}^{2}, is given by: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

Options

  1. AGm[A(1a+L−1a)+BL]Gm\left[A\left(\dfrac{1}{a + L} - \dfrac{1}{a}\right) + BL\right]
  2. BGm[A(1a+L−1a)−BL]Gm\left[A\left(\dfrac{1}{a + L} - \dfrac{1}{a}\right) - BL\right]
  3. CGm[A(1a−1a+L)−BL]Gm\left[A\left(\dfrac{1}{a} - \dfrac{1}{a + L}\right) - BL\right]
  4. DGm[A(1a−1a+L)+BL]Gm\left[A\left(\dfrac{1}{a} - \dfrac{1}{a + L}\right) + BL\right]

Correct answer

D. Gm[A(1a−1a+L)+BL]Gm\left[A\left(\dfrac{1}{a} - \dfrac{1}{a + L}\right) + BL\right]

Step-by-step solution

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

F=∫aa+LGmdMx2=Gm∫aa+L(A+Bx2)dxx2F = \mathop{∫}_{a}^{a + L}\dfrac{GmdM}{{x}^{2}} = Gm\mathop{∫}_{a}^{a + L}\dfrac{( A + B{x}^{2} ) dx}{{x}^{2}}

=Gm[∫aa+LAx2dx+∫aa+LBdx]= Gm\left[\mathop{∫}_{a}^{a + L}\dfrac{A}{{x}^{2}}dx + \mathop{∫}_{a}^{a + L}Bdx\right]

=Gm[A[−1x]aa+L+BL]= Gm [ A{[ \dfrac{- 1}{x} ]}_{a}^{a + L} + BL ]

= Gm\left[A\left(\dfrac{1}{a} - \dfrac{1}{a + L}\right) + BL\right]
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