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Gravitation question

2019 · 12 Jan · Shift 1 · Q51
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  5. /2019 · 12 Jan · Shift 1 · Q51

Gravitation question

2019 · 12 Jan · Shift 1 · Q51

JEE MainPhysicsGravitationMCQ+4 / −1
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 :
  1. A
    Gm[A(1a−1a+L)−BL]Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) - BL} \right]Gm[A(a1​−a+L1​)−BL]
  2. B
    Gm[A(1a−1a+L)+BL]Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]Gm[A(a1​−a+L1​)+BL]
  3. C
    Gm[A(1a+L−1a)+BL]Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) + BL} \right]Gm[A(a+L1​−a1​)+BL]
  4. D
    Gm[A(1a+L−1a)−BL]Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) - BL} \right]Gm[A(a+L1​−a1​)−BL]
View written solutionFree

Correct answer: B

  1. Set up the elemental force

A small element of the rod at position xxx of length dxdxdx has mass dm=λ(x) dx=(A+Bx2)dx.dm = \lambda(x)\,dx = (A + Bx^2)dx.dm=λ(x)dx=(A+Bx2)dx.

The rod lies from x=ax=ax=a to x=a+Lx=a+Lx=a+L, and the point mass mmm is at x=0x=0x=0.

The distance of the element from the mass mmm is xxx, so the magnitude of gravitational force due to dmdmdm is dF=Gm dmx2=Gm(A+Bx2)x2dx.dF = \frac{Gm\,dm}{x^2} = \frac{Gm(A+Bx^2)}{x^2}dx.dF=x2Gmdm​=x2Gm(A+Bx2)​dx.

So, dF=Gm(Ax2+B)dx.dF = Gm\left(\frac{A}{x^2}+B\right)dx.dF=Gm(x2A​+B)dx.

Since the rod is to the right of the mass, the force is towards the rod; here options are clearly giving the magnitude/sign in that direction.

  1. Integrate over the whole rod

F=Gm∫aa+L(Ax2+B)dx.F = Gm\int_a^{a+L}\left(\frac{A}{x^2}+B\right)dx.F=Gm∫aa+L​(x2A​+B)dx.

Split the integral: F=Gm[A∫aa+Lx−2dx+B∫aa+Ldx].F = Gm\left[ A\int_a^{a+L}x^{-2}dx + B\int_a^{a+L}dx\right].F=Gm[A∫aa+L​x−2dx+B∫aa+L​dx].

  1. Evaluate each integral

For the first integral, ∫x−2dx=−1x.\int x^{-2}dx = -\frac{1}{x}.∫x−2dx=−x1​.

Hence,

= A\left(-\frac{1}{a+L}+\frac{1}{a}\right) = A\left(\frac{1}{a}-\frac{1}{a+L}\right).$$ For the second integral, $$B\int_a^{a+L}dx = B[(a+L)-a]=BL.$$ 4. **Combine the results** Therefore, $$F = Gm\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+BL\right].$$ 5. **Match with the options** This matches **Option B**: $$Gm\left[ A\left(\frac{1}{a}-\frac{1}{a+L}\right)+BL\right].$$
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