A straight rod of length L extends from x = a | vedclass

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(D) Consider a small element of the rod of length $dx$ at a distance $x$ from the origin.The mass of this element is $dm = (A + Bx^2)dx$.The gravitati

Vedclass · Accepted Answer

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

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(D) Consider a small element of the rod of length $dx$ at a distance $x$ from the origin.The mass of this element is $dm = (A + Bx^2)dx$.The gravitational force exerted by this element on the point mass $m$ at $x = 0$ is given by $dF = \frac{G(dm)m}{x^2}$.Substituting $dm$,we get $dF = \frac{Gm(A + Bx^2)dx}{x^2} = Gm\left( \frac{A}{x^2} + B \right)dx$.To find the total force,we integrate from $x = a$ to $x = a + L$:$F = \int_a^{a+L} Gm\left( \frac{A}{x^2} + B \right)dx = Gm \left[ A \int_a^{a+L} x^{-2} dx + B \int_a^{a+L} dx \right]$.$F = Gm \left[ A \left( -\frac{1}{x} \right)_a^{a+L} + B(x)_a^{a+L} \right]$.$F = Gm \left[ A \left( -\frac{1}{a+L} + \frac{1}{a} \right) + B(a+L-a) \right]$.$F = Gm \left[ A \left( \frac{1}{a} - \frac{1}{a+L} \right) + BL \right]$.

$A$ 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 $A + Bx^2$,is given by

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