Question 26 of 223

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:

A$Gm\left[A\left(\dfrac{1}{a+L}-\dfrac{1}{a}\right)+BL\right]$B$Gm\left[A\left(\dfrac{1}{a+L}-\dfrac{1}{a}\right)-BL\right]$C$Gm\left[A\left(\dfrac{1}{a}-\dfrac{1}{a+L}\right)-BL\right]$D$Gm\left[A\left(\dfrac{1}{a}-\dfrac{1}{a+L}\right)+BL\right]$

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Answer: (D) $Gm\left[A\left(\dfrac{1}{a}-\dfrac{1}{a+L}\right)+BL\right]$

$F=\int _{a}^{a+L}\dfrac{Gm dM}{{x}^{2}}=Gm\int _{a}^{a+L​}\dfrac{(A+B{x}^{2})dx}{{x}^{2}}$

$=Gm\left[\int _{a}^{a+L}\dfrac{A}{{x}^{2}}dx+\int _{a}^{a+L}B dx\right]$

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

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