A straight rod of length L extends from x = a to x = L + a . The gravitational force it exerts on a point…

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

Solution

$F=\underset{a}{\overset{a+\mathrm{L}}{\int }}\frac{GmdM}{x^2}=\mathrm{Gm}\underset{a}{\overset{a+\mathrm{L}}{\int }}\frac{(A+B{x}^2)dx}{x^2}$

$=Gm\left[\underset{a}{\overset{a+\mathrm{L}}{\int }}\frac{A}{x^2}dx+\underset{a}{\overset{a+\mathrm{L}}{\int }}Bdx\right]$

$=Gm[A{\left[\frac{-1}{x}\right]}_a^{a+\mathrm{L}}+BL]$

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

Asked in: JEE Main 2019 (12 Jan Shift 1)

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