Physics · 2019
JEE Main · 12 January 2019, Shift 1 · Q8
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…
A straight rod of length L extends from $\displaystyle x=\mathrm{a}$ to $\displaystyle x=\mathrm{L}+\mathbf{a}$. The gravitational force it exerts on a point mass ' $\displaystyle m$ ' at $\displaystyle x=0$, if the mass per unit length of the rod is $\displaystyle \mathrm{A}+\mathrm{B} x^2$, is given by:
Official answer
(2)
$\displaystyle \operatorname{Gm}\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+B L\right]$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.