JEE Main 2019 — Physics solution | Fundamenthol

JEE Main2019Physics

Q.

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

A
$G m [A (\frac{1}{a + L} - \frac{1}{a}) + B L]$
B
$G m [A (\frac{1}{a + L} - \frac{1}{a}) - B L]$
C
$G m [A (\frac{1}{a} - \frac{1}{a + L}) + B L]$
D
$G m [A (\frac{1}{a} - \frac{1}{a + L}) - B L]$

Solution

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →