jee-main 2019 Q7

jee-main · India · session1_12jan_shift1 Not Maths
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:
(1) $G m \left[ A \left( \frac { 1 } { a + L } - \frac { 1 } { a } \right) + B L \right]$
(2) $G m \left[ A \left( \frac { 1 } { a + L } - \frac { 1 } { a } \right) - B L \right]$
(3) $G m \left[ A \left( \frac { 1 } { a } - \frac { 1 } { a + L } \right) - B L \right]$
(4) $G m \left[ A \left( \frac { 1 } { a } - \frac { 1 } { a + L } \right) + B L \right]$
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 $A + B x ^ { 2 }$, is given by:\\
(1) $G m \left[ A \left( \frac { 1 } { a + L } - \frac { 1 } { a } \right) + B L \right]$\\
(2) $G m \left[ A \left( \frac { 1 } { a + L } - \frac { 1 } { a } \right) - B L \right]$\\
(3) $G m \left[ A \left( \frac { 1 } { a } - \frac { 1 } { a + L } \right) - B L \right]$\\
(4) $G m \left[ A \left( \frac { 1 } { a } - \frac { 1 } { a + L } \right) + B L \right]$