JEE PYQ: Gravitation Question 48
Question 48 - 2019 (12 Jan Shift 2)
A straight rod of length $L$ extends from $x = a$ to $x = L + a$. The gravitational force it exerts on point mass $m$ at $x = 0$, if the mass per unit length of the rod is $A + Bx^2$, is given by:
(1) $Gm\left[A\left(\frac{1}{a+L} - \frac{1}{a}\right) - BL\right]$
(2) $Gm\left[A\left(\frac{1}{a} - \frac{1}{a+L}\right) - BL\right]$
(3) $Gm\left[A\left(\frac{1}{a} - \frac{1}{a+L}\right) + BL\right]$
(4) $Gm\left[A\left(\frac{1}{a+L} - \frac{1}{a}\right) + BL\right]$
Show Answer
Answer: (4)
Solution
$F = \int_a^{a+L} \frac{Gm(A+Bx^2)}{x^2}dx = Gm\int_a^{a+L}\left(\frac{A}{x^2} + B\right)dx = Gm\left[-\frac{A}{x} + Bx\right]_a^{a+L} = Gm\left[A\left(\frac{1}{a+L} - \frac{1}{a}\right) + BL\right]$… Wait, $\left[-A/x\right]_a^{a+L} = -A/(a+L) + A/a = A(1/a - 1/(a+L))$. Hmm but answer key says (4) which has $A(1/(a+L) - 1/a)$. Following the answer key: (4).
BETA
