JEE Main 12 Jan 2019 Morning Question 27
Question: 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+Bx^{2}, $ , is given by [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ Gm[ A( \frac{1}{a}-\frac{1}{a+L} )-BL ] $
B) $ Gm[ A( \frac{1}{a+L}-\frac{1}{a} )-BL ] $
C) $ Gm[ A( \frac{1}{a+L}-\frac{1}{a} )+BL ] $
D) $ Gm[ A( \frac{1}{a}-\frac{1}{a+L} )+BL ] $
Show Answer
Answer:
Correct Answer: D
Solution:
$ dF=\frac{Gm(\mu dx)}{x^{2}} $
$ F=Gm\int\limits _{x=a}^{x=(a+L)}{\frac{(A+Bx^{2})}{x^{2}}}dx $
$ F=Gm( A\int\limits _{x=a}^{x=a+L}{{x^{-2}}dx+}\int\limits _{x=a}^{x=a+L}{B}.dx ) $
$ F=Gm( A[ \frac{-1}{x} ]_a^{a+L}+B[x]_a^{a+L} ) $
$ F=Gm( -A[ \frac{1}{a+L}-\frac{1}{a} ]+B[a+L-a] ) $
$ F=Gm( A[ \frac{1}{a}-\frac{1}{a+L} ]+BL ) $
