A) \[Gm\left[ A\left( \frac{1}{a}-\frac{1}{a+L} \right)-BL \right]\]
B) \[Gm\left[ A\left( \frac{1}{a+L}-\frac{1}{a} \right)-BL \right]\]
C) \[Gm\left[ A\left( \frac{1}{a+L}-\frac{1}{a} \right)+BL \right]\]
D) \[Gm\left[ A\left( \frac{1}{a}-\frac{1}{a+L} \right)+BL \right]\]
Correct Answer: D
Solution :
\[dF=\frac{Gm(\mu dx)}{{{x}^{2}}}\] \[F=Gm\int\limits_{x=a}^{x=(a+L)}{\frac{(A+B{{x}^{2}})}{{{x}^{2}}}}dx\] \[F=Gm\left( A\int\limits_{x=a}^{x=a+L}{{{x}^{-2}}dx+}\int\limits_{x=a}^{x=a+L}{B}.dx \right)\] \[F=Gm\left( A\left[ \frac{-1}{x} \right]_{a}^{a+L}+B[x]_{a}^{a+L} \right)\] \[F=Gm\left( -A\left[ \frac{1}{a+L}-\frac{1}{a} \right]+B[a+L-a] \right)\] \[F=Gm\left( A\left[ \frac{1}{a}-\frac{1}{a+L} \right]+BL \right)\]You need to login to perform this action.
You will be redirected in
3 sec