Gravitation Question 122
Question:The gravitational field due to a mass distribution is $ [E=K/x^{3}] $ in the x-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is [MP PET 1994]
Options:
A) $ [K/x] $
B) $ [K/2x] $
C) $ [K/x^{2}] $
D) $ [K/2x^{2}] $
Show Answer
Answer:
Correct Answer: D
Solution:
Gravitational potential $ [=\int _{{}}^{{}}{I\ dx=\int _{x}^{\infty }{{}}\frac{K}{x^{3}}dx}] $ $ [=K,( \frac{x^{-3+1}}{-3+1} ) _{x}^{\infty }=| \frac{-K}{2x^{2}}| _{x}^{\infty }=\frac{K}{2x^{2}}] $
BETA
