Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
0 votes

The gravitational field due to mass distribution is $ E= k/x^3$ in the x direction where k is a constant. Taking gravitational potential to be zero at infinity, its value at a distance x is

\[(a)\;\frac{k}{x} \quad (b)\;\frac{k}{2x} \quad (c)\;\frac{k}{x^2}\quad (d)\;\frac{k}{2x^2} \]
Can you answer this question?

1 Answer

0 votes
Gravitational Potential $=\int \limits_x^{\infty} E dx$
$\qquad=\int \large \frac{k}{x^3} $
$\qquad= -k \large\frac{x^{-2}}{2} \bigg|_x ^ {\infty}$
magnitude is $\large\frac{k}{2x^2}$
Hence d is the correct answer.


answered Aug 26, 2013 by meena.p
edited Feb 17, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App