Ask Questions, Get Answers

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

A uniform rod of length $'l'$ whose mass per unit length is $\lambda$. The gravitational force on a particle of mass m located at a distance d from one end of the rod as shown is

\[(a)\;\frac{Gm\lambda l}{d(d+l)} \quad (b)\;\frac{Gm\lambda }{l(d+l)} \quad (c)\;\frac{Gml}{\lambda(d+l)} \quad (d)\;\frac{2Gm\lambda l}{d(d+l)} \]

Can you answer this question?

1 Answer

0 votes
Consider a small segement dx of the rod at a distance x from mass m
$F=\int \limits_d^{d+h} \large\frac{G(\lambda dx)m}{x^2}$
Where $\lambda dx$ is the mass of the small segment of the rod.
$\therefore F=Gm\lambda \bigg[\large\frac{-1}{x}\bigg]_d^{d+l}$
$F=Gm\lambda \bigg[\large\frac{l}{d}-\frac{l}{d+l}\bigg]$
$\quad=\large\frac{Gm\lambda l}{d(d+l)}$
Hence a is the correct answer.


answered Aug 29, 2013 by meena.p
edited Feb 18, 2014 by meena.p

Related questions

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