Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

\[(a)\;\sqrt {\frac{GM}{2l}} \quad (b)\;\sqrt {\frac{GM}{l}} \quad (c)\;\sqrt {\frac{2GM}{l}} \quad (d)\;\sqrt {\frac{GM}{3l}} \]

0 votes

The particle to move in a circular path without chaging distance 'l' will happen if the resultant gravitational force on each particle provides the centripetal force necessary for the motion

$\large\frac{mv^2}{r}$$=\sqrt {F^2+F^2+2 F^2 \cos 60}$

$\large\frac{mv^2}{r}$$=\sqrt {3 }\; F$

Radius of the path is

$r= \large\frac{\sqrt 3}{2}$$l \times \large\frac{2}{3}$

$\quad= \large\frac{l}{3}$

$\therefore v= \sqrt {\large\frac {GM}{l}}$

Hence b is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...