Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Questions  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Gravitation
Answer
Comment
Share
Q)

Three particles of equal mass m are situated at verticles of a equilateral triangle of side l. What should be velocity of each particle so that they move on a circular path without changing l

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

1 Answer

Comment
A)
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.

 

Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...