Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :Option: 1 <img alt="\sqrt{\frac{G}{2 R^{3}}}" src="/latex-image/?%5Csqrt

Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :
Option: 1 \sqrt{\frac{G}{2 R^{3}}}
Option: 2 \frac{1}{2} \sqrt{\frac{G}{R^{3}}}
Option: 3 \frac{1}{2 R} \sqrt{\frac{1}{G}}
Option: 4 \sqrt{\frac{2 G}{R^{3}}}

Answers (1)

best_answer

\begin{aligned} &F_{2}=F_{1}=\frac{G m_{1} m_{2}}{(2 R)^{2}} \end{aligned}

\begin{aligned} & F_{2}=F_{1}=m_{1} R \omega^{2}=m_{2} R \omega^{2}\\ \end{aligned}

\begin{aligned} &\frac{G}{4 R^{2}}=R \omega^{2}\\ \end{aligned}

\begin{aligned} &\omega^{2}=\frac{G}{4 R^{3}} \quad \omega=\left(\frac{G}{4 R^{3}}\right)^{1 / 2}\\ \end{aligned}

\begin{aligned} &\rightarrow \omega=\frac{1}{2} \sqrt{\frac{G}{R^{3}}} \end{aligned}

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 Registration Live: NTA session 1 application last date today; official website, syllabus
anam-khan

JEE Main 2025 session 1 registration ends today at jeemain.nta.nic.in; application correction from November 26
anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year

Latest Question