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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Gravitation

BITSAT 2023 · Paper 2 · Physics · Gravitation

Two identical particles each of mass ' m ' go round a circle of...

Two identical particles each of mass ' mm ' go round a circle of radius aa under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Options

  1. AGm2a3\sqrt{\frac{G m}{2 a^3}}
  2. BGm8a3\sqrt{\frac{\mathrm{Gm}}{8 \mathrm{a}^3}}
  3. CGm4a3\sqrt{\frac{G m}{4 a^3}}
  4. DGma3\sqrt{\frac{G m}{a^3}}

Correct answer

C. Gm4a3\sqrt{\frac{G m}{4 a^3}}

Step-by-step solution

Radius of circle, r=a\mathrm{r}=\mathrm{a}
Centrifugal force, F=mω2rF=m \omega^2 r
⇒Gmm(2a)2=mω2a(∵F=GMM2d)\begin{aligned} \\ & \Rightarrow \frac{\mathrm{Gmm}}{(2 \mathrm{a})^2}=\mathrm{m} \omega^2 \mathrm{a} \\ \\ & \left(\because \mathrm{F}=\frac{\mathrm{G} \mathrm{M} \mathrm{M}_2}{\mathrm{ d}}\right) \\ \end{aligned}
Here M1=m;M2=mandd=2aM_1=m ; M_2=m and d=2 a
\Rightarrow angular speed, ω=Gm4a3\omega=\sqrt{\frac{G m}{4 a^3}}
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