TS EAMCET · Physics · Gravitation
If \(A\) is the areal velocity of a planet of mass \(M\), its angular momentum is
- A \(\frac{M}{A}\)
- B \(2MA\)
- C \(A^2 M\)
- D \(A M^2\)
Answer & Solution
Correct Answer
(B) \(2MA\)
Step-by-step Solution
Detailed explanation
Areal velocity \(A=\frac{1}{2} R^2 \omega\) \(\therefore \quad\) Multiplying by \(M\) on both sides \[ \begin{aligned} M A & =\frac{1}{2} M R^2 \omega \\ M A & =\frac{1}{2} I \omega \end{aligned} \] \(I=\) moment of inertia. \[ M A=\frac{1}{2} L \] where, \(L=\) angular momentum…
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