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Two identical thin rings each of radius \mathrm{R} are coaxially placed at a distance \mathrm{R}. If the rings have a uniform mass distribution and each has mass \mathrm{m_1 \: and \: m_2} respectively, then the work done in moving a mass \mathrm{m} from centre of one ring to that of the other is
 

Option: 1

Zero
 


Option: 2

\frac{\mathrm{Gm}\left(\mathrm{m}_1-\mathrm{m}_2\right)(\sqrt{2}-1)}{\sqrt{2} \mathrm{R}}
 


Option: 3

\frac{\mathrm{Gm} \sqrt{2}\left(\mathrm{~m}_1+\mathrm{m}_2\right)}{\mathrm{R}}
 


Option: 4

\frac{\mathrm{Gmm}_1(\sqrt{2}+1)}{\mathrm{m}_2 \mathrm{R}}


Answers (1)

best_answer

\mathrm{ V_{\mathrm{A}}=\left(\begin{array}{c} \text { Potential at } \\ \text { A due to } A \end{array}\right)+\left(\begin{array}{c} \text { Potntial at } \\ \text { A due to } B \end{array}\right) }

\mathrm{ \Rightarrow V_A=-\frac{\mathrm{Gm}_1}{\mathrm{R}}-\frac{\mathrm{Gm}_2}{\sqrt{2} \mathrm{R}} \text { and }}
Similarly,
\mathrm{ \quad V_B=\left(\begin{array}{c} \text { Potential at } \\ \text { B due to } A \end{array}\right)+\left(\begin{array}{c} \text { Potntial at } \\ B \text { due to } B \end{array}\right) }

\mathrm{ \text { Since } W_{A \rightarrow B}=m\left(V_B-V_A\right) }

\mathrm{ \Rightarrow W_{A \rightarrow B}=\frac{G m\left(m_1-m_2\right)(\sqrt{2}-1)}{\sqrt{2} R}}

Hence option 2 is correct.


 

Posted by

Kuldeep Maurya

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