question_answer

Two identical thin rings each of radius R and placed coaxially at a distance R have a uniform mass distribution and mass \[{{m}_{1}}\] and \[{{m}_{2}}\] respectively. Work done in moving a particle of mass m from centre of one ring to the centre of other ring will be:

A)  \[\frac{Gm({{m}_{1}}-{{m}_{2}})(\sqrt{2}-1)}{R\sqrt{2}}\]      

B)  \[\frac{Gm({{m}_{1}}-{{m}_{2}})}{R\sqrt{2}}\]

C)  \[\frac{Gm({{m}_{2}}-{{m}_{1}})}{R\sqrt{2}}\]          

D)  \[\frac{Gm({{m}_{2}}-{{m}_{1}})}{R\sqrt{2}}\]

Correct Answer: A

Solution :

\[{{V}_{A}}=-\,\frac{G{{m}_{1}}}{R}-\,\frac{G{{m}_{2}}}{R\sqrt{2}}\] \[{{V}_{B}}=-\,\frac{G{{m}_{2}}}{R}-\frac{G{{m}_{1}}}{R\sqrt{2}}\] \[{{W}_{A-B}}=m({{V}_{B}}-{{V}_{A}})\] \[=\frac{gM({{m}_{1}}-{{m}_{2}})\,(\sqrt{2}-1)}{R\sqrt{2}}\]