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A point \mathrm{P(R \sqrt{3}, 0,0)} lies on the axis of a ring of mass \mathrm{M} and radius \mathrm{R}. The ring is located in \mathrm{y-z} plane with its centre at origin \mathrm{O}. A small particle of mass \mathrm{m} starts from \mathrm{P} and reaches \mathrm{O} under gravitational attraction only. Its speed at \mathrm{O} will be.
 

Option: 1

\mathrm{\sqrt{\frac{G M}{R}}}

 


Option: 2

\mathrm{\sqrt{\frac{G m}{R}}}
 


Option: 3

\mathrm{\sqrt{\frac{\mathrm{GM}}{\sqrt{2} \mathrm{R}}}}
 


Option: 4

\mathrm{\sqrt{\frac{G m}{\sqrt{2} R}}}


Answers (1)

best_answer

\mathrm{ \left(\begin{array}{c} \text { Total } \\ \text { Mechanical } \\ \text { Energy } \end{array}\right)_{\mathrm{P}}=\left(\begin{array}{c} \text { Total } \\ \text { Mechanical } \\ \text { Energy } \end{array}\right)_0 }

\mathrm{ \Rightarrow \frac{1}{2} \mathrm{~m}(0)^2-\frac{\mathrm{GMm}}{\sqrt{(\sqrt{3} \mathrm{R})^2+\mathrm{R}^2}}=\frac{1}{2} \mathrm{mv}^2-\frac{\mathrm{GMm}}{\mathrm{R}} }

\mathrm{ \Rightarrow-\frac{\mathrm{GMm}}{2 \mathrm{R}}=\frac{1}{2} \mathrm{mv}^2-\frac{\mathrm{GMm}}{\mathrm{R}} }

\mathrm{\Rightarrow \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}

Hence option 1 is correct.



 

Posted by

Gaurav

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