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  • A particle of mass 1 kg is placed at a distance of 4m from center on an axis of a uniform ring of mass 5 kg and radius 3 m. Calculate the potential difference to increase the distance of the partic

A particle of mass 1 kg is placed at a distance of 4m from center on an axis of a uniform ring of mass 5 kg and radius 3 m. Calculate the potential difference to increase the distance of the particle from 4m to 3\sqrt{3} m in terms of G will be.

Option: 1

\frac{G}{6}


Option: 2

\frac{5G}{6}


Option: 3

\frac{3G}{5}


Option: 4

\frac{2G}{5}


Answers (1)

 

 

 

Potential at point A

V_{A}=-\frac{GM}{(R^{2}+a^{2})^{\frac{1}{2}}}=\frac{G(5)}{\sqrt{(16+9)}}

V_{A}=\frac{5G}{5}=G

Potential at point B

V_{B}=-\frac{Gm}{\sqrt{(R^{2}+x^{2})}}=\frac{G(5)}{\sqrt{9+27}}=\frac{5G}{\sqrt{36}}=\frac{5G}{6}

Potential difference 

V_{A}-V_{B}=(G-\frac{5G}{6})

=(\frac{6G-5G}{6})=\frac{G}{6}

 

Posted by

Kshitij

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