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  • Infinity number of the masses, each 1 kg , are placed along the X-axis at <img alt="x= \pm1m ,\pm2m ,\pm4m ,\pm8m ,\pm16m,-----\infty" src="http://entrancecorner.oncodecogs.com/gif.latex?x%3D%

Infinity number of the masses, each 1 kg , are placed along the X-axis at x= \pm1m ,\pm2m ,\pm4m ,\pm8m ,\pm16m,-----\infty . The magnitude of the resultant gravitational potential in terms of gravitational constant G at the origin is n \times G. Then 'n' will be :

 

Option: 1

4


Option: 2

2


Option: 3

8


Option: 4

0.5


Answers (1)

best_answer

 

 

 

As we know V=\frac{-Gm}{r}

\left | V \right |=\left | \frac{-Gm}{r} \right |=\frac{Gm}{r}

Total Potential (v) =\frac{2Gm}{r} [because particle along both sides]

V=2Gm\left [ 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+-----------\infty \right ] 

=2Gm\left [ \frac{1}{1-\frac{1}{2}} \right ]\left [ As \ S_\infty =\frac{a}{1-r} \right ]

V=\frac{2Gm}{ \frac{1}{2}}=4Gm

As, m =1 kg

So, V= 4G

Hence n = 4.

Posted by

Divya Prakash Singh

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