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  • If the magnitude of gravitational potential due to uniform disc of radius r at its centre is V thenOption: 1 <span style="background-color:

If the magnitude of gravitational potential due to uniform disc of radius r at its centre is V then

Option: 1

V \ \alpha \ r
 


Option: 2

V \ \alpha \ r^2

 


Option: 3

V \ \alpha \ \frac{1}{ r^2}


Option: 4

V \ \alpha \ \frac{1}{ r}


Answers (1)

best_answer

As we learned

Gravitational Potential due to Uniform disc -

For Uniform disc

 

 

a \rightarrow Radius of disc

M-mass of disc

  • At the center of the disc 

              V=-\frac{2GM}{a}

  • At a point on its axis

                    V=-\frac{2GM}{a^2}(\sqrt{a^2+x^2}-x)

-

As |V| = \frac{2GM}{ r}

So V \ \alpha \ \frac{1}{ r}

Posted by

Divya Prakash Singh

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