Gravitation

JEE Main level, single correct, medium

141 of 223

Question 141 of 223

On the xx-axis and at a distance xx from the origin, the gravitational field due to a mass distribution is given by Ax(x2+a2)3/2\dfrac{Ax}{{\left({x}^{2}+{a}^{2}\right)}^{3/2}} in the xx-direction. The magnitude of the gravitational potential on the xx-axis at a distance xx, taking its value to be zero at infinity is:

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Answer: (A) A(x2+a2)1/2\dfrac{A}{{\left({x}^{2}+{a}^{2}\right)}^{1/2}}

Vx=−∫∞xAx(A2+x2)3/2(−dx){V}_{x}=-\int _{\infty }^{x}\dfrac{Ax}{{\left({A}^{2}+{x}^{2}\right)}^{3/2}}\left(-dx\right)
Vx=−AA2+x2{V}_{x}=-\dfrac{A}{\sqrt{{A}^{2}+{x}^{2}}}

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