JEE Mains · Physics · STD 11 - 7. gravitation
On the \(x-\)axis and a dsitance \(x\) from the origin, the gravitational field due to a mass distribution is given by \(\frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}}\) in the \(x-\)direction. The magnitude of gravitational potential on the \(x-\)axis at a distance \(x,\) taking its value to be zero at infinity, is
- A \(\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}\)
- B \(\frac{A}{\left(x^{2}+a^{2}\right)^{3 / 2}}\)
- C \(A \left( x ^{2}+ a ^{2}\right)^{3 / 2}\)
- D \(A\left(x^{2}+a^{2}\right)^{1 / 2}\)
Answer & Solution
Correct Answer
(A) \(\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}\)
Step-by-step Solution
Detailed explanation
Given \(E _{ G }=\frac{ Ax }{\left( x ^{2}+ a ^{2}\right)^{3 / 2}}, V _{\infty}=0\) \(\int_{V_{\infty}}^{V_{x}} d V=-\int_{\infty}^{x} \vec{E}_{G} \cdot \vec{d}_{x}\) \(V _{ x }- V _{\infty}=-\int_{\infty}^{ x } \frac{ Ax }{\left( x ^{2}+ a ^{2}\right)^{3 / 2}} d x\) put…
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