JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2020
- Hard
- MCQ
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
Pick an option to check your answer.
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 \(x^{2}+a^{2}=z\)
\(2 x dx = dz\)
\(V_{x}-0=-\int_{\infty}^{x} \frac{A d z}{2(z)^{3 / 2}}=\left[\frac{A}{z^{1 / 2}}\right]_{\infty}^{x}=\left[\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}\right]_{\infty}^{x}\)
\(V_{x}=\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}-0=\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}\)
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