Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

3) is your answer.

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Answer 2

Gravitational field intensity is given by, E = Ax/(A² + x²)³/² , then find out potential at x.

solution : using application,

$\int dV=-\int{E(x)}\,dx$

$\int\limits^{V_x}_{V_{\infty}}{dV}=-\int\limits^x_{\infty}{\frac{Ax}{(A^2+x^2)^{3/2}}}\,dx$

$V_x-V_{\infty}=-\int\limits^x_{\infty}{\frac{Ax}{(A^2+x^2)^{3/2}}}\,dx$

let A² + x² = p

differentiating both sides we get,

2x dx = dp

⇒x dx = dp/2

now $-\int\limits^x_{\infty}{\frac{Ax}{(A^2+x^2)^{3/2}}}\,dx$

$=-\int\limits^x_{\infty}{\frac{A}{2p^{3/2}}}\,dp$

$V_x-0=-\left[\frac{A}{p^{1/2}}\right]^x_{\infty}$

$V_x=-\left[\frac{A}{\sqrt{A^2+x^2}}\right]^x_{\infty}$

$V_x=-\frac{A}{\sqrt{A^2+x^2}}$

Therefore potential at x is $V_x=-\frac{A}{\sqrt{A^2+x^2}}$