JEE PYQ: Gravitation Question 23
Question 23 - 2020 (04 Sep Shift 1)
On the $x$-axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{Ax}{(x^2+a^2)^{3/2}}$ 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:
(1) $\frac{A}{(x^2+a^2)^{1/2}}$
(2) $\frac{A}{(x^2+a^2)^{3/2}}$
(3) $A(x^2+a^2)^{1/2}$
(4) $A(x^2+a^2)^{3/2}$
Show Answer
Answer: (1)
Solution
$V = -\int_\infty^x E_g,dx = -\int_\infty^x \frac{Ax}{(x^2+a^2)^{3/2}}dx = \frac{A}{(x^2+a^2)^{1/2}}$.
BETA
