JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 23)
On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is
given by $${{Ax} \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$ 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:
$${A{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}$$
$${A{{\left( {{x^2} + {a^2}} \right)}^{1/2}}}$$
$${A \over {{{\left( {{x^2} + {a^2}} \right)}^{1/2}}}}$$
$${A \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$
Explanation
Given $${E_x} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$$
$$ \therefore $$ $${{ - dV} \over {dx}} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$$
$$ \Rightarrow $$ $$\int\limits_0^V {dV} = - \int\limits_\infty ^x {{{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}dx} $$
$$ \Rightarrow $$ $$V = {A \over {{{({x^2} + {a^2})}^{1/2}}}}$$
$$ \therefore $$ $${{ - dV} \over {dx}} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$$
$$ \Rightarrow $$ $$\int\limits_0^V {dV} = - \int\limits_\infty ^x {{{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}dx} $$
$$ \Rightarrow $$ $$V = {A \over {{{({x^2} + {a^2})}^{1/2}}}}$$
Comments (0)
