A ring has a mass M and radius R. The distanc | vedclass

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(D) The gravitational field $E$ at a distance $x$ from the centre of a ring of mass $M$ and radius $R$ along its axis is given by:$E = \frac{GMx}{(R^2

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$\frac {R}{\sqrt 2}$

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(D) The gravitational field $E$ at a distance $x$ from the centre of a ring of mass $M$ and radius $R$ along its axis is given by:$E = \frac{GMx}{(R^2 + x^2)^{3/2}}$To find the point where the gravitational field is strongest,we differentiate $E$ with respect to $x$ and set it to zero:$\frac{dE}{dx} = 0$Using the quotient rule:$\frac{d}{dx} \left[ \frac{GMx}{(R^2 + x^2)^{3/2}} \right] = GM \left[ \frac{(R^2 + x^2)^{3/2}(1) - x \cdot \frac{3}{2}(R^2 + x^2)^{1/2}(2x)}{(R^2 + x^2)^3} \right] = 0$This simplifies to:$(R^2 + x^2)^{3/2} - 3x^2(R^2 + x^2)^{1/2} = 0$$(R^2 + x^2)^{1/2} [R^2 + x^2 - 3x^2] = 0$$R^2 - 2x^2 = 0$$2x^2 = R^2$$x = \frac{R}{\sqrt{2}}$Thus,the gravitational field is strongest at a distance of $\frac{R}{\sqrt{2}}$ from the centre.

$A$ ring has a mass $M$ and radius $R$. The distance of the point on its geometric axis from its centre at which the gravitational field is strongest is

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