A uniform ring of mass m is lying at a distance 1.73 a from the centre of a sphere of mass M just over the sphere where a is the small radius of the ring as well as that of the sphere. Then gravitational force exerted is

The correct option is 1.73 GMm/8a².

To find the gravitational force exerted between the sphere and the ring, we can use the formula for the gravitational force exerted by a point mass on a uniform ring along its axis. Since the sphere is spherically symmetric, its entire mass $M$ can be assumed to be concentrated at its center.

Let:
- Mass of the sphere concentrated at its center = $M$
- Mass of the ring = $m$
- Radius of the ring, $R=a$
- Distance of the ring's plane from the center of the sphere, $d=1.73a\approx \sqrt{3}a$

The gravitational force exerted by a point mass $M$ on a uniform ring of mass $m$ and radius $R$ placed at a distance $d$ along its axis is given by the formula:

$F=\frac{GMmd}{{\left(R^2+d^2\right)}^{3/2}}$

Substituting the values $R=a$ and $d=1.73a$ into the denominator term:

$R^2+d^2=a^2+{\left(1.73a\right)}^2$

Since $1.73\approx \sqrt{3}$, we have:

$R^2+d^2\approx a^2+3a^2=4a^2$

Now, raising this term to the power of $3/2$ gives:

${\left(R^2+d^2\right)}^{3/2}={\left(4a^2\right)}^{3/2}={\left({\left(2a\right)}^2\right)}^{3/2}={\left(2a\right)}^3$

Now, substitute this back into the force equation:

$F=\frac{GMm\left(1.73a\right)}{8a^3}$

Simplifying the expression by canceling $a$ in the numerator and denominator, we get:

$F=\frac{1.73GMm}{8a^2}$