One project after deviation from its path, starts moving round the earth in a circular path at radius equal to nine times the radius at earth R, its time period will be

The correct option is 27 x 2 π √(R/g).

To find the time period of the object moving in a circular orbit around the Earth, we can analyze the orbital mechanics from first principles.

For a circular orbit of radius $r$ around the Earth, the gravitational force acting on the object of mass $m$ provides the necessary centripetal force:
$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$
where $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $v$ is the orbital velocity of the object.

Solving for the orbital velocity $v$ gives:
$v=\sqrt{\frac{GM}{r}}$

The orbital period $T$ is the time taken to complete one full revolution of circumference $2\pi r$:
$T=\frac{2\pi r}{v}=2\pi \sqrt{\frac{r^3}{GM}}$

We can relate the term $GM$ to the acceleration due to gravity $g$ at the Earth's surface (radius $R$):
$g=\frac{GM}{R^2}\Rightarrow GM=g{R}^2$

Substituting $GM=g{R}^2$ back into the time period equation gives:
$T=2\pi \sqrt{\frac{r^3}{g{R}^2}}$

The problem states that the radius of the circular path is equal to nine times the radius of the Earth, so we substitute $r=9R$:
$T=2\pi \sqrt{\frac{{\left(9R\right)}^3}{g{R}^2}}$

Simplifying the mathematical expression:
$T=2\pi \sqrt{\frac{729R^3}{g{R}^2}}$
$T=2\pi \sqrt{729\frac{R}{g}}$

Since $\sqrt{729}=27$, we obtain:
$T=27\times 2\pi \sqrt{\frac{R}{g}}$