The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

To find the new period of rotation of the Sun after it expands, we can apply the principle of conservation of angular momentum. Since there is no external influence (external torque is zero), the angular momentum of the Sun must remain conserved.

The angular momentum $L$ of a rotating body is given by:
$L=I\omega$
where $I$ is the moment of inertia and $\omega$ is the angular velocity of the body.

The Sun is assumed to be a sphere of uniform density. The moment of inertia of a uniform solid sphere of mass $M$ and radius $R$ is:
$I=\frac{2}{5}M{R}^2$

The angular velocity $\omega$ is related to the time period of rotation $T$ by:
$\omega =\frac{2\pi }{T}$

Substituting these into the equation for angular momentum, we get:
$L=\left(\frac{2}{5}M{R}^2\right)·\left(\frac{2\pi }{T}\right)$

Since mass $M$ remains constant during the expansion and angular momentum $L$ is conserved, we have:
$I_1{\omega }_1=I_2{\omega }_2$
which simplifies to:
$\frac{{R_1}^2}{T_1}=\frac{{R_2}^2}{T_2}$

We are given:
Initial period of rotation, $T_1=27\text{days}$
Final radius, $R_2=2R_1$

Let us solve for the new period $T_2$:
$T_2=T_1·{\left(\frac{R_2}{R_1}\right)}^2$
$T_2=27·{\left(\frac{2R_1}{R_1}\right)}^2$
$T_2=27·2^2$
$T_2=27·4$
$T_2=108\text{days}$

Thus, the period of revolution of the Sun if it expands to twice its present radius is 108 days.