Answer:
0.2 milisecond
Explanation:
Let the mass of sun is M
mass of star = 2 M
initial radius of the star, R = radius of sun = 7 x 10^5 km
initial time period, T = 10 days
final radius, R' = 10 km
mass remains same.
As no external torque is acting on the star so its angular momentum remains constant.
So, Angular momentum before collapsing = Angular momentum after collapsing
I x ω = constant
[tex]\frac{2}{5}(2M)\times R^{2}\times \frac{2\pi }{T}=\frac{2}{5}(2M)\times R'^{2}\times \frac{2\pi }{T'}[/tex]
[tex]\frac{R^{2}}{T}=\frac{R'^{2}}{T'}[/tex]
[tex]T'=\frac{R'^{2}T}{R^{2}}[/tex]
[tex]T'=\frac{10\times 10\times10}{7\times10^{5}\times 7\times 10^{5}}[/tex]
T' = 2.04 x 10^-9 days
T' = 0.2 milisecond