A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second.

Yes

A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, $f_{\mathrm{g}}=\frac{\mathrm{G} M m}{R^{2}}$

Where,

$M=$ Mass of the star $=2.5 \times 2 \times 10^{30}=5 \times 10^{30} \mathrm{~kg}$

$m=$ Mass of the body

$R=$ Radius of the star $=12 \mathrm{~km}=1.2 \times 10^{4} \mathrm{~m}$

$\therefore f_{\mathrm{g}}=\frac{6.67 \times 10^{-11} \times 5 \times 10^{30} \times m}{\left(1.2 \times 10^{4}\right)^{2}}=2.31 \times 10^{11} \mathrm{~m} \mathrm{~N}$

Centrifuaal force, $f_{c}=m r \omega^{2}$

$\omega=$ Angular speed $=2 \pi v$

$v=$ Angular frequency $=1.2$ rev $s^{-1}$

$f_{\mathrm{c}}=m R(2 \pi v)^{2}$

$f_{\mathrm{c}}=m R(2 \pi v)^{2}$

$=m \times\left(1.2 \times 10^{4}\right) \times 4 \times(3.14)^{2} \times(1.2)^{2}=1.7 \times 10^{5} \mathrm{mN}$

Since $f_{g}>f_{c}$, the body will remain stuck to the surface of the star.