Kindly answer this question :

Dear Student,
The linear velocity of a rotating object is given by $v=\frac{\omega r}{\sin \theta }$. ω is angular velocity and r is the distance of the object from centre. Now if d is the distance from the spotlight to wall then distance from the spot of the light from the centre at an angle 45^o is
$r=\frac{d}{\cos \theta }=\frac{3}{\cos \left(45°\right)}$
So the velocity of the spot is $v=\frac{\omega r}{\sin \theta }=\frac{0.1\times {\displaystyle \frac{3}{\cos \left(45°\right)}}}{\sin \left(45°\right)}=\frac{0.1\times 3}{\sin \left(45°\right)\times \cos \left(45°\right)}=\frac{0.3}{\frac{1}{\sqrt{2}}\times {\displaystyle \frac{1}{\sqrt{2}}}}=2\times 0.3=0.6\text{m/s}$
option(A) is correct. Regards