b
Let extension in length of spring be \(x\).

Radius of circle \(r=0.2+ x\)

\(Kx = m \omega^2 r\)

\(7.5 x =\left(\frac{1}{10}\right)\left(5^2\right)(0.2+ x )\)

\(\Rightarrow \frac{15}{2} x =\frac{5}{2}\left( x +\frac{1}{5}\right)\)

\(\Rightarrow x =\frac{1}{10}\)

\(\therefore \text { Tension in spring }= kx =7.5 \times \frac{1}{10}=0.75\,N\)