c
Angular displacement in first one second \({\theta _1} = \frac{1}{2}\alpha \,{(1)^2} = \frac{\alpha }{2}\)  ......\((i)\)      [From \(\theta  = {\omega _1}t + \frac{1}{2}\alpha \,{t^2}\)]

Now again we will consider motion from the rest and angular displacement in total two seconds

\({\theta _1} + {\theta _2} = \frac{1}{2}\alpha \,{(2)^2} = 2\alpha \)                                   ......\((ii)\)

Solving \((i)\) and \((ii)\), we get \({\theta _1} = \frac{\alpha }{2}\) and \({\theta _2} = \frac{{3\alpha }}{2}\)    \(\therefore \)  \(\frac{{{\theta _2}}}{{{\theta _1}}} = 3\).