c
(c) Using relation \(\theta = {\omega _0}t + \frac{1}{2}a{t^2}\)

\({\theta _1} = \frac{1}{2}(\alpha ){(2)^2} = 2\alpha \)…\((i) \)  (As \({\omega _0} = 0,t = 2\,\sec \))

Now using same equation for \(t = 4 \,sec\), \(\omega_0 = 0\)

\({\theta _1} + {\theta _2} = \frac{1}{2}\alpha {(4)^2} = 8\alpha \)…\((ii)\)

From \((i)\) and \((ii),\)

\({\theta _1} = 2\alpha \)and\({\theta _2} = 6\alpha \)    \(\frac{{{\theta _2}}}{{{\theta _1}}} = 3\)