Each situation in Column-I gives graph of a particle moving in circular path. The variables $\omega ,\theta$ and t represent angular speed (at any time t), angular displacement (in time t) and time respectively. Column-II gives certain resulting interpretation.

Column 1	Column 2
(A)	P) Angular acceleration of particle is uniform
(B)	Q) Angular acceleration of particle is non-uniform
(C)	R) Angular acceleration of particle is directly  proportional to $t$
(D)	S) Angular acceleration of particle is directly proportional to $\theta$
	T) Angular acceleration of particle is directly proportional to slope of the curve.

From graph (A) $\Rightarrow \omega =\mathrm{k}\theta$

where k is positive constant

Angular acceleration$=\omega \frac{\mathrm{d}\omega }{\mathrm{d}\theta }=\mathrm{k}\theta \times \mathrm{k}={\mathrm{k}}^2\theta$

Angular acceleration is non-uniform and directly proportional to $\theta$      

$\therefore \mathrm{A}-\mathrm{q},\mathrm{s}$

From graph  $\text{ (B) }\Rightarrow {\omega }^2=\mathrm{k}\theta$

Differentiating both sides with respect to $\theta$.

$2\omega \frac{\mathrm{d}\omega }{\mathrm{d}\theta }=\mathrm{k}\text{ (or) }\omega \frac{\mathrm{d}\omega }{\mathrm{d}\theta }=\frac{\mathrm{k}}{2}$

$k$ is slope of curve hence angular acceleration is uniform. $\therefore \mathrm{B}-\mathrm{pt}$

From graph $\text{ (C) }\Rightarrow \omega =\mathrm{kt}$

Angular acceleration  $=\frac{\mathrm{d}\omega }{\mathrm{dt}}=\mathrm{k}$

$k$ is slope of curve hence angular acceleration is uniform. $\therefore \mathrm{C}-\mathrm{pt}$

From graph  $\text{ (D) }\Rightarrow \omega ={\mathrm{kt}}^2$

Angular acceleration  $=\frac{\mathrm{d}\omega }{\mathrm{dt}}=2\mathrm{kt}$

$k$  is slope of curve hence angular acceleration is non-uniform and directly proportional to
t. Slope of the curve is constant (can be seen  

in given graph) but $\alpha =\frac{\mathrm{d}\omega }{\mathrm{dt}}=2\mathrm{kt}$  increasing with time.  $\therefore \mathrm{D}-\mathrm{qr}$