A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2sec , it rotates through an angle θ1 . In the next 2sec , it rotates through an additional angle θ2 .The ratio of θ1/θ2 is

The correct option is 3 (which corresponds to the ratio $\frac{{\theta }_1}{{\theta }_2}=\frac{1}{3}$).

Let us solve the problem step-by-step:
We are given that a wheel starts from rest and is subjected to a uniform angular acceleration, which we can denote as $\alpha$.
Since the wheel starts from rest, its initial angular velocity is ${\omega }_0=0$.

Using the equation of rotational motion for the angular displacement:
$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

For the first time interval of $t=2\text{s}$, the angular displacement is ${\theta }_1$:
${\theta }_1=0+\frac{1}{2}\alpha \left(2^2\right)=2\alpha$

For the total time of the first $4\text{s}$ ($2\text{s}+2\text{s}$), the total angular displacement is:
${\theta }_{\text{total}}=0+\frac{1}{2}\alpha \left(4^2\right)=8\alpha$

The additional angle rotated in the next $2\text{s}$ is ${\theta }_2$, which is the difference between the total displacement after $4\text{s}$ and the displacement after the first $2\text{s}$:
${\theta }_2={\theta }_{\text{total}}-{\theta }_1$
${\theta }_2=8\alpha -2\alpha =6\alpha$

Now, we calculate the ratio of ${\theta }_1$ to ${\theta }_2$:
$\frac{{\theta }_1}{{\theta }_2}=\frac{2\alpha }{6\alpha }=\frac{1}{3}$

Thus, the ratio of the angles is 1:3, which corresponds to option 3.