( ) Given, $\omega=314 \mathrm{rad} / \mathrm{s}$ at $t=2 \mathrm{~s}$
Let $\alpha=$ constant angular acceleration
Then using $\omega_1=\omega_0+\alpha t$
We have,
$$
\begin{aligned}
& & \omega_1 & =\alpha t \\
\therefore & 314 & =\alpha(2) & \left(\because \omega_0=0\right) \\
\Rightarrow & & \alpha & =\frac{314}{2} \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$

Now, angle covered in 20 seconds is given by,
$$
\begin{array}{rlrl}
& & \theta_1=\omega_0 t+\frac{1}{2} \alpha t^2 \\
\Rightarrow & & \theta_1=\frac{1}{2} \alpha t^2 \\
& \text { so, } \quad & \theta_1=\frac{1}{2} \times \frac{314}{2} \times(20)^2=314 \mathrm{rad} \\
& \text { At } \quad & t=20 \mathrm{~s}, \text { angular speed of wheel, } \\
\Rightarrow \quad & \omega_2=\omega_0+\alpha t \\
\Rightarrow \quad & \omega_2=0+\frac{314}{2} \times 20 \\
\Rightarrow \quad & \omega_2=314 \mathrm{rad} / \mathrm{s}
\end{array}
$$

As, acceleration does not operates after $t=20 \mathrm{~s}$.
So, wheel now rotates freely (with constant angular speed) upto $40 \mathrm{~s}$.
Angular displacement in radians covered in another $20 \mathrm{~s}$ of freewheeling is
$$
\begin{aligned}
\theta_2 & =\omega_1 \times t \\
& =314 \times 20=628 \mathrm{rad}
\end{aligned}
$$

Total angle covered by wheel, $\theta=(314+628) \mathrm{rad}$
$$
=942 \mathrm{rad}
$$
$\therefore$ Number of revolutions of wheel
$$
\begin{aligned}
n & =\frac{\theta}{2 \pi}=\frac{942}{2 \pi} \\
& =\frac{314 \times 3}{2 \times 314}=150
\end{aligned}
$$