Answer
a) $1.745\times10^{-3 }\;\rm N\cdot m$
b) $\rm 50\;rev$
Work Step by Step
$\bf{Given:}$
$ \bullet$ $\omega_i=0\;\rm rad/s$
$ \bullet\;\omega_f=2000\;\rm rpm=\left(\dfrac{2000\;rev}{1\;min}\right)\left(\dfrac{2\pi\;rad}{1\;rev}\right)\left(\dfrac{1\;min}{60\;s}\right)=\color{red}{\dfrac{200\pi}{3}}\;rad/s$
$ \bullet$ $I_{disk}=2.5\times 10^{-5}\;\rm kg\cdot m^2$
$ \bullet$ $ \Delta t=3.0\;\rm s$
$ \bullet$ $ R_{disk}=\dfrac{R_{disk}}{2}=\dfrac{0.12}{2}=0.06\;\rm m$
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a) We know that the net torque is given by
$$\sum \tau=I\alpha$$
Thus, in our case,
$$\sum \tau=I_{disk}\alpha_{disk}\tag 1$$
So we need to find the angular acceleration which we are told is constant.
We can use the rotational kinematic equation of
$$\omega_f=\omega_i+\alpha t$$
$$\omega_f=0+\alpha t$$
Hence,
$$\alpha_{disk}=\dfrac{\omega_f}{ t}\tag 2$$
Plugging into (1);
$$\sum \tau=I_{disk}\dfrac{\omega_f}{ t}$$
Plugging the known;
$$\sum \tau=(2.5\times 10^{-5})\dfrac{\frac{200\pi}{3}}{3}$$
$$\sum \tau=\color{red}{\bf 1.745\times10^{-3}}\;\rm N\cdot m$$
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b) We can use the rotational kinematic formula of $\theta$ to find the angular position in rad and then convert it to revolutions.
$$\theta_f=\theta_i+\omega_i t+\frac{1}{2}\alpha t^2$$
$$\theta_f=0+(0)t+\frac{1}{2}\alpha t^2$$
Plugging from (2);
$$\theta_f= \frac{1}{2}\dfrac{\omega_f}{ \color{red}{\bf\not} t} t^{ \color{red}{\bf\not}2}=\dfrac{\omega_f t}{2 }=\dfrac{\frac{200\pi }{3}\times 3}{2}=\bf100\pi \;\rm rad $$
Hence, the number of revolutions is given by
$$N= 100\pi \;\rm rad \left(\dfrac{1\;rev}{2\pi\;rad}\right)$$
$$N= \color{red}{\bf 50}\;\rm rev$$
