Solution

It is known that the induced EMF of an electric motor is directly proportional to the magnetic flux and its frequency. Since only the motor's frequency varies, we could write the following:

\begin{equation} \mathcal{E}_i=k\omega \end{equation}

where $\mathcal{E}_i$ is the induced EMF, $\omega$ is angular frequency of the motor and $k$ is a constant.

For the first case we have:

\begin{equation} \mathcal{E}_1+k\omega_1=IR \end{equation}

where $\mathcal{E}_1=24V$, $I=0.2$A,$R=20$$\Omega$ and $\omega_1=600$ rpm

Similarly,for the second case we could write:

\begin{equation} \mathcal{E}_2+k\omega_2=0 \end{equation}

where $\omega_2=1200$ rpm

By substituting $k$ from eq(2) to eq(3) we get the answer for $\mathcal{E}_2$:

\begin{equation} \mathcal{E}_2=(\mathcal{E}_1-IR)\frac{\omega_2}{\omega_1}=40V \end{equation}