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+### Statement
+
+$11.1.28.$ [Insert the problem statement]
+
+### 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 $\omega_1=600\min^{-1}$
+
+
+
+Similarly,for the second case we could write:
+
+
+
+\begin{equation}
+\mathcal{E}_2+k\omega_2=0
+\end{equation}
+
+
+where $\omega_2=1200\min^{-1}$
+
+
+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}
+
+
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]