In an experiment with sonometer when a mass of $180 \mathrm{~g}$ is attached to the string, it vibrates with fundamental frequency of $30 \mathrm{~Hz}$. When a mass $\mathrm{m}$ is attached, the string vibrates with fundamental frequency of $50 \mathrm{~Hz}$. The value of $\mathrm{m}$ is ___________ g.

We can use the fact that the ratio of frequencies is equal to the square root of the ratio of tensions: <br/><br/> $\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}$ <br/><br/> In the first case, the mass attached to the string is $180 \mathrm{~g}$ and the frequency is $30 \mathrm{~Hz}$, so we have: <br/><br/> $\frac{f_2}{30~\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$ <br/><br/> In the second case, the frequency is $50 \mathrm{~Hz}$, so we have: <br/><br/> $\frac{50~\mathrm{Hz}}{30~\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$ <br/><br/> Simplifying, we get: <br/><br/> $\frac{5}{3}=\sqrt{\frac{T_2}{T_1}}$ <br/><br/> Squaring both sides, we get: <br/><br/> $\frac{25}{9}=\frac{T_2}{T_1}$ <br/><br/> Since the tension in the string is proportional to the mass attached to it, we can write: <br/><br/> $\frac{m}{180~\mathrm{g}}=\frac{T_2}{T_1}=\frac{25}{9}$ <br/><br/> Solving for $m$, we get: <br/><br/> $m=\frac{25}{9}(180~\mathrm{g})=\boxed{500~\mathrm{g}}$ <br/><br/> Therefore, the mass attached to the string in the second case is $500 \mathrm{~g}$.