To solve this problem, we need to understand the resonance condition for an air column closed at one end.$\\$Such a column supports standing waves with nodes at the closed end and antinodes at the open end.$\\$The fundamental frequency (first harmonic) occurs when the length of the air column $L_1$ is $\frac{\lambda}{4}, \\$where $\lambda$ is the wavelength of the sound wave.$\\$Given:$\\ L_1 = 50$ cm$\\$For the first harmonic:$\\ L_1 = \frac{\lambda}{4} \\$The next resonance occurs at the third harmonic, where the length of the air column $L_2$ is $\frac{3\lambda}{4}. \\$Thus, $L_2 = \frac{3\lambda}{4} \\$Since $L_1 = \frac{\lambda}{4},$ we can express $\lambda$ as:$\\ \lambda = 4L_1 = 4 \times 50$ cm$= 200$ cm$\\$Substitute $\lambda$ into the expression for $L_2: \\ L_2 = \frac{3 \times 200}{4} \\ L_2 = 150$ cm$\\$Therefore, the next larger length of the column resonating with the same tuning fork is $150$ cm.$\\$This corresponds to Option 1.