Step 1: Understand the resonance conditions for a closed-end tube. In a resonance tube (which is typically a closed-end tube due to the water column), resonance occurs when the length of the air column ($L$) is an odd multiple of a quarter wavelength. The formula for resonance lengths is: $$L_n = \frac{(2n-1)\lambda}{4}$$ where $n = 1, 2, 3, \dots$ represents the order of resonance. For the first resonance ($n=1$): $L_1 = \frac{(2(1)-1)\lambda}{4} = \frac{\lambda}{4}$ For the second resonance ($n=2$): $L_2 = \frac{(2(2)-1)\lambda}{4} = \frac{3\lambda}{4}$ For the third resonance ($n=3$): $L_3 = \frac{(2(3)-1)\lambda}{4} = \frac{5\lambda}{4}$ Step 2: Use the given first resonance position to find the wavelength ($\lambda$). The problem states that the first position of resonance is $16.50 \, \text{cm}$ from the open end. So, $L_1 = 16.50 \, \text{cm}$. Using the formula for the first resonance: $$L_1 = \frac{\lambda}{4}$$ $$16.50 \, \text{cm} = \frac{\lambda}{4}$$ Now, solve for $\lambda$: $$\lambda = 4 \times 16.50 \, \text{cm}$$ $$\lambda = 66.00 \, \text{cm}$$ Step 3: Calculate the distance to the next resonance position. The "next position where resonance occurs" refers to the second resonance ($L_2$). Using the formula for the second resonance: $$L_2 = \frac{3\lambda}{4}$$ Substitute the calculated value of $\lambda$: $$L_2 = \frac{3 \times 66.00 \, \text{cm}}{4}$$ $$L_2 = 3 \times \left(\frac{66.00}{4}\right) \, \text{cm}$$ $$L_2 = 3 \times 16.50 \, \text{cm}$$ $$L_2 = 49.50 \, \text{cm}$$ The distance from the open end to the next position where resonance occurs is $\boxed{\text{49.50 cm}}$.