To solve the problem, we need to understand how the pressure changes in the J-tube when additional mercury is added to one side. Let's break down the steps to find the volume of air trapped above Y after pouring 15 cm of mercury into the limb above X. ### Step-by-Step Explanation 1. **Understanding the Initial Conditions**: - Initially, there is 30 cm³ of air trapped above Y. - The atmospheric pressure is given as 75 cm Hg. This means that the pressure exerted by the air column above Y is equal to the atmospheric pressure. 2. **Using Boyle's Law**: - Boyle's Law states that for a given mass of gas at constant temperature, the product of pressure (P) and volume (V) is constant: \( P_1 V_1 = P_2 V_2 \). - Here, \( P_1 \) is the initial pressure, \( V_1 \) is the initial volume, \( P_2 \) is the final pressure, and \( V_2 \) is the final volume. 3. **Calculating Initial Pressure**: - The initial pressure above Y (where the air is trapped) is the atmospheric pressure, which is 75 cm Hg. 4. **Adding Mercury**: - When we pour 15 cm Hg of mercury into the limb above X, we are effectively increasing the pressure on the air trapped above Y. - The new pressure above Y becomes: \[ P_2 = P_{\text{atmospheric}} + P_{\text{mercury}} = 75 \text{ cm Hg} + 15 \text{ cm Hg} = 90 \text{ cm Hg} \] 5. **Applying Boyle's Law**: - Now we can apply Boyle's Law to find the new volume of air trapped above Y. - Let \( V_2 \) be the new volume of air above Y. We know: - \( P_1 = 75 \text{ cm Hg} \) - \( V_1 = 30 \text{ cm}^3 \) - \( P_2 = 90 \text{ cm Hg} \) - Using Boyle's Law: \[ P_1 V_1 = P_2 V_2 \] \[ 75 \text{ cm Hg} \times 30 \text{ cm}^3 = 90 \text{ cm Hg} \times V_2 \] - Rearranging to solve for \( V_2 \): \[ V_2 = \frac{75 \text{ cm Hg} \times 30 \text{ cm}^3}{90 \text{ cm Hg}} = \frac{2250 \text{ cm}^3 \text{ cm Hg}}{90 \text{ cm Hg}} = 25 \text{ cm}^3 \] ### Conclusion The volume of air trapped above Y after pouring 15 cm Hg of mercury into the limb above X is **25 cm³**. ### Explanation of Other Options - **Option A (15 cm³)**: This option is incorrect because it underestimates the volume of air remaining after the increase in pressure. The volume cannot decrease to this extent given the initial conditions. - **Option C (35 cm³)**: This option is incorrect as it suggests that the volume increased, which contradicts Boyle's Law since the pressure increased. - **Option D (45 cm³)**: This option is also incorrect because it implies a significant increase in volume, which is not possible under increased pressure conditions. ### Revision Summary - Boyle's Law relates pressure and volume of a gas at constant temperature: \( P_1 V_1 = P_2 V_2 \). - Adding mercury increases the pressure on the trapped air, reducing its volume. - The new volume can be calculated by rearranging Boyle's Law. - The correct answer is 25 cm³, as calculated from the initial and final pressures.