To analyze the pressures acting at a fixed point below a liquid surface, let's break down the concepts involved and understand why the correct answer is D (P1 = P2). ### Step-by-Step Explanation 1. **Understanding Pressure in Fluids**: - Pressure in a fluid at rest is defined as the force exerted per unit area. It is given by the formula: \[ P = \frac{F}{A} \] - In a liquid, pressure increases with depth due to the weight of the liquid above. The pressure at a depth \( h \) in a liquid of density \( \rho \) is given by: \[ P = P_0 + \rho g h \] where \( P_0 \) is the atmospheric pressure at the surface, \( g \) is the acceleration due to gravity, and \( h \) is the depth. 2. **Analyzing the Situation**: - At a fixed point below the liquid surface, we have two pressures to consider: - **P1**: The pressure acting downwards at that point, which is the pressure due to the weight of the liquid above it. - **P2**: The pressure acting upwards at that point, which is the pressure exerted by the liquid below it (or any other fluid or gas present). 3. **Equilibrium Condition**: - In a static fluid, the forces acting on any small volume of fluid must be in equilibrium. This means that the upward pressure (P2) must balance the downward pressure (P1) at that point. - If P1 were greater than P2 (option A), there would be a net downward force, causing the fluid to accelerate downwards, which contradicts the assumption of static equilibrium. - If P1 were less than P2 (option C), there would be a net upward force, causing the fluid to accelerate upwards, again contradicting static equilibrium. - If P1 were greater than or equal to P2 (option B), it would imply that there could be a situation where the fluid is not in equilibrium, which is not the case in a static fluid. 4. **Conclusion**: - The only scenario that satisfies the condition of equilibrium in a static fluid is when the downward pressure (P1) equals the upward pressure (P2). Therefore, we conclude that: \[ P1 = P2 \] ### Why Other Options Are Incorrect - **Option A (P1 > P2)**: - This suggests that the downward pressure is greater than the upward pressure, leading to a net downward force. This would cause the fluid to accelerate downwards, which contradicts the assumption of static equilibrium. - **Option B (P1 ≥ P2)**: - While this option allows for the possibility of P1 being equal to P2, it also includes the case where P1 is greater than P2. As explained, if P1 is greater than P2, it would lead to a net force and thus a non-static situation. - **Option C (P1 < P2)**: - This implies that the upward pressure is greater than the downward pressure, leading to a net upward force. This would cause the fluid to rise, again contradicting the static condition. - **Option D (P1 = P2)**: - This is the only option that maintains the equilibrium condition necessary for a static fluid, making it the correct answer. ### Revision Summary - Pressure in a fluid at rest increases with depth due to the weight of the liquid above. - At a fixed point in a static fluid, the downward pressure (P1) must equal the upward pressure (P2) for equilibrium. - The correct relationship is P1 = P2, ensuring no net force acts on the fluid. - Any deviation from this equality (P1 > P2 or P1 < P2) would result in fluid motion, contradicting the static condition.