Final answer:
Using the principle of conservation of mass and the continuity equation, the velocity of the plunger of the syringe is much less than the velocity of the water being ejected due to the larger cross-sectional area of the syringe compared to the nozzle. Without specific dimensions, we can only indicate that the answer is a lower velocity option, given that the ejection velocity is 10 m/s.
Step-by-step explanation:
To determine the speed at which the plunger of the syringe is being depressed, we can use the principle of conservation of mass and the continuity equation for incompressible fluids. The continuity equation states that the product of the cross-sectional area (A) and the velocity (v) of the fluid must remain constant throughout its flow. This implies that A1 * v1 = A2 * v2, where A1 and v1 are the area and velocity at one point and A2 and v2 are the area and velocity at another point in the system.
Assuming the syringe and the nozzle are cylindrical, we can calculate the cross-sectional areas using their diameters. Let's say the syringe has a larger cross-sectional area (A1) compared to the nozzle's smaller cross-sectional area (A2). Using the formula for the area of a circle (A = π * d^2 / 4), we can calculate both areas. Since we know the velocity of the water ejected (v2 = 10 m/s), we can rearrange the equation to solve for the velocity of the plunger (v1): v1 = (A2 * v2) / A1.
Without the specific dimensions of the syringe and nozzle, it is not possible to give a numerical answer. However, since A1 is larger than A2, the velocity of the plunger (v1) will be much less than the velocity of the water being ejected (v2). Therefore, the correct answer would be one of the lower velocities listed, either A, B, or C, depending on the exact dimensions of the syringe and nozzle.