A force of 250 N is applied to a hydraulic jack piston that is 0.01 m in diameter. If the piston that supports the load has a diameter of 0.10 m, approximately how much mass can be lifted by the jack? Ignore any difference in height between the pistonsGroup of answer choices2550 kg255 kg500 kg800 kg

To solve this problem, we need to use the principle of Pascal's law, which states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container.

Step 1: Calculate the area of the small piston (A1) using the formula for the area of a circle, A = πr². The radius is half the diameter, so r1 = 0.01 m / 2 = 0.005 m. Therefore, A1 = π * (0.005 m)² = 0.0000785 m².

Step 2: Calculate the area of the large piston (A2) in the same way. The radius is r2 = 0.10 m / 2 = 0.05 m. Therefore, A2 = π * (0.05 m)² = 0.00785 m².

Step 3: Use Pascal's law to set up the equation F1/A1 = F2/A2, where F1 is the force applied to the small piston, A1 is the area of the small piston, F2 is the force applied to the large piston (which we're trying to find), and A2 is the area of the large piston.

Step 4: Substitute the known values into the equation: 250 N / 0.0000785 m² = F2 / 0.00785 m². Solve for F2 to find F2 = 250 N * 0.00785 m² / 0.0000785 m² = 25000 N.

Step 5: The force F2 is the weight of the mass that can be lifted, so use the equation F = mg to solve for m, where g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, m = F2 / g = 25000 N / 9.8 m/s² = 2551 kg.

So, the hydraulic jack can lift approximately 2550 kg.