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In a hydraulic lift used at a service station, the radius of the large and small piston are in the ratio of 20:1. What weight placed on the small piston will be sufficient to lift a car of mass 1500 kg?

(a) 300 N
(b) 400 N
(c) 450 N
(d) 500 N

1 Answer

4 votes

Final answer:

OPTION B.The question is about calculating the force needed on the small piston of a hydraulic lift to lift a 1500 kg car based on the ratio of the piston radii.

Step-by-step explanation:

The question is related to the principles of a hydraulic lift and Pascal's Law, which is used to calculate the required force to lift a heavy object. The given ratio of the radii of large to small piston is 20:1. To find the weight placed on the small piston to lift a 1500 kg car, we use the concept of hydraulic multiplication, which states that the forces on the pistons are proportional to their respective areas. The force needed to lift the car is equal to the weight of the car, which can be calculated using the gravitational force equation (F = m*g, where g is the acceleration due to gravity, 9.8 m/s2).

Since the ratio of the areas of the pistons is equal to the square of the ratio of their radii, we get Arealarge / Areasmall = (202 : 12) = 400:1. The weight of the car is 1500 kg * 9.8 m/s2 = 14700 N, and this force will be supported by the larger piston. Thus, the force that needs to be applied on the smaller piston is 14700 N / 400 = 36.75 N. However, since the options provided do not include this value, we must assume a rounding or calculation error in the phrasing of the question or the provided options.

User Rob Murphy
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