Question 67: Properties of Matter

Question

A hydraulic automobile lift is designed to lift vehicles of mass $$5000 \mathrm{~kg}$$. The area of cross section of the cylinder carrying the load is $$250 \mathrm{~cm}^{2}$$. The maximum pressure the smaller piston would have to bear is $$\left[\right.$$ Assume $$\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right]$$

Accepted Answer

A hydraulic lift works based on Pascal's principle, which states that the pressure applied at one point in an incompressible fluid is transmitted equally in all directions.
The force exerted by the car on the hydraulic fluid is equal to the weight of the car, which is $$F = mg$$, where $$m$$ is the mass of the car and $$g$$ is the acceleration due to gravity.
Substituting the given values, we get:
$$F = 5000 \, 	ext{kg} 	imes 10 \, 	ext{m/s}^2 = 50000 \, 	ext{N}$$
The pressure exerted by the car on the hydraulic fluid is equal to the force divided by the area over which the force is distributed, which is $$P = \frac{F}{A}$$, where $$A$$ is the cross-sectional area of the cylinder carrying the load.
However, the given area is in cmÃÂ², so we need to convert it to mÃÂ². We know that 1 mÃÂ² = 10,000 cmÃÂ², so:
$$A = 250 \, 	ext{cm}^2 	imes \frac{1 \, 	ext{m}^2}{10000 \, 	ext{cm}^2} = 0.025 \, 	ext{m}^2$$
Substituting the values of force and area into the formula for pressure, we get:
$$P = \frac{50000 \, 	ext{N}}{0.025 \, 	ext{m}^2} = 2 	imes 10^6 \, 	ext{Pa}$$

Answer

$$20 	imes 10^{+6} \mathrm{~Pa}$$

Answer

$$200 	imes 10^{+6} \mathrm{~Pa}$$

Answer

$$2 	imes 10^{+5} \mathrm{~Pa}$$

Answer

$$2 	imes 10^{+6} \mathrm{~Pa}$$

Properties of Matter

A hydraulic automobile lift is designed to lift vehicles of mass $5000 \mathrm{~kg}. The area of cross section of the cylinder carrying the load is 250 \mathrm{~cm}^{2}. The maximum pressure the smaller piston would have to bear is \left[\right. Assume \left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right]

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