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 $\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right]$

<p>A hydraulic lift works based on Pascal&#39;s principle, which states that the pressure applied at one point in an incompressible fluid is transmitted equally in all directions. </p> <p>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.</p> <p>Substituting the given values, we get:</p> <p>$F = 5000 \, \text{kg} \times 10 \, \text{m/s}^2 = 50000 \, \text{N}$</p> <p>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.</p> <p>However, the given area is in cm², so we need to convert it to m². We know that 1 m² = 10,000 cm², so:</p> <p>$$A = 250 \, \text{cm}^2 \times \frac{1 \, \text{m}^2}{10000 \, \text{cm}^2} = 0.025 \, \text{m}^2$$</p> <p>Substituting the values of force and area into the formula for pressure, we get:</p> <p>$P = \frac{50000 \, \text{N}}{0.025 \, \text{m}^2} = 2 \times 10^6 \, \text{Pa}$</p>