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

We have a vehicle of mass $$m = 5000$$ kg and the area of cross section of the load-carrying cylinder is $$A = 250$$ cm$$^2 = 250 \times 10^{-4}$$ m$$^2$$, with $$g = 10$$ m/s$$^2$$.

In a hydraulic lift, the pressure is the same throughout the fluid (by Pascal's law). So the maximum pressure the smaller piston would have to bear equals the pressure in the larger cylinder:

$$P = \frac{F}{A} = \frac{mg}{A} = \frac{5000 \times 10}{250 \times 10^{-4}} = \frac{50000}{0.025} = 2000000 \text{ Pa} = 2 \times 10^6 \text{ Pa}$$

Hence, the correct answer is Option 4.