A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg the area of cross section of the piston carrying the load is 425 cm. What maximum pressure would the smaller piston have to bear. 27. State and explain equation of continuity.
Hello candidate,
The equation of continuity suggests that any liquid flowing in a medium if it is incompressible and has a continuous inform velocity while flowing, then the volumetric flow rate at two points in the motion is constant.
The Mass is of 3000 kg, and the force needed to lift it is equal to 3000*9.8= 29400 Newton.
Hope this information was helpful for you!!
Good luck.
Hi,
The question you asked is from Class 11th syllabus and is from fluid mechanics and can be solved using Pascal's law. Here is the solution for your question:
Pressure on the piston P= F/A
Force F=m×a
=3000×9.8
=29400 N
Area of cross section A=425×10 ^-4 sqm
Therefore the pressure P=3000×9.8/ 425×10 ^−4
=6.92×10 ^5 Pa.
I hope this helped you.
