Question

In a hydraulic lift, there is a piston at one end and a car of mass 400 kg balanced on the load side. If the cross sectional area of load side is 20 $m^2$ and that of the piston is 5 $m^2$, what is the force applied by the piston? [Take $g=10m{s}^{-2}$]

• A. 1000 N
• B. 1600 N
• C. 4000 N
• D. None of these

Answer 1

The correct option is A

1000 N

Mass of the car is $m=400kg$

So, weight of the car is $F_1=m\times g=400\times 10=4000N$

Cross sectional area of load side is $A_1=20m^2$

Cross sectional area of the piston is $A_2=5m^2$

Let the force applied by the piston be $F_2$.

Now, according to Pascal's law:

$\frac{\text{Force applied by the car}}{\text{Cross sectional area of load side}}=\frac{\text{Force applied by the piston}}{\text{Cross sectional area of the piston}}$

$\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\Rightarrow \frac{F_1}{F_2}=\frac{A_1}{A_2}\Rightarrow \frac{F_1}{4000}=\frac{5}{20}\Rightarrow F_1=4000\times \frac{5}{20}=1000N$