Question

The small piston of a hydraulic lift has an area of $0.20\,{m^2}$. A car weighing $1.20 \times {10^4}\,N$ sits on a rack mounted on the large piston. The large piston has an area of $0.90\,{m^2}$. How large a force must be applied to the small piston to support the car?
A) $2.67 \times {10^3}\,N$

B) $2.67 \times {10^4}\,N$

C) $2.67 \times {10^5}\,N$

D) $2.67 \times {10^6}\,N$

Answer 1

Hint:Here we have to apply Pascal's law to get the measure of force applied to the piston.

Pascal’s law states that the pressure applied to the confined fluid can be transferred to any point of the fluid and to the walls of the container without any difference of magnitude.
Pressure at every point of the fluid is identical in both directions.

Complete step by step solution:
Hydraulic lift is sometimes referred to as the hydraulic jack. Its working is based on the principle of Pascal’s law.

Hydraulic jack is used for the raising of heavy vehicles such as cars, trucks etc. Hydraulic jack consists of two cylindrical vessels which are connected to each other by a tube having a pipe. One cylindrical vessel is shorter than the other. The piston of the smaller cylindrical vessels is connected to the lever of the piston of the larger cylindrical vessels and has a platform for the raising of the vehicles.

As the handle of the lever is pushed down by applying force, the valve in the tube is opened by increasing the force of the narrower cylinder. The liquid flows out of the smaller to the broader cylinder. As a consequence, the piston in the larger cylindrical vessel lifts the objects positioned in the platform.

When the objects reach the target height, the handle of the lever is no longer pushed. The valve is locked so that the liquid does not spill out from the wider cylindrical vessels to the narrower ones.

Thus, we can see how Pascal law is applied on the hydraulic pressure.
Given,
${F_1} = 1.2 \times {10^4}\,N$

${a_1} = 0.90\,{m^2}$

${a_2} = 0.2\,{m^2}$

The Pascal law is mathematically given by:
$
  P = \dfrac{{{F_1}}}{{{a_1}}} = \dfrac{{{F_2}}}{{{a_2}}} \\
   \Rightarrow \dfrac{{1.2 \times {{10}^4}}}{{0.90}} = \dfrac{{{F_2}}}{{0.2}} \\
  {F_2} = \dfrac{{0.24 \times {{10}^4}}}{{0.90}} = 2.67 \times {10^3}\,N \\
$

Hence, option A is correct.

Note:Here we may get confused as all the options may appear the same. But the powers are different. So, while writing the final answer, we have to be careful. Also, if we incorrectly put the values in the formula, the answer may be wrong.