Question

In a brahma press a force of 10N is applied on a piston of cross section $5m^2$. Then the upward thrust exerted by the piston of the area of the cross section $10m^2$ is __________.

Answer 1

Hint: In brahma press, there are two sides of different cross-sectional areas. On one side we have a larger area and on the other side, we have a smaller area. This system works on the basis of Pascal’s law. Pascal’s law states that pressure is transmitted equally to every direction in a fluid at rest, irrespective of the distance between two parts of the container.
Formula used:
$Pressure =\dfrac{F (thrust)}{A}$

Complete answer:

As shown in the above figure, we can see two limbs having area $A_1$ and $A_2$. We apply load ‘P’ at $A_1$ and get the lift ‘Q’ at $A_2$. According to Pascal’s law, pressure inside the fluid is transmitted equally in every direction. Thus pressure at P is equal to pressure at Q.
Hence, $\dfrac{P}{A_1} = \dfrac{Q}{A_2}$
Given, $A_1 = 10m^2$and $A_2 = 5m^2$, also Q = 10N (applied at smaller end).
Hence using $\dfrac{P}{A_1} = \dfrac{Q}{A_2}$
$P = \dfrac{QA_1}{A_2} = \dfrac{10\times 10}{5} = 20N$
Hence we get an upward thrust of 20N at the bigger end.

Additional Information:
This is the advantage of brahma press. We apply a smaller load, and get a hugely increased load at the bigger end. This could be used to lift heavy objects like trucks and cars. This is based on the fundamental principle of Pascal’s law.

Note:
Pressure is a scalar quantity. Hence its nothing to do with its direction. It gets transmitted in every direction equally when the fluid is at rest. The phenomenon of getting an additional or increased force on the other end of the instrument is also called mechanical advantage. But this phenomenon has a limitation either. Practically there is a limit after which the fluid gets compressed as no fluid is ideally incompressible.