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Question: The areas of the piston in hydraulic machine are \(10c{{m}^{2}}\) and \(225c{{m}^{2}}\) the force re...

The areas of the piston in hydraulic machine are 10cm210c{{m}^{2}} and 225cm2225c{{m}^{2}} the force required on the smaller piston to support a load of 1000N on the larger piston.
A. 44.44N B. 55.55N C. 33.33N D. 4.44N \begin{aligned} & A.\text{ }44.44N \\\ & B.\text{ }55.55N \\\ & C.\text{ }33.33N \\\ & D.\text{ }4.44N \\\ \end{aligned}

Explanation

Solution

This question is related to Pascal's law; in which we will equalize pressure on smaller piston with pressure on larger piston to get the relevant answer or to find remaining quantities.

Formula used:
P=FAP=\dfrac{F}{A}

Complete step by step solution:
According to Pascal's law pressure on smaller piston should be equal to pressure on larger piston.

If P1{{P}_{1}} is pressure on the smaller piston and P2{{P}_{2}} be the pressure on the larger piston then,
P1=P2....(1){{P}_{1}}={{P}_{2}}....\left( 1 \right)

Now we know that formula for pressure,
P=FA....(2)P=\dfrac{F}{A}....\left( 2 \right)
P = pressure,
F = force
A = area

Now given that,
F2=1000N A2=225cm2 A1=10cm2 F1=? \begin{aligned} & {{F}_{2}}=1000N \\\ & {{A}_{2}}=225c{{m}^{2}} \\\ & {{A}_{1}}=10c{{m}^{2}} \\\ & {{F}_{1}}=? \\\ \end{aligned}

Now from the equation (1) and the equation (2)
F1A1=F2A2....(3)\dfrac{{{F}_{1}}}{{{A}_{1}}}=\dfrac{{{F}_{2}}}{{{A}_{2}}}....\left( 3 \right)

Now put all the values in equation (3)
F110=1000225 F1=1000×10225 F1=44.44N \begin{aligned} & \Rightarrow \dfrac{{{F}_{1}}}{10}=\dfrac{1000}{225} \\\ & \Rightarrow {{F}_{1}}=\dfrac{1000\times 10}{225} \\\ & \therefore {{F}_{1}}=44.44N \\\ \end{aligned}
Here force required on the smaller piston is 44.44N

Hence, the option (A) is correct.

Additional information:
According to Pascal’s principle in a hydraulic system a pressure erected on a piston produces equal increases in pressure on another piston in the system as we used in the above question. Pascal also discovered that the pressure at a point in a fluid is the same in all directions; the pressure would be the same on all planes passing through a specific point; this fact is also known as Pascal’s principle or Pascal’s law.
Real life examples of Pascal’s law
It is broadly used in the automobile industry it is used in hydraulic brake systems, hydraulic jack, hydraulic presses and hydraulic machines.

Note:
In this question we have used Pascal’s law in which pressure should be equal on both pistons so that by putting pressure equation P=FAP=\dfrac{F}{A}and equalizing the equation we will get required force on the smaller piston.