Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A hydraulic machine exerts a force of 900 N on a piston of diameter 1.80 cm. The output force is exerted on a piston of diameter 36 cm. What will be the output force?A. $12 \times {10^4}\;{\rm{N}}$B. $16 \times {10^4}\;{\rm{N}}$C. $36 \times {10^4}\;{\rm{N}}$D. $38 \times {10^4}\;{\rm{N}}$

Last updated date: 17th Jun 2024
Total views: 385.2k
Views today: 3.85k
Verified
385.2k+ views
Hint: According to Pascal law, the pressure at the input and output of the hydraulic machine is the same. Obtain the area of the cross section of the input and output of the hydraulic machine and the pressure can be written as force divided by area of cross section.

Formula used:
According to the Pascal law: $\dfrac{{{F_{\rm{1}}}}}{{{A_{\rm{1}}}}} = \dfrac{{{F_{\rm{2}}}}}{{{A_{\rm{2}}}}}$
Here, ${F_1}$ and ${F_2}$ are the forces at point 1 and 2 and ${A_1}$ and ${A_2}$ are the cross sectional area of point 1 and 2.

From the question, we know that the magnitude of the input force is ${F_{{\rm{in}}}} = 900\;{\rm{N}}$, the diameter of piston at input is ${d_{{\rm{in}}}} = 1.80\;{\rm{cm}}$ and the diameter of piston at output is ${d_{{\rm{out}}}} = 36\;{\rm{cm}}$.
The cross sectional area of the piston at input,
${A_{{\rm{in}}}} = \dfrac{\pi }{4}d_{{\rm{in}}}^2$
Substitute the value of as 1.80 cm.
${A_{{\rm{in}}}} = \dfrac{\pi }{4}{\left( {1.80} \right)^2}$
Similarly, the cross sectional area of the piston at output,
${A_{{\rm{out}}}} = \dfrac{\pi }{4}{\left( {36} \right)^2}$

We know that the hydraulic machine works on the principle of Pascal law. So, the pressure at the input is equal to the pressure at output of the hydraulic machine.
${P_{{\rm{in}}}} = {P_{{\rm{out}}}}$
Now we write the pressure in terms of force and area of the input and output of the hydraulic machine.
$\dfrac{{{F_{{\rm{in}}}}}}{{{A_{{\rm{in}}}}}} = \dfrac{{{F_{{\rm{out}}}}}}{{{A_{{\rm{out}}}}}}$
Now we substitute the given values in the above equation.
$\dfrac{{900}}{{\dfrac{\pi }{4}{{\left( {1.80} \right)}^2}}} = \dfrac{{{F_{{\rm{out}}}}}}{{\dfrac{\pi }{4}{{\left( {36} \right)}^2}}}$
Now we simplify and rewrite the above equation, we get,
${F_{{\rm{out}}}} = \dfrac{{900 \times {{\left( {36} \right)}^2}}}{{{{\left( {1.80} \right)}^2}}}\\ \Rightarrow{F_{{\rm{out}}}} = 360000\;{\rm{N}}\\ \therefore{F_{{\rm{out}}}} = {\rm{36}} \times {\rm{1}}{{\rm{0}}^4}\;{\rm{N}}$

Thus, the output force of the hydraulic is ${\rm{36}} \times {\rm{1}}{{\rm{0}}^4}\;{\rm{N}}$ and option C is correct.

Note: The hydraulic machine is an application of Pascal law, which states that the pressure exerted in a fluid at rest is equal in all directions. The other applications of Pascal law are hydraulic brakes, lifts, syringe piston etc.These are machinery and tools that use fluid power for its functioning. In these machines, a large amount of power is transferred through small tubes and hoses.