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A. $12 \times {10^4}\;{\rm{N}}$

B. $16 \times {10^4}\;{\rm{N}}$

C. $36 \times {10^4}\;{\rm{N}}$

D. $38 \times {10^4}\;{\rm{N}}$

Answer

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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}}

$

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