Answer:
4 times
Explanation:
Given:
diameter of the input piston, [tex]d_i=2\ cm=0.02\ m[/tex]
diameter of the output piston, [tex]d_o=4\ cm=0.04\ m[/tex]
In such mechanical arrangements the fluid acts as an incompressible link and the pressure remains uniform throughout the fluid bulk.
Then, by the Pascal's law:
[tex]\frac{F_i}{A_i} =\frac{F_o}{A_o}[/tex]
where:
[tex]F_i\ \&\ F_o[/tex] are the input and the output forces respectively.
[tex]A_i\ \&\ A_o[/tex] are the input and the output area of pistons respectively.
[tex]\frac{F_i\times 4}{\pi.d_i^2} =\frac{F_o\times 4}{\pi.d_o^2}[/tex]
[tex]\frac{F_i}{d_i^2} =\frac{F_o}{d_o^2}[/tex]
[tex]\frac{F_o}{0.04^2}=\frac{F_i}{0.02^2}[/tex]
[tex]F_o=4\times F_i[/tex]
