Answer:
[tex]A_{1}[/tex] = 0.2 [tex]m^{2}[/tex]
Explanation:
The pressure on the pistons is given as;
Pressure = [tex]\frac{Force}{Area}[/tex]
So that,
Pressure on the small piston = [tex]\frac{F_{1} }{A_{1} }[/tex] and Pressure on the large piston = [tex]\frac{F_{2} }{A_{2} }[/tex]
Thus,
[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
Given that: [tex]F_{1}[/tex] = 100 N, [tex]F_{2}[/tex] = 2000 N, [tex]A_{2}[/tex] = 4 [tex]m^{2}[/tex].
[tex]\frac{100}{A_{1} }[/tex] = [tex]\frac{2000}{4}[/tex]
[tex]A_{1}[/tex] = [tex]\frac{100*4}{2000}[/tex]
= [tex]\frac{400}{2000}[/tex]
= 0.2
[tex]A_{1}[/tex] = 0.2 [tex]m^{2}[/tex]
The area of the small piston is 0.2 [tex]m^{2}[/tex].