Answer:
600 N
119366.20731 Pa
Explanation:
[tex]F_1[/tex] = Force on the smaller piston
[tex]A_1[/tex] = Area of the smaller piston = [tex]\pi 0.04^2[/tex]
[tex]F_2[/tex] = Force on the larger piston = 15000 N
[tex]A_2[/tex] = Area of the larger piston = [tex]\pi 0.2^2[/tex]
From Pascal's law we have
[tex]\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_1=\dfrac{F_2A_1}{A_2}\\\Rightarrow F_1=\dfrac{15000\pi 0.04^2}{\pi 0.2^2}\\\Rightarrow F_1=600\ N[/tex]
The force that must be applied to the small piston is 600 N
Pressure would be
[tex]P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{600}{\pi \times 0.04^2}\\\Rightarrow P=119366.20731\ Pa[/tex]
The pressure applied to the fluid in the lift is 119366.20731 Pa
