Answer:
[tex]2F_{1}[/tex]
Explanation:
F₁ = Force on one side of the jack
A₁ = Area of cross-section of one side of the jack
F₂ = Force on second side of the jack
A₂ = Area of cross-section of second side of the jack = 2 A₁
Using pascal's law
[tex]\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}[/tex]
[tex]\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}[/tex]
[tex]F_{1}= \frac{F_{2}}{2}\\[/tex]
[tex]F_{2}= 2F_{1}[/tex]
