To solve this problem it is necessary to apply the Pascal principle for a hydraulic system. Of which you have to
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
Where
[tex]F_{1,2}[/tex]= Force applied to master cylinder and created at each of the slave cylinder
[tex]A_{1,2}[/tex] = Cross sectional area of master cylinder and slave cylinder.
The cork works as master cylinder and jug as slave cylinder:
[tex]F_1 = 120N[/tex]
Each sectional area would be
[tex]A_1 = \pi r_1^2 = \pi (1)^2 = \pi[/tex]
[tex]A_2 = \pi r_2^2 = \pi 7^2 = 49\pi[/tex]
From this Pascal law we have
[tex]F_2 = \frac{A_2}{A_1} F_1[/tex]
[tex]F_2 = \frac{49\pi}{\pi}(120)[/tex]
[tex]F_2 = 5880N[/tex]
Therefore the extra force is
[tex]F = F_2 - F_1[/tex]
[tex]F = 5880-120[/tex]
[tex]F = 5760N[/tex]
Therefore the extra force against the bottom is 5760N
