Answer:
[tex]D = 55.2 cm[/tex]
Explanation:
As we know that the total mass of the all four players is given as
[tex]M = 4\times 110[/tex]
[tex]M = 440 kg[/tex]
diameter of the piston of cheer leader is given as
[tex]d_1 = 16 cm[/tex]
are of cross-section is given as
[tex]A_1 = \pi r^2[/tex]
[tex]A_1 = \pi(0.08)^2 = 0.02 m^2[/tex]
mass of the cheer leader is given as
m = 55 kg
so the pressure due to cheer leader is given as
[tex]P_{in} = \frac{mg}{A_1}[/tex]
[tex]P_{in} = \frac{55 \times 9.81}{0.02}[/tex]
[tex]P_{in} = 26835 Pa[/tex]
Now on the other side pressure must be same
so we have
[tex]\frac{Mg}{A} + \rho gH = P_{in}[/tex]
[tex]\frac{440 \times 9.8}{A} + (900)(9.8)(1) = 26835[/tex]
[tex]A = 0.24 m^2[/tex]
[tex]\pi r^2 = 0.24[/tex]
[tex]r = 0.276 m[/tex]
so diameter on the other side is given as
[tex]D = 2 r[/tex]
[tex]D = 55.2 cm[/tex]
