Curious Balance

The system is at rest, so the sum of all forces must be zero:

$P_{atm} \cdot S_1 + T - P_1 \cdot S_1= 0 \\ P_{atm} \cdot S_2 + T - P_2 \cdot S_2= 0$

Because of hydrostatic pressure, $P_2=P_1+\rho g l$

Solving for $P_1, P_2$ and $T$ is quite straightforward:

$P_1=P_{atm}+\rho g l \cdot \dfrac{S_2}{S_1-S_2} \\ P_2=P_{atm}+\rho g l \cdot \dfrac{S_1}{S_1-S_2} \\ T=\rho g l \cdot \dfrac{S_1\cdot S_2}{S_1-S_2}$

Note that $T$ is not affected by the atmospheric pressure and that the string makes $P_1$ and $P_2$ dependent on all the parameters of the system.

The numerical values given yield $T=10^3 \frac{\text{kg}}{\text{m}^3} \cdot 9.81\frac{\text{m}}{\text{s}^2}\cdot 0.35 \text{ m}\cdot \dfrac{0.04\text{ m}^2 \cdot 0.01\text{ m}^2}{0.04\text{ m}^2-0.01\text{ m}^2} = \boxed{45.78 \text{ N}}$

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When $S_2 \ll S_1 \implies P_1 \approx P_{atm},\, P_2 \approx P_{atm}+ \rho g l$ and $\,T \approx \rho g l S_2$ . In this case, the upper piston is unaffected by the pull of the string and the tension must balance only the force exerted by the hydrostatic pressure on the lower piston.