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(A) The pressure at the level of the horizontal tube (where the piston is located) must be equal to the sum of the pressures exerted by the liquid col

Vedclass · Accepted Answer

$(\rho_1 h_1 + \rho_2 h_2 + \rho_3 h_3)g$

Vedclass · Answer

(A) The pressure at the level of the horizontal tube (where the piston is located) must be equal to the sum of the pressures exerted by the liquid columns above it.Let $P_0$ be the atmospheric pressure. The pressure at the piston is $P = P_0 + \frac{F}{A}$.According to the principle of hydrostatics,the pressure at the same horizontal level in a continuous liquid is the same.Considering the horizontal level at the piston,the pressure is due to the three liquid columns:$1$. The pressure due to liquid of density $\rho_1$ over length $h_1$ is $\rho_1 g h_1$.$2$. The pressure due to liquid of density $\rho_2$ over vertical height $h_2$ is $\rho_2 g h_2$.$3$. The pressure due to liquid of density $\rho_3$ over vertical height $h_3$ is $\rho_3 g h_3$.Summing these,the total pressure at the piston level is $P = P_0 + (\rho_1 h_1 + \rho_2 h_2 + \rho_3 h_3)g$.Equating the two expressions for pressure at the piston,we get $\frac{F}{A} = (\rho_1 h_1 + \rho_2 h_2 + \rho_3 h_3)g$ (assuming $P_0$ is negligible or acting on both sides).

$A$ tube is bent into an $L$ shape and kept in a vertical plane. If these three liquids are kept in equilibrium by the piston of area $A$,the value of $\frac{F}{A}$ is:

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