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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

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

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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 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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