A hydraulic lift as shown in the figure is us | vedclass

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(B) According to Pascal's law, the pressure applied at both pistons must be equal for equilibrium.$P_1 = P_2$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$Here,

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$10$

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(B) According to Pascal's law, the pressure applied at both pistons must be equal for equilibrium.$P_1 = P_2$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$Here, $F_1 = M_1 g = 1000 	imes g$ and $F_2 = M_2 g$.Given $A_1 = 1 \ m^2$ and $A_2 = 0.01 \ m^2$.Substituting the values:$\frac{1000 	imes g}{1} = \frac{M_2 	imes g}{0.01}$$1000 = \frac{M_2}{0.01}$$M_2 = 1000 	imes 0.01 = 10 \ kg$.Thus, a mass of $10 \ kg$ needs to be placed on piston $(P_2)$ to lift the $1000 \ kg$ mass.

$A$ hydraulic lift as shown in the figure is used to lift a mass of $1000 \ kg$, which is placed on a piston $(P_1)$ of area $1 \ m^2$. If the cross-section area of the piston $(P_2)$ at the other end is $0.01 \ m^2$, then how much mass needs to be put on it to lift the $1000 \ kg$ (in $kg$)?

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