A hydraulic press can lift 100, kg when a mas | vedclass

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(B) According to Pascal's law,the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the

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$25600$

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(B) According to Pascal's law,the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. For a hydraulic lift,the pressure at both pistons is equal:$\frac{F_1}{A_1} = \frac{F_2}{A_2}$In the first case,let $A_1$ be the area of the smaller piston and $A_2$ be the area of the larger piston. The force on the smaller piston is $mg$ and on the larger piston is $100g$:$\frac{mg}{A_1} = \frac{100g}{A_2} \implies \frac{100}{m} = \frac{A_2}{A_1} \quad .......(1)$Since $A = \frac{\pi d^2}{4}$,the ratio of areas is proportional to the square of the ratio of diameters: $\frac{A_2}{A_1} = \left(\frac{d_2}{d_1}ight)^2$.In the second case,the new diameter of the larger piston is $d_2' = 4d_2$ and the new diameter of the smaller piston is $d_1' = \frac{d_1}{4}$.Let the new mass lifted be $M_0$. The new areas are $A_2' = (4)^2 A_2 = 16A_2$ and $A_1' = (\frac{1}{4})^2 A_1 = \frac{A_1}{16}$.Applying Pascal's law again:$\frac{mg}{A_1'} = \frac{M_0g}{A_2'} \implies \frac{M_0}{m} = \frac{A_2'}{A_1'} = \frac{16A_2}{A_1/16} = 256 \left(\frac{A_2}{A_1}ight)$Substituting $\frac{A_2}{A_1} = \frac{100}{m}$ from equation $(1)$:$\frac{M_0}{m} = 256 \left(\frac{100}{m}ight)$$M_0 = 256 	imes 100 = 25600 \, kg$.

$A$ hydraulic press can lift $100\, kg$ when a mass $'m'$ is placed on the smaller piston. It can lift ......... $kg$ when the diameter of the larger piston is increased by $4$ times and that of the smaller piston is decreased by $4$ times,keeping the same mass $'m'$ on the smaller piston.

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