Hydraulic press is used to lift a load of 500 Kg when the ratio of the diameter of pump plunger and press plunger is $1:5$ . The effort applied on piston of pump plunger. If the mechanical advantage of handle of pump plunger is 4 , the effort applied on handle of pump plunger is ,

Pump plunger                          Press plunger

${\text{d}}_{\text{1}}=x$            ${\text{d}}_{\text{2}}=5x$

${\text{A}}_{\text{1}}=\frac{\pi {\text{d}}_1{}^2}{4}=\frac{\pi x^2}{4}$                         ${\text{A}}_{\text{2}}=\frac{\pi {\text{d}}_2{}^2}{4}=\frac{\pi 25x^2}{4}$

${\text{F}}_1=?$             ${\text{F}}_{\text{2}}=500\text{kgf}$

We know for hydraulic press $\frac{{\text{F}}_{\text{1}}}{{\text{A}}_{\text{1}}}=\frac{{\text{F}}_{\text{2}}}{{\text{A}}_{\text{2}}}$

$\frac{F}{\frac{\pi x^2}{4}}=\frac{500}{\frac{\pi 25x^2}{4}}$

$\Rightarrow F=\frac{500}{25}\times 1=20kgf$

Given that mechanical advantage of pump plunger = 4

            $i.e.,MA=\frac{F}{F_A}=4$

$F_A=\frac{F}{4}=\frac{20kgf}{4}=5kgf$