A hydraulic press with the larger piston of diameter 35 cm at a height of 1.5 m relative to the smaller piston of diameter 10 cm and mass 20 Kg . Then force exerted on the load by the large piston is, [The density of oil on press is $750\text{kg}{\text{m}}^{-3}$ ]

Diameter of load piston A = 35 cm.

Radius $\text{r}=17.5\times {10}^{-2}\text{m}$

Diameter of effort piston = 10 cm.

Radius $=5\times {10}^{-2}\text{m}$

We have

${\text{P}}_{\text{s}}={\text{P}}_{\text{L}}+\rho \text{gh}$

$\frac{{\text{F}}_{\text{s}}}{\text{a}}\text{=}\frac{{\text{F}}_{\text{L}}}{\text{A}}+\rho \text{gh}$

$\frac{20}{\pi \times {\left(5\times {10}^{-2}\right)}^2}=\frac{{\text{F}}_{\text{L}}}{\pi \times {\left(17.5\times {10}^{-2}\right)}^2}+750\times 1.5\times 10$

$\Rightarrow \frac{20\times {10}^4}{\pi \times 25}=\frac{{\text{F}}_{\text{L}}\times {10}^4}{\pi \times 306.25}+11250$  $\Rightarrow 25480=10.4{\text{F}}_{\text{L}}+11250$

${\text{F}}_{\text{L}}=1.34\times {10}^3\text{N}$