The hydraulic press shown in the figure is used to raise the mass M through a height of 5.0 mm by performing 500 J of work at the small piston. The diameter of the large piston is 10 cm while that of the smaller one is 2 cm. The mass M is

Since, M moves 5.0 mm and volume of liquid remains constant  
$\therefore \mathrm{\pi }\times {\left(5\times {10}^{-2}\right)}^2\left(5\times {10}^{-3}\right)=\mathrm{\pi }\times {\left(1\times {10}^{-2}\right)}^2\mathrm{x}$
$\therefore \mathrm{x}=125\times {10}^{-3}\mathrm{m}$
Suppose force F does 500 J work on small piston.
Fx = 500
$\mathrm{F}=\frac{500}{125\times {10}^{-3}}N=4\times {10}^3\mathrm{N}$
Using Pascal’s law , $\mathrm{Mg}=\frac{\mathrm{F}{\left(5\times {10}^{-2}\right)}^2}{{\left(1\times {10}^{-2}\right)}^2}$
$\mathrm{Mg}=100\times {10}^3={10}^5\mathrm{N}\Rightarrow \mathrm{M}={10}^4\mathrm{kg}$