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.

Explanation: 

here is a hydraulic press lift being used. we are calculating the change in mass that a hydraulic press lift can lift after changing its diameters.

here pascal's law can be used so that according to this   \(\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}\)  ----(1)

  \(\frac{Mg}{\pi \frac{d_{1}^{2}}{4}} = \frac{mg}{\pi \frac{d_{2}^{2}}{4}}\)  \( \Rightarrow M \times (\frac{d_{2}}{d_{1}})^{2} = m \)   ----(2)

Calculation:  

Given: 

A hydraulic press can lift when "m" mass on small piston (M) = 100 kg

Smaller piston diameter decreased by 4 times = D1/4

The larger piston is increased by  4 times = 4D2

Using equation second and putting the value of M = 100kg then,

\(\Rightarrow 100 \times (\frac{d_{2}}{d_{1}})^{2} = m \)   ---- (3)

here, 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. so that we can write it as:

\((\frac{d_{2}}{d_{1}})^{2} = (\frac{4d_{2}}{d_{1}/4})^{2} \Rightarrow 256\times (\frac{d_{2}}{d_{1}})^{2}\)   ---- (4)

from equations (3) and (4) we get: 

\((\frac{d_{2}}{d_{1}})^{2} = \frac{25600}{m}\)   ---- (5)

again using pascals law from equation (2) we get :

 \(\therefore (\frac{d_{1}}{d_{2}})^{2} = \frac{M}{m} \)  then we get M = 25600 kg

Hence the hydraulic press can lift up to 25600 kg weight.