Question

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.

Answer 1

Atmospheric pressure $P_{0}$ will be acting on both the limbs of hydraulic lift.
Applying pascal's law for same liquid level
$\Rightarrow P_{0}+\frac{m g}{A_{1}}=P_{0}+\frac{\left( 100 \right) g}{A_{2}}$
$\Rightarrow \frac{m g}{A_{1}}=\frac{\left( 100\right) g}{A_{2}}\Rightarrow \frac{m}{100}=\frac{A_{1}}{ A_{2}}$ ..... (1)
Diameter of piston on side of $100\,kg$ is increased by $4$ times so new area $=16\,A_{2}$
Diameter of piston on side of $\left(m\right)kg$ is decreasing
$A_{2}=\frac{A_{1}}{16}$
(In order to increasing weight lifting capacity, diameter of smaller piston must be reduced)
Again, $\frac{m g}{\left(\frac{A_{1}}{16}\right)}=\frac{M^{'} g}{16 A_{2}}\Rightarrow \frac{256\, m}{M^{'}}=\frac{A_{1}}{ A_{2}}$
From equation $\left(1\right)=\frac{256 m}{M^{'}}=\frac{m}{100}$
$\Rightarrow \therefore M^{'}=25600\,kg$