Question

A hydraulic press can lift $100kg$ 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 ${}^{\prime }{m}^{\prime }$ on the smaller piston.

Answer 1

Answer: 25600

Atmospheric pressure $P_0$ will be acting on both the limbs of hydraulic lift.
Applying pascal's law for same liquid level
$\Rightarrow P_0+\frac{mg}{A_1}=P_0+\frac{\left(100\right)g}{A_2}$
$\Rightarrow \frac{mg}{A_1}=\frac{\left(100\right)g}{A_2}\Rightarrow \frac{m}{100}=\frac{A_1}{A_2}$ ..... (1)
Diameter of piston on side of $100kg$ is increased by $4$ times so new area $=16A_2$
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, $\frac{mg}{\left(\frac{A_1}{16}\right)}=\frac{M^{{}^{\prime }}g}{16A_2}\Rightarrow \frac{256m}{M^{{}^{\prime }}}=\frac{A_1}{A_2}$
From equation $\left(1\right)=\frac{256m}{M^{{}^{\prime }}}=\frac{m}{100}$
$\Rightarrow \therefore M^{{}^{\prime }}=25600kg$