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 ‘$m$’ on the smaller piston.

Answer 1

Step 1: Given data and assumption.

Mass on the larger piston $m_2=100kg$

Mass on the smaller piston $m_1=m$

Let mass $m$ also lift the mass in second case be $M_0$

Step 2: Finding the mass $M_0$ on the larger piston in second case.

Using Pascal Law:

${\rho }_0+\frac{m_{1}g}{A_1}={\rho }_0+\frac{m_{2}g}{A_2}$

Where, ${\rho }_0=$atmospheric pressure, $A_1=A_2=$area of piston, $g=$acceleration due to gravity.

Now,

From figure:$\left(a\right)$

${\rho }_0+\frac{mg}{A_1}={\rho }_0+\frac{100\times g}{A_2}$

$\Rightarrow$ $\frac{mg}{A_1}=\frac{100\times g}{A_2}$ ….$\left(i\right)$

Again,

From figure:$\left(b\right)$

Mass $m$ on smaller piston can lift $M_0$ on the larger piston in second case

Diameter of larger piston is increased by 4 times then new area is,

$A_2=16A_2$ $\left[\because A=\frac{\pi d^2}{4}\right]$

Diameter of smaller piston is decreased by $4$ times then new area is,

$A_1=\frac{A_1}{16}$

Therefore,

${\rho }_0+\frac{mg}{{\displaystyle \raisebox{1ex}{$A_1$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}}={\rho }_0+\frac{M_0\times g}{16A_2}$

$\Rightarrow$ $\frac{mg}{{\displaystyle A_1}}=\frac{M_0\times g}{16\times 16A_2}$ ….$\left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right)$ we get.

$\frac{M_0\times g}{16\times 16A_2}=\frac{100\times g}{{\displaystyle A_2}}$

$\Rightarrow$ $M_0=25600kg$

Hence, It can lift $\mathbf{25600}$ $\mathit{k}\mathit{g}$