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.

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Solution

From Pascal's law,

$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$

$\Rightarrow \frac{4 m_{1} g}{π d_{1}^{2}} = \frac{4 m_{2} g}{π d_{2}^{2}}$

$\Rightarrow \frac{100}{d_{1}^{2}} = \frac{m}{d_{2}^{2}}$

$\Rightarrow \frac{d_{1}^{2}}{d_{2}^{2}} = \frac{100}{m} . . . (1)$

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,

$\frac{{d_{1}^{^{'}}}^{2}}{{d_{2}^{^{'}}}^{2}} = \frac{(4 d_{1})^{2}}{(d_{2} / 4)^{2}} = 256 \times \frac{d_{1}^{2}}{d_{2}^{2}}$

$\Rightarrow \frac{{d_{1}^{^{'}}}^{2}}{{d_{2}^{^{'}}}^{2}} = \frac{25600}{m} . . . (2)$
$[from (1)]$

Again from Pascal's law,
$\frac{{d_{1}^{^{'}}}^{2}}{{d_{2}^{^{'}}}^{2}} = \frac{M}{m}$

$\Rightarrow \frac{25600}{m} = \frac{M}{m}$
$[from (2)]$

$\Rightarrow M = 25600 kg$

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