A hydraulic press can lift 100 kg when a mass ‘m’ is placed on the smaller piston. It can lift _______ $\times 10^{4} k g$ 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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answer is 2.56.

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Detailed Solution

$\begin{array}{l} U \sin g P a s c a l' s L a w, s a m e P r e s s u r e i s t r a n s m i t t e d . \\ \Rightarrow \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \\ Initially, \frac{100 g}{A_{1}} = \frac{mg}{A_{2}} \dots \dots \dots . (i) \\ h e r e, A_{1} is area of larger piston, which will become 16 A_{1} and \\ A_{2} is area of smaller piston, which will become \frac{A_{2}}{16} \\ Finally, \frac{Mg}{16 A_{1}} = \frac{mg}{(\frac{A_{2}}{16})} \dots \dots \dots (i i) \\ \Rightarrow \frac{100 \times 16}{M} = \frac{1}{16} \\ \Rightarrow M = 25600 kg \end{array}$