JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Fluid Mechanics

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: 25600

Solution

For a hydraulic press, Pascal's law gives:

\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}

Initially:

\frac{m g}{A _{1}} = \frac{100 g}{A _{2}}

, so

\frac{A _{2}}{A _{1}} = \frac{100}{m}

After changes:

New diameter of larger piston = $4 d_{2}$ , so new area $A_{2}^{'} = π (2 d_{2})^{2} = 16 A_{2}$
New diameter of smaller piston = $\frac{d _{1}}{4}$ , so new area $A_{1}^{'} = π (\frac{d _{1}}{8})^{2} = \frac{A _{1}}{16}$

New ratio:

\frac{A _{2}^{'}}{A _{1}^{'}} = \frac{16 A _{2}}{A _{1} /16} = 256 \times \frac{A _{2}}{A _{1}}

New load:

M = 256 \times 100 = 25600

kg

Answer: 25600

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Numerical answer

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JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Fluid Mechanics

Solution

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