﻿ A force of 2x102 N is applied on one of the piston of area of cross section 100cm2of hydraulic lift to support a car placed on the second piston of lift of area of cross section 1960cm2. Find the mass of the car. from Physics Mechanical Properties of Fluids Class 11 Manipur Board

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Mechanical Properties of Fluids

Physics Part II

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A force of 2x102 N is applied on one of the piston of area of cross section 100cm2of hydraulic lift to support a car placed on the second piston of lift of area of cross section 1960cm2. Find the mass of the car.

Given,
Force, F = 2$×$102
Area of cross-section of the first piston = 100 cm2
Area of cross-section of the second piston = 1960 cm

Weight of the car = ?

Let, M be the mass of car.

The pressure on the piston on which force is applied is,

...(1)

The pressure on the second piston on which car is placed is,

${\mathrm{P}}_{2}=\frac{\mathrm{Mg}}{\mathrm{A}}=\frac{\mathrm{M}×9.8}{1960}=\frac{\mathrm{M}}{200}\mathrm{N}/\mathrm{cm}$    ...(2)

Now, According to Pascal law we have

P=  P2

$\therefore$ we get from equation (1) and (2),

Mass of the car, M=400  kg
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