Question

a effort of 100n can raise a load of 2000n in a hydraulic press. calculate the cross-sectional area of a small piston in it. The cross-sectional area of a large piston is 4m^s​

Answer 1

`A_(1)` = 0.2
`m^(2)`

Step-by-step explanation:

The pressure on the pistons is given as;

Pressure =
`(Force)/(Area)`

So that,

Pressure on the small piston =
`(F_(1) )/(A_(1) )` and Pressure on the large piston =
`(F_(2) )/(A_(2) )`

Thus,

`(F_(1) )/(A_(1) )` =
`(F_(2) )/(A_(2) )`

Given that:
`F_(1)` = 100 N,
`F_(2)` = 2000 N,
`A_(2)` = 4
`m^(2)`.

`(100)/(A_(1) )` =
`(2000)/(4)`

`A_(1)` =
`(100*4)/(2000)`

=
`(400)/(2000)`

= 0.2

`A_(1)` = 0.2
`m^(2)`

The area of the small piston is 0.2
`m^(2)`.