Question

In a hydraulic press, the cross sectional area of the big cylinder is 0.5m^2 and a load of 6000N can be lifted by it using 500N effort.Calculate the cross-sectional area of the small cylinder please give the answer with full process.

Answer 1

Answer:

In hyraulic press , pressure applied at one end will transfer to other end

Pressure 1 = Pressure2

$\frac{f1}{ a1} = \frac{f2}{a2}$

$\frac{6000}{0.5} = \frac{5000}{a}$

on solving

a = 0.41 m2

Answer 2

The cross-sectional area of the small cylinder is 6 m².

Given: The cross-sectional area of the big cylinder is 0.5 m² and a load of 6000 N can be lifted by it using 500 N effort.

To Find: The cross-sectional area of the small cylinder.

Solution:

• We know that, as per Pascal's Law, the pressure on both sides must be equal, so we can solve the question by the formula,

        F1 / A1 = F2 / A2                                 ....(1)

Coming to the numerical, we are given;

Both the bigger cylinder, F1 = 500 N, A1 = 0.5 m².

For the smaller cylinder, F2 = 6000 N.

Placing respective values in the formula (1), we can say;

     F1 / A1 = F2 / A2          

⇒  500 / 0.5 = 6000 / A2

⇒ A2 = 6000 / 1000

          = 6 m²

Hence, the cross-sectional area of the small cylinder is 6 m².

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