Question

hydraulic lift is designed to lift heavy objects of maximum mass 2000 kg the area of cross section of piston carrying the load is 2.25 into 10 raise to minus 2 square what is the maximum pressure the smaller Piston would have to bear ?​

Answer 1

Given:

Hydraulic lift is designed to lift heavy objects of maximum mass 2000 kg the area of cross section of piston carrying the load is $\mathbf{2.25\times 10^{-2} \ m^{2}}$.

To Find:

What is the maximum pressure the smaller Piston would have to bear ?​

Solution:

Here relative height of load and effort is not given:

So we can take both are on same level of height of load and effort:

Area of effort piston is in smaller in area:

Pressure is maximum for smaller area:

From law of Pascal's Law intensity of pressure is same in same level of static fluid:

Means:

Pressure at effort = Pressure at load

We can also write:

Pressure at smaller area piston = Pressure at larger area piston

$\mathbf{\textrm{Pressure at smaller area piston }=\dfrac{Weight}{Area}}$

$\mathbf{\textrm{Pressure at smaller area piston }=\dfrac{M\times g}{Area}}$

$\mathbf{\textrm{Pressure at smaller area piston }=\dfrac{2000\times 10}{2.25\times 10^{-2}}}$

On simplify:

$\mathbf{\textrm{Maximum pressure at smaller piston }=8.89 \times 10^{5} \ \ \ \dfrac{N}{m^{2}}}$

Answer 2

Answer:

Correct option is

A

0.8711×106 N/m2

Pressure on the piston=A/F

Force F= m×a

              =2000×9.8 

          

Area of cross section A= 2.25×10^-2m²

Therefore the pressure P= 2000×9.8/2.25×10^-2m²

P= 0.8711×106m2N