Question

3) The hydraulic lift has a large cross
section and a small cross section.
Large cross-sectional area is 20
times the small cross-sectional area.
If on the small cross section is given
an input force of 25 N, then determine
the output force.​

Answer 1

Answer:

500 N

Explanation:

Size difference between two pistons = 20 times

lets take the smaller piston as p1 and the bigger one as p2

lets take the area on smaller piston as "x".

lets solve the p1 first:

force applied  = 25 N

area = x

pressure  = force/area

                = 25 N / x = 25/"x"N/m2

we all know that the liquids exert same pressure on all sides

now lets calculate the pressure on piston 2 (bigger piston - p2)

area = 20x

force = ?

as pressure will be exerted same in all directions the pressure apllied will be same on all pistons:

pressure = 25/x  N/m2

so force = Area multiplied by pressure

               = 25 multipled by 20

               = 500 N

Thus we get the answer = 500 N

Hope this helps :)

please mark my answer as the brainliest answer!!

Thank you!!

Answer 2

GiveN :

• In a hydraulic lift large area of cross section is equal to 20 times the small cross sectional.
• Force applied on the smaller area is 25 N.

To FinD :

• Output Force

SolutioN :

As we know that Hydraulic lift works on the principle of Pascal's Law. So, pressure at both the ends of the lift will remains constant.

Let, smaller area be x m².

Then, bigger area will be 20x m².

• Smaller Area $\sf{(A_1) \: = \: x \: m^2}$
• Input Force $\sf{(F_1) \: = \: 25 \: N}$
• Larger Area $\sf{(A_2) \: = \: 20x \: m^2}$
• Output Force $\sf{(F_2) \: = \: ?}$

Pressure will be Constant Here,

P = P_{input} = P_{Output}

$\longrightarrow \sf{P \: = \: \dfrac{F_1}{A_1} \: = \: \dfrac{F_2}{A_2}}$

$\longrightarrow \sf{\dfrac{25}{x} \: = \: \dfrac{F_2}{20x}}$

$\longrightarrow \sf{F_2 \: = \: \dfrac{25 \: \times \: 20 \cancel{x}}{\cancel{x}}}$

$\longrightarrow \sf{F_2 \: = \: 20 \: \times \: 25}$

$\longrightarrow \sf{F_2 \: = \: 450}$

∴ Output force is 450 N.