Question

in a hydraulic lift,the area of crosses section of large and small piston are 1000cm2 and 10 cm2 respectively find the force applied on small piston to lift a mass of 2500kg with the large piston​

Answer 1

Answer: F = 250 N

Large Piston:

• Area, A = 1000 cm²

Here, a mass of 2500 kg is put on the large piston which means the force applied on the large piston will be the weight of that body which is given by,

⇒ Weight = Mass × acceleration due to gravity

[ Generally, g is taken as 10 m²/s ]

⇒ Weight = 2500 × 10

⇒ Weight = 2.5 × 10⁴ N

Now, let's convert the area of piston into m², we know, 1 m² = 10⁴ cm²

So, area in m² will be,

⇒ Area = 1000 / 10⁴

⇒ Area = 10³ / 10⁴

⇒ Area = 0.1 m²

Because the pressure will be same everywhere as it is a hydraulic lift. So, pressure on the large piston will be equal to the pressure on small piston. Let's find the pressure,

⇒ Pressure = Force / Area

⇒ P = 2.5 × 10⁴ / 0.1

⇒ P = 2.5 × 10⁵ Pa

Now, small piston

• Area, a = 10 cm² = 0.001 m²

Here, pressure on small piston is the same as on large piston i.e., 2.5 × 10⁵ Pa

⇒ Pressure = Force / Area

⇒ 2.5 × 10⁵ = F / 0.001

⇒ F = 2.5 × 10⁵ × 10⁻³

⇒ F = 2.5 × 10²

⇒ F = 250 N

Hence, the force that should be applied in order to lift a mass of 2500 kg is 250 N.

Answer 2

Answer:

Given :-

Area of large piston = 1000 cm²

Area of small piston = 10 cm

Mass = 2500 kg

To Find :-

Force applied

Solution :-

We know that

$\bf \red{Weight = Mass \times Acceleration}$

Weight = 2500 × 10

Weight = 25000 kg

Or,

Weight = 2.5 × 10⁴ kg

Now,

Area = 1000/10⁴

Area = 0.1 m²

Now,

We know that.

P = F/A

Where,

P = Pressure

F = Force

A = Area

$\sf \: P = \dfrac{2.5 \times 10 {}^{4} }{0.1}$

$\sf \: P = 25 \times{ 10}^{4}$

$\sf \: P = 2.5 \times {10}^{5}$

Now,

Again by using same

$\sf \: \: 2.5 \times {10}^{5} = f \times 0.001$

$\sf \: f \: = 2.5 \times 0.001 \times {10}^{ 5}$

$\sf \: f = 0.0025 \times {10}^{5}$

$\sf \: f = 250 \: newton$