8. एडटा हाइड्रोलिक लिफ्टको प्रयोग गरी [tex]$2500 \, \text{N}$[/tex] को तौल उचाल्नु छ। यदि उक्त लिफ्टको सानो पिस्टनको क्रस सेक्सनल क्षेत्रफल ठुलो पिस्टनको भन्दा [tex]$\frac{1}{10}$[/tex] छ भने सानो पिस्टनमा चाहिने बल कति हुन्छ, पत्ता लगाउनुहोस्।

By using a hydraulic lift, a load of [tex]$2500 \, \text{N}$[/tex] needs to be lifted. If the cross-sectional area of the small piston is [tex]$\frac{1}{10}$[/tex] of the big piston, calculate the force required to apply on the small piston.



Answer :

Certainly! Let's solve the problem step by step:

1. Understanding the given information:

- The weight (force) to be lifted using the hydraulic lift is [tex]\(2500 \, \text{N}\)[/tex].

- The cross-sectional area of the small piston is [tex]\(\frac{1}{10}\)[/tex] of the cross-sectional area of the big piston.

2. Hydraulic lift principle:

- A hydraulic system works on the principle of Pascal's law which states that pressure applied at any point in a confined fluid is transmitted equally in all directions throughout the fluid.

- Therefore, the pressure exerted by the big piston and the small piston in the hydraulic lift must be the same.

3. Relationship between force and area:

- Let [tex]\(F_{\text{big}}\)[/tex] be the force applied by the big piston, and [tex]\(A_{\text{big}}\)[/tex] be the area of the big piston.

- Similarly, let [tex]\(F_{\text{small}}\)[/tex] be the force applied by the small piston, and [tex]\(A_{\text{small}}\)[/tex] be the area of the small piston.

4. Setting up the equations based on the principle of equal pressure:

- According to Pascal's law:
[tex]\[ \text{Pressure}_{\text{big}} = \text{Pressure}_{\text{small}} \][/tex]

- Thus,
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{F_{\text{small}}}{A_{\text{small}}} \][/tex]

5. Expressing [tex]\(A_{\text{small}}\)[/tex] in terms of [tex]\(A_{\text{big}}\)[/tex]:

- We are given that [tex]\(A_{\text{small}} = \frac{1}{10} A_{\text{big}}\)[/tex].

6. Substituting [tex]\(A_{\text{small}}\)[/tex] in the pressure equation:

- Substituting [tex]\(A_{\text{small}} = \frac{1}{10} A_{\text{big}}\)[/tex] into the pressure equation, we get:
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{F_{\text{small}}}{\frac{1}{10} A_{\text{big}}} \][/tex]

- Simplifying, we find:
[tex]\[ \frac{F_{\text{big}}}{A_{\text{big}}} = \frac{10 F_{\text{small}}}{A_{\text{big}}} \][/tex]

7. Simplifying the equation to find [tex]\(F_{\text{small}}\)[/tex]:

- Eliminating [tex]\(A_{\text{big}}\)[/tex] from both sides,
[tex]\[ F_{\text{big}} = 10 F_{\text{small}} \][/tex]

- Solving for [tex]\(F_{\text{small}}\)[/tex],
[tex]\[ F_{\text{small}} = \frac{F_{\text{big}}}{10} \][/tex]

8. Substituting the known value of [tex]\(F_{\text{big}}\)[/tex]:

- We know [tex]\(F_{\text{big}} = 2500 \, \text{N}\)[/tex], so:
[tex]\[ F_{\text{small}} = \frac{2500 \, \text{N}}{10} = 250 \, \text{N} \][/tex]

Therefore, the force required to be applied on the small piston is [tex]\(250 \, \text{N}\)[/tex].