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प्रश्न
A lever of length 100 cm has the effort of 15 kgf at a distance of 40 cm from the fulcrum at one end. What load can be applied at its other end?
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उत्तर
Given,
Effort = 15 kgf
Load arm = 100 cm
Effort arm = 40 cm
Mechanical advantage = `"Effort arm "/"Load arm" = 40 /100 = 0.4`
By definition,
Mechanical advantage =`"Load"/"Effort"`
∴ 0.4 = `"Load"/"15 kgf"`
⇒ Load = 15 kgf × 0.4
∴ Load = 6 kgf
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संबंधित प्रश्न
Give one example of a class I lever in a case where the mechanical advantage is less than 1?
Name the class to which the following lever belong:
a see-saw
Draw diagram to illustrate the position of fulcrum, load, and effort, of the following:
a see-saw.
How can you increase the mechanical advantage of a lever?
State three differences between the three classes of levers.
Name the machine to which the following belong:
Lemon crusher
Answer the following in short.
What is the basis of the classification of levers?
What is the use of the lever if its mechanical advantage is
- more than 1,
- equal to 1, and
- less than 1?
Explain why the mechanical advantage of the class III type of lever is always less than 1.
Name the load, effort, and fulcrum.
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