A nut cracker takes an effort of $4 kgf$ for a load of $6 kgf$ . What is its mechanical advantage?
1.5

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Solution

The correct option is A 1.5
Given:
The load, $L = 6 kgf$
The effort, $E = 4 kgf$
Mechanical advantage, $M.A. = \frac{L}{E} = \frac{6}{4}$
$= \frac{3}{2} = 1.5$

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