Class II levers are designed to have:

M . A . = V . R .

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M . A . < V . R .

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M . A . > 1

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M . A . < 1

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Solution

The correct option is C $M . A . > 1$
M.A $=$ $\frac{d_{1}}{d_{2}}$
since, $d_{1}$ > $d_{2}$ (according to class 2 lever diagram)
so,the value of M.A will be greater than 1.

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Similar questions

Class II levers are designed to have :

(a) M.A = V. R. (b) M.A > V. R.

The correct relationship betwen the mechanical advantage (M.A.), the velocity ratio (V. R.) and the efficiency $η$ is :

(a) $M . A = η \times V . R .$ (b) $V . R = η \times M . A$

The figure below shows a lever in use.

(a) To which class of lever does it belong?

(b) If AB = 1m, AF = 0.4 m, find its mechanical advantage.

How is the mechanical advantage (M.A) related to the velocity ratio(V.R) for an ideal machine?

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