A mechanic can open a nut by applying a force of 'F' N, while using a lever handle of 'h' cm length. What should be the length of the handle required if he wants to open it, by applying a force of only $(2 F / 3) N$ ?

\frac{h}{4} c m

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\frac{3 h}{2} c m

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\frac{3 h}{4} c m

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\frac{h}{2} c m

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Solution

The correct option is B $\frac{3 h}{2} c m$
From Equiliblium condition we get,
$F_{1} \times R_{1} = F_{2} \times R_{2} ∴ F \times h = \frac{2 F}{3} \times R_{2} ∴ R_{2} = \frac{3}{2} h$
$∴$ lenght of the handle will be $\frac{3}{2} h$ cm.

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A mechanic can open a nut by applying a force of 'F' N, while using a lever handle of 'h' cm length. What should be the length of the handle required if he wants to open it, by applying a force of only (2F/3)N?

The correct option is B 3h2cmFrom Equiliblium condition we get,F1×R1=F2×R2∴F×h=2F3×R2∴R2=32h∴ lenght of the handle will be 32h cm.

A mechanic can open a nut by applying a force of 'F' N, while using a lever handle of 'h' cm length. What should be the length of the handle required if he wants to open it, by applying a force of only $(2 F / 3) N$ ?

The correct option is B $\frac{3 h}{2} c m$
From Equiliblium condition we get,
$F_{1} \times R_{1} = F_{2} \times R_{2} ∴ F \times h = \frac{2 F}{3} \times R_{2} ∴ R_{2} = \frac{3}{2} h$
$∴$ lenght of the handle will be $\frac{3}{2} h$ cm.