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**Hint:**The amount of torque required to open the nut remains constant but the amount of force to be applied by the mechanic varies in the two cases. Therefore, the perpendicular distance of force from the position vector must also vary. We shall equate the torque in both cases since it is the constant quantity and find the new length of the handle.

**Complete answer:**

Let us suppose the body is fixed at a particular point. If we apply force on the body at some distance from this fixed point, then the body will rotate about the fixed point. This force that is applied a distance away from the pivot point (or we could say the axis of rotation or centre of mass) is called torque.

Torque is a vector quantity. It consists of both magnitude and direction.

Thus, torque is defined as force times distance. It is the force that is perpendicular to the object, that is, the component of force is perpendicular to the position vector.

$\tau =F\times r$

The unit of torque is Newton. Meters, $\left( Nm \right)$.

Torque, $\tau =$ constant

$\Rightarrow {{F}_{1}}\times {{r}_{1}}={{F}_{2}}\times {{r}_{2}}$

Where,

${{F}_{1}}=$ force applied in the first case

${{F}_{2}}=$ force applied in the second case

${{r}_{1}}=$ length of lever handle in first case

${{r}_{2}}=$ length of lever handle in second case

Now, we have ${{F}_{1}}=F,{{F}_{2}}=\dfrac{2F}{3},{{r}_{1}}=h$,

$\begin{align}

& \Rightarrow F\times h=\dfrac{2F}{3}\times {{r}_{2}} \\

& \Rightarrow {{r}_{2}}=\dfrac{3h}{2} \\

\end{align}$

Therefore, the length of the lever required is $\dfrac{3h}{2}$.

**Hence, the correct option is (B) $\dfrac{3h}{2}$.**

**Note:**

The centre of mass of a body or a system of bodies is expressed as the point where all the mass of the body seems to be concentrated and is at equilibrium. Thus, if we apply force on the centre of mass of a body, the whole body accelerates in the direction of the force applied.

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