Question

A metallic rod of mass $20kg$ and of uniform thickness rests against a wall while the lower end of the rod is in contact with the rough floor . The rod makes an angle ${{60}^{0}}$ with floor .if the weight of rod produces a torque $150Nm$ above its lower end the length of rod $(g=10m{{s}^{-2}})$ .
A) $1.5m$
B) $2m$
C) $3m$
D) $4m$

Answer 1

Hint: Torque is the turning or twisting effectiveness of a force and it has both magnitude and direction. The direction of torque is obtained by using the right hand grip rule and torque is equivalent to linear force. Torque is also referred to as rotational force or turning effect and SI unit of torque is newton –meter (N-m).

Complete step by step answer:
Torque is a force applied to an object to make it rotate and it is a vector quantity. The vector quantity of a torque is equal to the vector product of the vector pointing from the axis to the point of applied force and the vector of force. Torque depends on the magnitude of the force and the perpendicular distance between the points at which torque has to be calculated.
Torque is mathematically represented as
$T=F\times r\sin \theta $
From the data we known the values of angle which was made by the rod
So the torque equation becomes as
$T=mg\times \dfrac{l}{2}\cos {{60}^{0}}$
On rearranging
$L=\dfrac{2T}{mg}{{\cos }^{-1}}{{60}^{0}}$ $(1)$
From the data we known the values of T (torque) , mass(m)and g(length of rod)
On substituting in equation (1)
$\begin{align}
  & L=\dfrac{2\times 150}{20\times 10\times {{10}^{-3}}}{{\cos }^{-1}}{{60}^{^{0}}} \\
 & L=1.5\times {{\cos }^{-1}}{{60}^{0}} \\
 & L=1.5\times 2 \\
 & L=3m \\
\end{align}$

So, the correct answer is “Option C”.

Note: Students the magnitude of torque of a rigid body depends on the force applied, the lever arm vectors and force is a vector quantity. The rotational speed is measured in revolutions per unit time and torque and angular displacement are in the same direction then the scalar product reduces to a product of magnitude.