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A uniform rod of mass $m$ and length $l$ is in equilibrium under the constraints of horizontal and vertical rough surfaces. Then :

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the net torque of normal reaction about $O$ is equal to $\frac{m g l}{2} \cos θ$

the net torque due to normal reactions is numerically equal to the net torque due to the frictional force about the $CM$ of the rod

the net torque due to friction about $O$ is zero

answer is A, B, C, D.

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Detailed Solution

$N_{1} = m g - f_{2} \dots \dots (1); N_{2} = f_{1} . . . . . . (2)$

Now taking moment about A

$N_{2} \times l \sin θ + f_{2} \times l \cos θ = m g \frac{l}{2} \times \cos θ$ …..(3)

Take moment about point B

$N_{1} \times l \cos θ = m g \times \frac{l}{2} \cos θ + f_{1} \times l \sin θ$

$τ_{0} (of normals) =$

$N_{1} l \cos θ + N_{2} \times l \sin θ = \frac{m g}{2}$

$l \cos θ + \frac{m g}{2} \cot θ = 2 m g l \cos θ / 2$

$B, C & D$ are easily explain

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