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Correct option is D)

$R_{2}=$Reactional force on rod by ground

$R_{1}=$ Reactional force on rod by wall

$f=$ frictional force $=μR_{2}$

$θ$ is the maximum angle the rod can make with horizontal.

Equating the forces on verticle direction we get

$R_{2}=mg−R_{1}cosθ−−−−−−−−(1)$

Equating the forces on horizontal direction we get

$R_{1}sinθ=f=μR_{2}$

Now the torque about point B should be balanced

$R_{1}cosθ×AB+R_{1}Sinθ×h=mg2l cosθ$

$⟹R_{1}h[sinθcos_{2}θ+sin_{2}θ ]=2mgl cosθ$

$⟹R_{1}=2hmgl cosθsinθ$

$⟹R_{1}cosθ=2hmgl cos_{2}θsinθ$

So, $R_{2}=mg−2hmgl cos_{2}θsinθ=2hmg (2h−lcos_{2}θsinθ)$

And $μ=R_{2}R_{1}sinθ =2hmg (2h−lcos_{2}θsin_{2}θ)2hmgh cosθsin_{2}θ $

$μ=2h−lcos_{2}θsinθlcosθsin_{2}θ $

Solve any question of Laws of Motion with:-

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