The right angle rule shown in figure hangs at rest from a peg thin rod. Mass distribution inside thin rod is uniform if linear mass density is λ, then the angleα at which it will hang is :-


a) tan-1(1/4)
b) tan-1(1/3)
c) tan-1(1/2)
d( cot-1(1/4)

Dear Student,
Please find below the solution to the asked query:

the torque acting due to the both arms about the hinge should be equal.

Now, 

The whole mass of weight of the side L will lie at the COM of that side ,i.e, at a distance of L2.  Now,The perpendicular distance of the COM  from the  hinge = x (say)Now,xL2 = cos αx = L2 cos αNow, toreque at hinge due to shorter side =τ1 =  λLg× xτ1 =  λLg × L2 cos α      ----- (1)Similarly the perpendicular distance of the COM of the larger arm = y (say)yL = sinα y = L sinα.Now,torque at the hinge due to longer arm = τ2 = 2λgL × y τ2 = 2λgL × L sinα    ----(2)For the rule to be in equilibrium eq 1 = eq 22λgL × L sinα = λLg × L2 cos α sinα cos α  = 14tan α = 14α = tan-114
Therefore, the correct option is (a)


Hope this information will clear your doubts about System of Particles and Rotational Motion.
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Regards
Satyendra Singh

 

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