A rod of mass 5 kg is connected to the string at point B. The span of rod is along horizontal. The other endof the rod is hinged at point A If the string is massless, then the reaction of hinge at the instant when stringis cut, is (Take, 9 = 10 ms-2) [UP CPMT 2015]

When string is cut, the weight of the rod constitutes a torque about the hinge point A.So,  τA=mgl2          …(i)Also, from Newton's law,   τA=Iα              …(ii)where, α = angular acceleration of the rodand I = moment of inertia of rod.From Eqs. (i) and (ii), we getIα=mgl2⇒α=mgl2/IAs,   I=ml23So,    α=mg(l/2)ml23=3g2lNow, acceleration of centre of mass of rod,  aCM=α⋅rHere, r = distance between hinge point A and centre of the rod.⇒             aCM=32gll2=3g4⇒ mg−RA=maCMwhere, RA = reaction at A.⇒ mg−RA=m×3g4⇒RA=mg4=5×104=12.5N