PrepRung
Sign inSign up free

JEE Main 2023 · Physics · Rotational Motion · hard

A uniform rod of mass M and length L carries a point mass M fixed at one end. The rod is hinged at its other end and released from the horizontal…

Single correct option

A uniform rod of mass MM and length LL carries a point mass MM fixed at one end. The rod is hinged at its other end and released from the horizontal position. The magnitude of the vertical force exerted by the hinge on the rod, just after release, is
  1. A. Mg4\dfrac{Mg}{4}
  2. B. 3Mg8\dfrac{3Mg}{8}
  3. C. 11Mg16\dfrac{11Mg}{16}
  4. D. 5Mg16\dfrac{5Mg}{16}
Show answer and solution

Answer: (D) 5Mg16\dfrac{5Mg}{16}

Moment of inertia about the hinge: I=13ML2+ML2=43ML2I=\tfrac13ML^2+ML^2=\tfrac43ML^2. Torque of weights: τ=MgL2+MgL=32MgL\tau=Mg\tfrac L2+MgL=\tfrac32MgL, so α=9g8L\alpha=\dfrac{9g}{8L}. Just after release ω=0\omega=0, so every point accelerates only tangentially (vertically). The centre of mass of the combined system is at M(L/2)+ML2M=3L4\dfrac{M(L/2)+ML}{2M}=\dfrac{3L}{4} from the hinge, so acm=3L4α=27g32a_{cm}=\dfrac{3L}{4}\alpha=\dfrac{27g}{32}. Newton's law for the system: 2Mg−N=2Macm2Mg-N=2M a_{cm}, giving N=2Mg(1−2732)=5Mg16N=2Mg\left(1-\dfrac{27}{32}\right)=\dfrac{5Mg}{16}.
Report a mistake

Original question written in the JEE Main 2023 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake