Q109 / 200 · NEET 2007 · Physics

A uniform rod $\mathrm{AB}$ of length $I$ and mass $m$ is free to rotate about point $A$. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $\mathrm{A}$ is $m l^2 / 3$, the initial angular acceleration of the rod will be Figure

  • A
    $\frac{m g l}{2}$
  • B
    $\frac{3}{2} g l$
  • Correct answer
    C
    $\frac{3 g}{2 l}$
  • D
    $\frac{2 g}{3 l}$

Solution

Given that moment of inertia about \(A\) is \(I=m l^3 / 3\) Now, torque about \(A\) is given as \(\begin{aligned} & \tau=F \times r \\ & \tau=m g \frac{1}{2} \\ & I \alpha=m g \frac{1}{2} \\ & \frac{m l^2}{3} \alpha=m g \frac{1}{2} \\ & \alpha=\frac{3 g}{2 l} \end{aligned}\)

Hence, the correct answer is option: 'C'.

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