Question

A thin uniform rod AB of mass '\(m\)' and length '\(l\)' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its B end strikes the ground is
(\(g\) = acceleration due to gravity)

• A. \(\sqrt{\frac{2g}{l}}\)
• B. \(\sqrt{\frac{3g}{l}}\)
• C. \(\sqrt{\frac{mg}{l}}\)
• D. \(\sqrt{\frac{mg}{3l}}\)

Hint:

Use energy conservation with I = m l^2 / 3 about the hinge.

Answer 1

The Correct Option is B

Step 1: Torque Route Check:
Energy method is simplest, but we can sanity-check with the formula for a falling rod: at the horizontal position the rod loses $mgl/2$.

Step 2: Compute:
Kinetic energy $=\tfrac12\cdot\tfrac13ml^2\cdot\omega^2=\tfrac16ml^2\omega^2$. Equate to $\tfrac12mgl$: $\omega^2=\dfrac{3g}{l}$.

Step 3: Answer:
Option (B). The mass cancels, so options with $m$ are ruled out.

Final Answer:
Option (B). \[ \boxed{\text{(B) } \sqrt{\frac{3g}{l}}} \]