Question

A thin uniform rod of length \(2\) m cross-sectional area 'A' and density 'd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity \(ω\). If the value of \(ω\) in terms of its rotational kinetic energy \(E\) is \((αE/Ad)^{1/2}\), then the value of \(α\) is

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

Moment of inertia of a rod about its centre is \(\frac{ML^2}{12}\), with \(M = Ad L\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Rotational kinetic energy is \(E = \frac12I\omega^2\). For a thin uniform rod of mass \(M\) and length \(L\) about its centre, \(I = \frac{ML^2}{12}\).

Step 2: Find the mass and I:
\(L = 2\) m, so \(M = \text{volume}\times\text{density} = A\cdot2\cdot d = 2Ad\).
\[ I = \frac{2Ad\times4}{12} = \frac{2Ad}{3} \]

Step 3: Solve for omega:
\(E = \frac12\cdot\frac{2Ad}{3}\omega^2 = \frac{Ad\,\omega^2}{3}\), so \(\omega^2 = \frac{3E}{Ad}\), and \(\omega = \left(\frac{3E}{Ad}\right)^{1/2}\).
Comparing with \(\left(\frac{\alpha E}{Ad}\right)^{1/2}\): \(\alpha = 3\).

Final Answer:
The value of \(\alpha\) is \(3\), option (B). \[ \boxed{3} \]