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 \omega . If value of \omega in terms of its rotational kinetic energy E is \sqrt\frac\alpha EA d then value of \alpha is \underline\hspace2cm .

Rotational Motion PYQ — JEE Main 2023 · 30 Jan · Shift 1 · Q63 with Solution

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Correct answer: 3

We need to express $\omega$ in the form

\omega = \sqrt{\frac{\alpha E}{Ad}}

and find $\alpha$ .

Mass $m$ is

m = (\text{density})\times(\text{volume}) = d \times A \times L

Since $L=2$ ,

m = d \cdot A \cdot 2 = 2Ad

For a thin uniform rod about an axis through its centre and perpendicular to its length,

I = \frac{1}{12}mL^2

Substitute $L=2$ :

I = \frac{1}{12}m(2)^2 = \frac{1}{12}\cdot 4m = \frac{m}{3}

Now substitute $m=2Ad$ :

I = \frac{2Ad}{3}

Rotational kinetic energy is

E = \frac{1}{2}I\omega^2

Substitute $I = \frac{2Ad}{3}$ :

E = \frac{1}{2}\cdot \frac{2Ad}{3}\,\omega^2 = \frac{Ad}{3}\omega^2

\omega^2 = \frac{3E}{Ad}

So,

\omega = \sqrt{\frac{3E}{Ad}}

Comparing with

\omega = \sqrt{\frac{\alpha E}{Ad}}

we get

\alpha = 3

\boxed{3}

The derived answer matches the stored correct answer.