Verified Solution • 2025–26 Session moderate

QJEEM23Ph6d7a: JEE Main 2023 Physics - Class 11 JEEM23Ph6d7a Solution: A thin uniform rod of length , cross sectional area ' ' and…

A thin uniform rod of length 2 m2 \mathrm{~m} , cross sectional area ' AA ' and density ' d\mathrm{d} ' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity ω\omega . If value of ω\omega in terms of its rotational kinetic energy EE is αEAd\sqrt{\frac{\alpha E}{A d}} then value of α\alpha is ______________.

Step-by-Step Verified Solution

Kinetic energy of rod E=12ml212ω2E = {1 \over 2}{{m{l^2}} \over {12}}{\omega ^2}

or ω=24Eml2=24Ed×A×l3\omega = \sqrt {{{24E} \over {m{l^2}}}} = \sqrt {{{24E} \over {d \times A \times {l^3}}}}

⇒ω=24EdA23\Rightarrow \omega = \sqrt {{{24E} \over {dA{2^3}}}}

=3EAd= \sqrt {{{3E} \over {Ad}}}

So, α=3\alpha = 3

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