Q104 / 150 · MHT CET 2026 (13 April Shift 2) · Physics
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 the value of $\omega$ in terms of its rotational kinetic energy $E$ is $(\alpha E/Ad)^{1/2}$, then the value of $\alpha$ is
- A$2$
- Correct answerB$3$
- C$4$
- D$5$
Solution
Mass of the rod, $M = \text{Volume} \times \text{density} = A \times L \times d$
Given $L = 2$, $M = 2Ad$
Moment of inertia of the rod about an axis passing through its centre and perpendicular to its length is $I = \dfrac{ML^2}{12}$
Substituting the values of $M$ and $L$, we get $I = \dfrac{2Ad \times 2^2}{12} = \dfrac{2Ad}{3}$
Rotational kinetic energy is given by $E = \dfrac{1}{2}I\omega^2$
Substituting the value of $I$, $E = \dfrac{1}{2} \left( \dfrac{2Ad}{3} \right) \omega^2 = \dfrac{Ad}{3} \omega^2$
Rearranging for $\omega$, we get $\omega = \left( \dfrac{3E}{Ad} \right)^{1/2}$
Comparing this with the given expression $\omega = \left( \dfrac{\alpha E}{Ad} \right)^{1/2}$, we find $\alpha = 3$.
Answer: $3$
Given $L = 2$, $M = 2Ad$
Moment of inertia of the rod about an axis passing through its centre and perpendicular to its length is $I = \dfrac{ML^2}{12}$
Substituting the values of $M$ and $L$, we get $I = \dfrac{2Ad \times 2^2}{12} = \dfrac{2Ad}{3}$
Rotational kinetic energy is given by $E = \dfrac{1}{2}I\omega^2$
Substituting the value of $I$, $E = \dfrac{1}{2} \left( \dfrac{2Ad}{3} \right) \omega^2 = \dfrac{Ad}{3} \omega^2$
Rearranging for $\omega$, we get $\omega = \left( \dfrac{3E}{Ad} \right)^{1/2}$
Comparing this with the given expression $\omega = \left( \dfrac{\alpha E}{Ad} \right)^{1/2}$, we find $\alpha = 3$.
Answer: $3$
Hence, the correct answer is option: 'B'.