A molecule consists of two atoms each of mass ' $\mathrm{m}$ ' and separated by a distance ' $d$ '. At room…

A molecule consists of two atoms each of mass ' $\mathrm{m}$ ' and separated by a distance ' $d$ '. At room temperature, if the average rotational kinetic energy is ' $\mathrm{E}$ ' then the angular frequency is
  1. $\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{E}}{\mathrm{m}}}$
  2. $\frac{\mathrm{d}}{2} \sqrt{\frac{\mathrm{m}}{\mathrm{E}}}$
  3. $\sqrt{\frac{\mathrm{Ed}}{\mathrm{m}}}$
  4. $\sqrt{\frac{\mathrm{m}}{\mathrm{Ed}}}$

Solution

$\begin{aligned} & E=\frac{1}{2} I \omega^2 \\ & I=2 m\left(\frac{d}{2}\right)^2=\frac{m^2}{2} \\ & \therefore E=\frac{1}{2} \times \frac{m^2}{2} \cdot \omega^2=\frac{m^2}{4} \cdot \omega^2 \\ & \therefore \omega^2=\frac{4 E}{m d^2} \\ & \therefore \omega=\frac{2}{d} \sqrt{\frac{E}{m}}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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