MHT CET · Physics · Rotational Motion
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
- A \(\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{E}}{\mathrm{m}}}\)
- B \(\frac{\mathrm{d}}{2} \sqrt{\frac{\mathrm{m}}{\mathrm{E}}}\)
- C \(\sqrt{\frac{\mathrm{Ed}}{\mathrm{m}}}\)
- D \(\sqrt{\frac{\mathrm{m}}{\mathrm{Ed}}}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{E}}{\mathrm{m}}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & E=\frac{1}{2} I \omega^2 \\ & I=2 m\left(\frac{d}{2}\right)^2=\frac{m^2}{2} \\ & \therefore E=\frac{1}{2} \times \frac{m^2}{2} \cdot \omega^2=\frac{m^2}{4} \cdot \omega^2 \\ & \therefore \omega^2=\frac{4 E}{m d^2} \\ & \therefore \omega=\frac{2}{d} \sqrt{\frac{E}{m}}\end{aligned}\)
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