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Q6.17: NCERT Physics - Part 1 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.17 Solution: The oxygen molecule has a mass of 5.30 _ 10___ kg and a moment…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently reverse the statement “rotational kinetic energy is two thirds of translational kinetic energy” and write $K_t=\frac{2}{3}K_r$ instead of the correct $K_r=\frac{2}{3}K_t$, producing an angular velocity error of a factor $\sqrt{3/2}$. Always translate the wording into an equation before substituting values.

The oxygen molecule has a mass of 5.30 _ 10___ kg and a moment of inertia of 1.94 _ 10___ kg m_ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Step-by-Step Verified Solution

**Answer:** ω=6.79×1012\omega = 6.79 \times 10^{12} rad/s. **Step-by-Step Explanation:** Translational KE: Kt=12mv2=12(5.30×10−26)(500)2=6.625×10−21 JK_t = \frac{1}{2}mv^2 = \frac{1}{2}(5.30 \times 10^{-26})(500)^2 = 6.625 \times 10^{-21} \text{ J}. Rotational KE: Kr=23Kt=4.417×10−21 JK_r = \frac{2}{3}K_t = 4.417 \times 10^{-21} \text{ J}. Using Kr=12Iω2K_r = \frac{1}{2}I\omega^2:
ω=2KrI=2×4.417×10−211.94×10−46=6.79×1012 rad/s\omega = \sqrt{\frac{2K_r}{I}} = \sqrt{\frac{2 \times 4.417 \times 10^{-21}}{1.94 \times 10^{-46}}} = 6.79 \times 10^{12} \text{ rad/s}

All Questions in this Chapter

Q6.1 Q6.2 Q6.3 Q6.4 Q6.5 Q6.6 Q6.7 Q6.8 Q6.9 Q6.10 Q6.11 Q6.12 Q6.13 Q6.14 Q6.15 Q6.16 Q6.17

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