System of particles and Rotational motion NCERT Solutions Class 11 Physics - Solved Exercise Question 7.20

Question 7.20:
The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94×10–46 kg m2 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.
Solution:
Mass of an oxygen molecule, m = 5.30 × 10–26 kg
Moment of inertia, I = 1.94 × 10–46 kg m2
Velocity of the oxygen molecule, v = 500 m/s
The separation between the two atoms of the oxygen molecule = 2r
Mass of each oxygen atom = m/2
Hence, moment of inertia I, is calculated as:
(m/2)r2 + (m/2)r2 = mr2
r = ( I / m)1/2
(1.94 X 10-46 / 5.36 X 10-26 )1/2  =  0.60 X 10-10 m
It is given that:
KErot = (2/3)KEtrans
(1/2) I ω2 = (2/3) X (1/2) X mv2
mr2ω2 = (2/3)mv2
ω = (2/3)1/2 (v/r)
= (2/3)1/2 (500 / 0.6 X 10-10) = 6.80 X 1012 rad/s

3 comments:

  1. The given solution is wrong.check the line where you are calculating the moment of inertia.

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    Replies
    1. You should write mr^2 as I. To avoid confusion.

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    2. You should write mr^2 as I. To avoid confusion.

      Delete