Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A uniform cylinder of steel of mass M, radius R is placed on frictionless bearings and set to rotate about its vertical axis with angular velocity\[{{\omega }_{0}}\]. After the cylinder has reached the specified state of rotation. It is heated without any mechanical contact from temperature\[{{T}_{0}}\]to\[{{T}_{0}}+\Delta T\]. If\[\frac{\Delta I}{I}\]is the fractional change in moment of inertia of the cylinder and \[\frac{\Delta \omega }{{{\omega }_{0}}}\]be the fractional change in the angular velocity of the cylinder and a be the coefficient of linear expansion, then (1) \[\frac{\Delta I}{I}=\frac{2\Delta R}{R}\] (2) \[\frac{\Delta I}{I}=\frac{\Delta \omega }{{{\omega }_{0}}}\] (3) \[\frac{\Delta \omega }{{{\omega }_{0}}}=-2\alpha \Delta T\] (4) \[\frac{\Delta I}{I}=\frac{2\Delta R}{R}\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: A
Solution :
Since no mechanical contact is there, so angular momentum is conserved. \[I\omega =\]constant \[\Delta (I\omega )=0\] \[I\Delta \omega +\omega \Delta I=0\] \[\frac{\Delta \omega }{\omega }=-\frac{\Delta I}{I}\] \[\frac{\Delta \omega }{\omega }=-\frac{\Delta I}{I}=-2\alpha \Delta T\]\[\left\{ \because \frac{\Delta I}{I}=\frac{2\Delta R}{R}=2\alpha \Delta T \right\}\]
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