A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

The correct option is: its speed of rotation decreases

Step-by-Step Explanation:

1. Thermal Expansion of the Rod:
When a uniform metallic rod is heated, it undergoes thermal expansion. The increase in length depends on the initial length, the coefficient of linear expansion, and the rise in temperature. The new length of the rod is given by:
$L\prime =L(1+\alpha \Delta T)$
where $L$ is the initial length, $\alpha$ is the coefficient of linear expansion, and $\Delta T$ is the increase in temperature. Since the temperature increases, the length of the rod increases ($L\prime >L$).

2. Effect on the Moment of Inertia:
The moment of inertia ($I$) of a uniform rod of mass $M$ and length $L$ rotating about its perpendicular bisector (center) is given by:
$I=\frac{1}{12}M{L}^2$
Since the mass $M$ of the rod remains constant during heating, the moment of inertia is directly proportional to the square of its length ($I\propto L^2$). Because the length of the rod increases due to heating, its moment of inertia also increases ($I\prime >I$).

3. Conservation of Angular Momentum:
Since there is no external torque acting on the rotating rod, the total angular momentum ($J$) of the system must be conserved:
$J=I\omega =\text{constant}$
where $\omega$ is the angular speed of rotation.

Therefore, we can write the relationship between the initial and final states as:
$I\omega =I\prime \omega \prime$
Since the moment of inertia increases ($I\prime >I$), the angular speed of rotation must decrease ($\omega \prime <\omega$) to keep the angular momentum constant. Thus, the speed of rotation decreases.