A rigid spherical body is spinning around an axis without any external torque. Due to a change in temperature, the volume increases by $1\% $. Its angular speed will be
A) Increases approximately $1\% $
B) Decreases approximately by $1\% $
C) Decreases approximately by $0.67\% $
D) Decreases approximately by $0.33\% $

Hint:
The angular momentum of a rigid spherical body is defined as the product of the moment of inertia and its angular velocity. Angular momentum has both magnitude and direction, so it is a vector quantity. The angular momentum of a rigid body is the same at every point on the sphere.

Complete step by step answer:
Step I:
According to the Principle of conservation of angular momentum, if no external torque is acting on the body, then the angular momentum of the body remains constant. Angular momentum is given by
$L = I.\omega $ ---(i)
$L$ is the angular momentum
$I$ is the moment of inertia
$\omega $ is the angular speed
Step II:
Volume of a sphere is given by $V = \dfrac{4}{3}\pi {R^3}$
This shows that if the radius is increased three times, then there will be change in the volume.
$\dfrac{{\Delta V}}{V}\% = \dfrac{{3\Delta R}}{R}\% $ ---(ii)
Step III:
Given that the volume increases by $1\% $ therefore, equation (i) can be written as
$\Rightarrow \dfrac{{3\Delta R}}{R} \times 100 = 1$
$\Rightarrow \dfrac{{\Delta R}}{R} \times 100 = \dfrac{1}{3}$
$\Rightarrow \dfrac{{\Delta R}}{R} \times 100 = 0.333\% $ ---(iii)
Step IV:
It is known that the moment of inertia is the product of mass and the square of the radius of the body. The moment of inertia varies directly with the radius.
$I = M{R^2}$
I is the moment of inertia
M is the mass
R is the radius
Step V:
The change in inertia of the body is given by
$\Rightarrow\dfrac{{\Delta I}}{I} \times 100 = \dfrac{{2\Delta R}}{R} \times 100$
Substitute the value from equation (iii) to (iv),
$\Rightarrow \dfrac{{\Delta I}}{I} \times 100 = 2 \times 0.333$ ---(iv)
$\Rightarrow \dfrac{{\Delta I}}{I} \times 100 = 0.67\% $
The change in inertia is increasing by $0.67\% $
Step VI:
From equation (i) it is clear that ‘w’ is the angular speed.
$W=\dfrac{L}{I}$
‘w’ varies inversely with the value of inertia. If the moment of inertia is increasing by $0.67\% $, then the angular speed will decrease by $0.67\% $.

Therefore option C is the correct answer.

Note:
It is to be noted that the angular momentum is the same at every point on the sphere, but when moved towards the center then the angular speed will increase. The direction of angular momentum can be known by applying the Right-hand thumb rule. When wrapped a right hand around the axis of rotation, the thumb points in the direction of angular velocity. Whereas fingers point in the direction of the axis of rotation.