A solid sphere of lithium is rotating with angular frequency $\omega $ about an axis passing through its diameter. If its temperature is raised by ${50^\circ }C$, then what will be its new frequency? (${\alpha _{lithium}} = 6 \times {10^{ - 5}}{C^{ - 1}}$).
A. $0.99\omega $
B. $0.73\omega $
C. $0.83\omega $
D. $0.94\omega $

Hint
When the temperature of a metallic object is changed, its dimensions are bound to change due to thermal activity of molecules. This change in shape and size also affects the motion of the object.
$\Rightarrow {R_t} = {R_0}(1 + \alpha (\Delta T))$, where R denotes the length of the object in consideration, $\alpha $ is the coefficient of linear expansion, and $\Delta T$ is the change in temperature. The subscripts t and 0, denote the points in time when change in temperature occurs.

Complete step by step answer
In this question, we are provided with a sphere of lithium rotating with some angular frequency, and are asked to find the new angular frequency when the temperature of the sphere changes. The data given to us includes the following:
Initial temperature of lithium ${T_i} = 0^\circ C$
Final temperature of lithium ${T_f} = 50^\circ C$
Coefficient of thermal expansion for lithium $\alpha = 6 \times {10^{ - 5}}{C^{ - 1}}$
Hence, the change in temperature will be:
$\Rightarrow \Delta T = {T_f} - {T_i} = 50 - 0 = 50^\circ C$
We know that the change in the size of an object is dependent on the coefficient of thermal expansion as:
$\Rightarrow {R_t} = {R_0}(1 + \alpha (\Delta T))$
Squaring both sides will give us:
$\Rightarrow R_t^2 = R_0^2{(1 + \alpha (\Delta T))^2} = R_0^2(1 + 2\alpha (\Delta T) + {\alpha ^2}{(\Delta T)^2})$
Neglecting the higher terms:
$\Rightarrow R_t^2 = R_0^2(1 + 2\alpha \Delta T)$
Since we are provided with a solid sphere, R will be the radius of the sphere for us. Putting the known values in this equation, we get:
$\Rightarrow R_t^2 = R_0^2(1 + 2 \times 6 \times {10^{ - 5}} \times 50)$
$\Rightarrow R_t^2 = R_0^2(1 + 0.006) = R_0^2(1.006)$ [Eq. 1]
We also know that the law of conservation of angular momentum is given as:
$\Rightarrow {I_t}{\omega _t} = {I_0}{\omega _0}$ where I is the moment of inertia which is given for an axis passing through the centre of a solid sphere as:
$\Rightarrow I = \dfrac{2}{5}M{R^2}$
Based on the above equation, the conservation of angular momentum for our case would be given as:
$\Rightarrow \dfrac{2}{5}MR_t^2{\omega _t} = \dfrac{2}{5}MR_0^2{\omega _0}$
$\Rightarrow R_t^2{\omega _t} = R_0^2{\omega _0}$
Putting the value from Eq. 1 gives us the new angular frequency as:
$\Rightarrow {\omega _t} = \dfrac{{R_0^2{\omega _0}}}{{R_t^2}} = \dfrac{{R_0^2\omega }}{{R_0^2(1.006)}}$
$\Rightarrow {\omega _t} = \dfrac{\omega }{{(1.006)}} = 0.99\omega $
Hence, the correct answer is option (A).

Note
In the given question, we already knew the thermal expansion coefficient. We know it depends on the temperature change, and the change in size of the object, but is independent of mass. This is why in the law of conservation of angular momentum, we only needed to take care of the radius of the sphere and not the mass, because it remains constant.