$\text{A conducting solid sphere of radius}R\text{and mass}M\text{carries a charge}Q.\text{The sphere is rotating about an axis passing}$

$\text{through its center with a uniform angular speed}\mathrm{ \omega }.\text{The ratio of the magnitudes of the magnetic dipole moment}$

$\text{to the angular momentum about the same axis is given as}\alpha \frac{Q}{2M}.\text{The value of}\alpha \text{is \_\_\_\_\_}$

The correct answer is 1.67.

To find the value of $\alpha$, we need to calculate the magnetic dipole moment and the angular momentum of the rotating sphere and then find their ratio.

Step 1: Calculate the Magnetic Dipole Moment ($\mu$)
For a conducting solid sphere, any net charge resides entirely on its outer surface. Therefore, the charge $Q$ is distributed uniformly on the surface of the sphere of radius $R$, effectively forming a spherical shell of charge.
When the sphere rotates about an axis passing through its center with an angular speed $\omega$, this rotating surface charge creates a magnetic dipole moment.

The magnetic dipole moment of a uniformly charged thin spherical shell of radius $R$ and total charge $Q$ rotating with angular speed $\omega$ is given by:
$\mu =\frac{1}{3}Q{R}^2\omega$

Step 2: Calculate the Angular Momentum ($L$)
The mass $M$ is distributed uniformly throughout the volume of the solid sphere. The moment of inertia of a uniform solid sphere of mass $M$ and radius $R$ about an axis passing through its center is:
$I=\frac{2}{5}M{R}^2$

The angular momentum $L$ of the rotating solid sphere is:
$L=I\omega =\frac{2}{5}M{R}^2\omega$

Step 3: Find the Ratio of Magnetic Dipole Moment to Angular Momentum
Taking the ratio of the magnitude of the magnetic dipole moment $\mu$ to the angular momentum $L$:
$\frac{\mu }{L}=\frac{\frac{1}{3}Q{R}^2\omega }{\frac{2}{5}M{R}^2\omega }$

Simplifying the fraction:
$\frac{\mu }{L}=\frac{5}{6}\frac{Q}{M}$

We can rewrite this expression to match the given form $\alpha \frac{Q}{2M}$:
$\frac{\mu }{L}=\frac{5}{3}\left(\frac{Q}{2M}\right)$

By comparing the two expressions, we find:
$\alpha =\frac{5}{3}\approx 1.67$