1. Given data

• Mass of sphere: $m = 500\,\text{g} = 0.5\,\text{kg}$
• Radius: $R = 5\,\text{cm} = 0.05\,\text{m}$
• Angular speed about diameter: $\omega = 10\,\text{rad s}^{-1}$

We need to use:

• Moment of inertia about a tangent
• Angular momentum about the diameter

The statement says:

$I_{\text{tangent}} = x \times 10^{-2} \times L_{\text{diameter}}$

where $L_{\text{diameter}}$ is the angular momentum about the diameter.

2. Moment of inertia about the diameter

For a solid sphere about any diameter,

$I_{\text{diameter}} = \frac{2}{5}mR^2$

So,

$I_{\text{diameter}} = \frac{2}{5}(0.5)(0.05)^2$

$= \frac{2}{5}(0.5)(0.0025)$

$= 0.0005\,\text{kg m}^2$

3. Angular momentum about the diameter

$L_{\text{diameter}} = I_{\text{diameter}}\,\omega$

$L_{\text{diameter}} = 0.0005 \times 10 = 0.005\,\text{kg m}^2\text{/s}$

4. Moment of inertia about a tangent

Using parallel axis theorem,

$I_{\text{tangent}} = I_{\text{diameter}} + mR^2$

$= \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$

Now,

$I_{\text{tangent}} = \frac{7}{5}(0.5)(0.05)^2$

$= \frac{7}{5}(0.5)(0.0025)$

$= 0.00175\,\text{kg m}^2$

5. Relate with given expression

Given,

$I_{\text{tangent}} = x \times 10^{-2} \times L_{\text{diameter}}$

Substitute values:

$0.00175 = x \times 10^{-2} \times 0.005$

$0.00175 = x \times 0.00005$

$x = \frac{0.00175}{0.00005} = 35$

6. Final answer

$\boxed{x = 35}$

The derived answer matches the stored correct answer.