A solid sphere of mass 500 g radius 5 cm is rotated about one of its diameter with angular speed of 10 rad s$$^{-1}$$. If the moment of inertia of the sphere about its tangent is $$x \times 10^{-2}$$ times its angular momentum about the diameter. Then the value of $$x$$ will be _______

We have a solid sphere of mass $$M = 500$$ g $$= 0.5$$ kg, radius $$R = 5$$ cm $$= 0.05$$ m, spinning with angular speed $$\omega = 10$$ rad/s.

The moment of inertia about a diameter is:

$$I_{diameter} = \frac{2}{5}MR^2 = \frac{2}{5}(0.5)(0.05)^2 = 5 \times 10^{-4} \text{ kg m}^2$$

So the angular momentum about the diameter is:

$$L = I_{diameter} \times \omega = 5 \times 10^{-4} \times 10 = 5 \times 10^{-3} \text{ kg m}^2\text{/s}$$

Now, using the parallel axis theorem, the moment of inertia about a tangent parallel to the diameter is:

$$I_{tangent} = I_{diameter} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$

$$I_{tangent} = \frac{7}{5}(0.5)(0.0025) = 1.75 \times 10^{-3} \text{ kg m}^2$$

We are given that $$I_{tangent} = x \times 10^{-2} \times L$$. Substituting:

$$1.75 \times 10^{-3} = x \times 10^{-2} \times 5 \times 10^{-3}$$

$$1.75 \times 10^{-3} = x \times 5 \times 10^{-5}$$

$$x = \frac{1.75 \times 10^{-3}}{5 \times 10^{-5}} = 35$$

So, the answer is $$35$$.