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Step-by-Step Solution
Step 1: Understand the Problem
We have a hollow cylinder of inner radius $r = 10 \text{ cm}$ and outer radius $R = 20 \text{ cm}$. Its moment of inertia about its axis is $I$. We want a thin (single radius) cylinder of the same mass $m$ whose moment of inertia about the same axis is also $I$. We need to find that single radius $r_0$.
Step 2: Express the Moment of Inertia for the Hollow Cylinder
Consider a ring element at a distance $x$ from the axis, with thickness $dx$. The mass per unit area is
$\sigma = \frac{m}{\pi\,(R^2 - r^2)}.$
The mass of this ring element is
$dm = \sigma \cdot 2\pi x\, dx.$
The moment of inertia $dI$ of this ring element about the axis is
$dI = x^2\,dm = x^2 \left( \sigma \cdot 2\pi x\, dx \right).$
Step 3: Integrate to Find Total Moment of Inertia
Summing (integrating) over the entire cross section from $x = r$ to $x = R$, we get
$I = \int dI = \sigma \cdot 2\pi \int_{r}^{R} x^3\, dx.$
Performing the integral,
$\int_{r}^{R} x^3\, dx = \left[\frac{x^4}{4}\right]_{r}^{R} = \frac{R^4 - r^4}{4}.$
So,
$I = \sigma \cdot 2\pi \cdot \frac{R^4 - r^4}{4}.$
Substituting $\sigma = \frac{m}{\pi\,(R^2 - r^2)}$, we get
$I = \frac{m}{\pi\,(R^2 - r^2)} \cdot 2\pi \cdot \frac{R^4 - r^4}{4}
= \frac{m}{2} \cdot \frac{R^4 - r^4}{R^2 - r^2}.$
Using the identity $R^4 - r^4 = (R^2 - r^2)(R^2 + r^2)$, it simplifies to
$I = \frac{m}{2} \left( R^2 + r^2 \right). \quad (1)
Step 4: Express the Moment of Inertia for the Thin Cylinder
For a thin cylinder of radius $r_0$ with mass $m$, its moment of inertia about its axis is
$I = m\,r_0^2. \quad (2)
Step 5: Equate the Two Moments of Inertia
Since both cylinders have the same moment of inertia $I$, we set equation (1) equal to equation (2):
$m\,r_0^2 = \frac{m}{2} \left( R^2 + r^2 \right).
The mass $m$ cancels from both sides:
$r_0^2 = \frac{1}{2}\left( R^2 + r^2 \right).
Substitute $R = 20 \text{ cm}$ and $r = 10 \text{ cm}$:
$r_0^2 = \frac{1}{2}\left( (20)^2 + (10)^2 \right) = \frac{1}{2}\bigl(400 + 100\bigr) = \frac{500}{2} = 250.
Step 6: Solve for the Radius of the Thin Cylinder
Taking the square root gives
$r_0 = \sqrt{250} = 15.811\ldots \approx 16 \text{ cm}.
Thus, the radius of the thin cylinder is $16 \text{ cm}$.
Step 7: Final Answer
Therefore, the correct answer is 16 cm.
