Two iron solid discs of negligible thickness have radii \( R_1 \) and \( R_2 \) and moment of inertia \( I_1 \) and \( I_2 \), respectively. For \( R_2 = 2R_1 \), the ratio of \( I_1 \) and \( I_2 \) would be \( \frac{1}{x} \), where \( x \) is:
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Correct Answer: 16
Approach Solution - 1
The moment of inertia of a solid disc about its central axis (perpendicular to its plane) is: \[ I = \frac{1}{2}MR^2 \]
For the first disc: \[ I_1 = \frac{1}{2} M_1 R_1^2 \] For the second disc: \[ I_2 = \frac{1}{2} M_2 R_2^2 \]
It is given that: \[ R_2 = 2R_1 \]
Since the discs are made of iron (same material), the mass is proportional to the area (assuming thickness is negligible). So: \[ \frac{M_2}{M_1} = \left( \frac{R_2}{R_1} \right)^2 = \left( \frac{2R_1}{R_1} \right)^2 = 4 \] \[ M_2 = 4M_1 \]
Substituting the values: \[ I_1 = \frac{1}{2} M_1 R_1^2 \] \[ I_2 = \frac{1}{2} \times 4M_1 \times (2R_1)^2 = \frac{1}{2} \times 4M_1 \times 4R_1^2 = 8M_1R_1^2 \]
Ratio \( \frac{I_1}{I_2} \): \[ \frac{I_1}{I_2} = \frac{\frac{1}{2} M_1 R_1^2}{8M_1R_1^2} = \frac{1}{2 \times 8} = \frac{1}{16} \]
Therefore, \[ \boldsymbol{x = 16} \]
Approach Solution -2
We are given two solid iron discs with radii \( R_1 \) and \( R_2 \), and corresponding moments of inertia \( I_1 \) and \( I_2 \). Both have negligible thickness, meaning they can be treated as uniform solid discs. It is given that \( R_2 = 2R_1 \), and we need to find the ratio of \( I_1 \) to \( I_2 \), which is \( \frac{1}{x} \).
Step 2: Recall the moment of inertia of a solid disc.
The moment of inertia of a solid disc about its central axis (perpendicular to the plane of the disc) is given by:
\[ I = \frac{1}{2} M R^2 \] where \( M \) is the mass of the disc and \( R \) is its radius.
Step 3: Express the mass in terms of density and radius.
Since both discs are made of the same material and have negligible thickness, the mass is proportional to the area (since thickness and density are constant):
\[ M \propto R^2 \] Hence, we can write:
\[ I \propto M R^2 \propto R^2 \times R^2 = R^4 \] So, the moment of inertia of a solid disc varies as the fourth power of its radius.
Step 4: Find the ratio of \( I_1 \) and \( I_2 \).
\[ \frac{I_1}{I_2} = \left( \frac{R_1}{R_2} \right)^4 \] Substitute \( R_2 = 2R_1 \):
\[ \frac{I_1}{I_2} = \left( \frac{R_1}{2R_1} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16} \] Hence, the ratio is \( \frac{1}{x} = \frac{1}{16} \).
Step 5: Conclusion.
\[ x = 16 \]
Final Answer:
\[ \boxed{x = 16} \]
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