Question

Two rings of radius 'R' and 'nR' made of same material have the ratio of moment of inertia about an axis passing through its centre and perpendicular to the plane is $1 : 8$. The value of 'n' is (mass per unit length = $\lambda$)

• A. 2
• B. 4
• C. 1
• D. 3

Hint:

For objects where mass scales with size (like wires, rings, or chains), the mass itself introduces an extra power of length ($M \propto R$). Combined with the standard $R^2$ term in the moment of inertia, the scaling relationship always becomes $I \propto R^3$. Recognizing this cubic relationship lets you solve for $n = \sqrt[3]{8} = 2$ mentally!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two separate circular rings manufactured from the same material (meaning they share a constant linear mass density $\lambda$). The first ring has a radius $R_1 = R$, and the second ring has a larger radius $R_2 = nR$. The ratio of their moments of inertia about their respective central perpendicular axes is $1 : 8$. We need to calculate the scaling value $n$.

Step 2: Key Formula or Approach:
The moment of inertia of a thin circular ring about its central perpendicular axis is: $$I = M R^2$$ Since the mass of a ring is equal to its circumference multiplied by its linear mass density ($M = \lambda \cdot 2\pi R$), we can express the moment of inertia solely in terms of its radius: $$I = (\lambda \cdot 2\pi R) \cdot R^2 = 2\pi\lambda R^3$$ This shows that the moment of inertia is directly proportional to the cube of the radius: $I \propto R^3$.

Step 3: Detailed Explanation:
Let's write down the ratio of the moments of inertia for the two rings using our proportional relationship: $$\frac{I_1}{I_2} = \left( \frac{R_1}{R_2} \right)^3$$ Substitute the given values into this equation ($\frac{I_1}{I_2} = \frac{1}{8}$, $R_1 = R$, and $R_2 = nR$): $$\frac{1}{8} = \left( \frac{R}{nR} \right)^3$$ The radius variable $R$ cancels out completely: $$\frac{1}{8} = \left( \frac{1}{n} \right)^3 = \frac{1}{n^3}$$ Cross-multiply to isolate $n^3$: $$n^3 = 8$$ Take the cube root of both sides to find $n$: $$n = \sqrt[3]{8} = 2$$ This matches option (A).

Step 4: Final Answer:
The scaling value $n$ is 2, which corresponds to option (A).