Question

A wire of length \(L\) and mass \(100\,\text{g}\) is bent in the form of a circular ring. If the moment of inertia of the ring about its diameter is \[ 98\times10^{-5}\,\text{kg m}^2, \] then the value of \(L\) is

• A. \(176\,\text{cm}\)
• B. \(88\,\text{cm}\)
• C. \(44\,\text{cm}\)
• D. \(22\,\text{cm}\)

Hint:

Remember the standard moments of inertia of a thin ring: \[ \boxed{ I_{\text{diameter}}=\frac12MR^2, \qquad I_{\text{centre, perpendicular}}=MR^2. } \] The circumference of the ring is \[ \boxed{L=2\pi R.} \]

Answer 1

The Correct Option is B

Step 1: Use the moment of inertia of a ring about its diameter. For a thin circular ring, \[ I=\frac12MR^2. \] Given, \[ I=98\times10^{-5}\,\text{kg m}^2, \] and \[ M=100\,\text{g}=0.1\,\text{kg}. \] Hence, \[ 98\times10^{-5} = \frac12(0.1)R^2. \]

Step 2: Find the radius. Thus, \[ R^2 = \frac{98\times10^{-5}}{0.05} = 196\times10^{-4}, \] \[ R = 14\times10^{-2} = 0.14\,\text{m}. \]

Step 3: Calculate the length of the wire. Since the wire forms a complete circle, \[ L=2\pi R. \] Taking \[ \pi=\frac{22}{7}, \] \[ L = 2\times\frac{22}{7}\times0.14 = 0.88\,\text{m} = 88\,\text{cm}. \] Hence, \[ \boxed{L=88\,\text{cm}.} \] Therefore, the correct option is \(\boxed{(B)}\).