JEE Mains 2025 Moment of Inertia of Ring – 8 April Shift 2 PYQ Solved

This post explains a key concept from the JEE Mains 2025 Physics paper held on 8 April Shift 2. The question is based on Rotational Motion and involves calculating the moment of inertia of a ring formed by bending a rod. This is a must-practice question for students preparing for JEE and other competitive exams.

📘 Question:

A rod of linear mass density λ and length L is bent to form a ring of radius R. What is the moment of inertia of the ring about any of its diameters?

🧮 Detailed Step-by-Step Solution:

Step 1: Mass of the rod

The linear mass density λ means mass per unit length. So the total mass of the rod is:

M = λ × L

Step 2: Finding the radius of the ring

When the rod is bent into a ring, the length of the rod becomes the circumference of the ring:

2πR = L ⇒ R = L / 2π

Step 3: Moment of inertia of a ring about a diameter

The standard formula for the moment of inertia of a uniform ring about its diameter is:

I = (1/2) M R²

Step 4: Substituting the values

Now, substitute M = λL and R = L / 2π into the formula:

I = (1/2) × λL × (L / 2π)²
  = (1/2) × λL × (L² / 4π²)
  = λL³ / 8π²

✅ Final Answer:

Option (2): λL³ / 8π²

🎥 Watch Full Video Solution:

📌 Why This Question is Important

It combines formula manipulation with physical understanding.
Uses two important JEE topics: linear density and rotational motion.
Highlights the correct application of the moment of inertia formula.

🔍 Keywords:

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