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The moment of inertia of a thin uniform rod of mass 'M' and length 'L' about an axis passing through a point at a distance L by 4 from one of its ends and perpendicular to the length of the rod is
(A) M L squared by 48
(B) 7 M L squared by 48
(C) 5 M L squared by 48
(D) 9 M L squared by 48

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🚀 Step-by-Step Solution to Solve This Question 🚀
📌 Chapter: Rotational Dynamics
📌 Topic: Moment of Inertia of a Rod using Parallel Axis Theorem

🔹 Step 1: Recall the standard moment of inertia
Moment of inertia of a uniform rod about its center is one twelfth M L squared.

🔹 Step 2: Use parallel axis theorem
Since the axis is at a distance L by 4 from one end, first find the distance between the center of mass and the axis:
The center of the rod is at L by 2 from the end, and the axis is at L by 4 from the same end.
So, distance d = L by 2 minus L by 4 = L by 4

🔹 Step 3: Apply parallel axis theorem
I = I_center + M d squared
I = one twelfth M L squared + M × (L by 4) squared
I = M L squared by 12 + M L squared by 16
Take L squared common:
I = M L squared × (1 by 12 + 1 by 16)
L.C.M. of 12 and 16 is 48
= M L squared × (4 by 48 + 3 by 48) = M L squared × 7 by 48

✅ Final Answer: (B) 7 M L squared by 48 🚀
🔥 Pro Tip: Always break down the position of the axis relative to the center and use the parallel axis theorem precisely. Geometry matters!


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