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NEET PHYSICSRotational motionMedium

Question

Four identical thin rods each of mass MM and length tt, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is:

A

43Mt2\frac{4}{3}Mt^2

B

23Mt2\frac{2}{3}Mt^2

C

133Mt2\frac{13}{3}Mt^2

D

13Mt2\frac{1}{3}Mt^2

Step-by-Step Solution

Let us consider one of the four rods. The moment of inertia of a thin rod of mass MM and length tt about an axis passing through its own centre of mass and perpendicular to its length is ICM=Mt212I_{CM} = \frac{Mt^2}{12}. The perpendicular distance from the centre of the square to the centre of each rod is d=t2d = \frac{t}{2}. According to the parallel axis theorem, the moment of inertia of one rod about the axis passing through the centre of the square and perpendicular to its plane is: I1=ICM+Md2=Mt212+M(t2)2=Mt212+Mt24=4Mt212=Mt23I_1 = I_{CM} + Md^2 = \frac{Mt^2}{12} + M\left(\frac{t}{2}\right)^2 = \frac{Mt^2}{12} + \frac{Mt^2}{4} = \frac{4Mt^2}{12} = \frac{Mt^2}{3} Since the square frame consists of 4 identical rods, the total moment of inertia of the frame is: Itotal=4×I1=4×Mt23=43Mt2I_{total} = 4 \times I_1 = 4 \times \frac{Mt^2}{3} = \frac{4}{3}Mt^2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionidenticallengthsquaremomentinertia

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