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NEET 2000 (AIPMT) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from NEET 2000 (AIPMT)

Q200NEET 2000 · AIPMTEasy

Moment of Inertia & Radius of Gyration

For the adjoining diagram, the correct relation between I1, I2, and I3 is, (I-moment of inertia)

Figure

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Answer: (b)

  1. For a triangular lamina about one side, I=Mh26I = \frac{Mh^2}{6}, where hh is the perpendicular distance of the opposite vertex from that side.
  2. About ABAB (I1I_1): h=BC=3⇒I1=9M6=1.5Mh = BC = 3 \Rightarrow I_1 = \frac{9M}{6} = 1.5M
  3. About ACAC (I2I_2): h=3×45=2.4⇒I2=5.76M6=0.96Mh = \frac{3 \times 4}{5} = 2.4 \Rightarrow I_2 = \frac{5.76M}{6} = 0.96M
  4. About BCBC (I3I_3): h=AB=4⇒I3=16M6≈2.67Mh = AB = 4 \Rightarrow I_3 = \frac{16M}{6} \approx 2.67M
  5. So I3>I1>I2I_3 > I_1 > I_2: options (a), (c) and (d) hold, and (b) I2>I1I_2 > I_1 does not.
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