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JEE Main 2021 (26 Aug, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2021 (26 Aug, Shift 2)

Q129JEE Main 2021 · 26 AUG · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m2.

Figure

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Answer: (d)

  1. CD is parallel to the cylinder axis AB at distance L2=0.4 m\frac{L}{2} = 0.4\text{ m}.
  2. I=12Mr2+M(L2)2=M(0.02+0.16)=0.18M=2.7⇒M=15 kgI = \frac{1}{2}Mr^2 + M\left(\frac{L}{2}\right)^2 = M(0.02 + 0.16) = 0.18M = 2.7 \Rightarrow M = 15\text{ kg}
  3. V=πr2L=π(0.04)(0.8)≈0.1005 m3V = \pi r^2 L = \pi(0.04)(0.8) \approx 0.1005\text{ m}^3
  4. ρ=150.1005≈149 kg/m3\rho = \frac{15}{0.1005} \approx 149\text{ kg/m}^3
  5. So the answer is (d) 1.49×102 kg/m31.49 \times 10^2\text{ kg/m}^3.
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