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JEE Main 2024 (31 Jan, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2024 (31 Jan, Shift 2)

Q74JEE Main 2024 · 31 JAN · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

Two identical spheres each of mass 2 kg2\text{ kg} and radius 50 cm50\text{ cm} are fixed at the ends of a light rod so that the separation between the centers is 150 cm150\text{ cm}. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is x20 kg m2\frac{x}{20}\text{ kg m}^2, where the value of xx is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 53

  1. Each sphere: I=25mr2+md2I = \frac{2}{5}mr^2 + md^2 with r=0.5 mr = 0.5\text{ m}, d=0.75 md = 0.75\text{ m} from the middle.
  2. I1=25(2)(0.25)+2(0.5625)=0.2+1.125=1.325 kg m2I_1 = \frac{2}{5}(2)(0.25) + 2(0.5625) = 0.2 + 1.125 = 1.325\text{ kg m}^2
  3. I=2I1=2.65 kg m2=5320 kg m2I = 2I_1 = 2.65\text{ kg m}^2 = \frac{53}{20}\text{ kg m}^2
  4. So x=53x = 53.
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