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JEE Main 2026 (21 Jan, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2026 (21 Jan, Shift 1)

Q14JEE Main 2026 · 21 JAN · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

Two identical thin rods of mass M kgM\text{ kg} and length L mL\text{ m} are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point PP and perpendicular to the plane of the rods is x12ML2 kg m2\frac{x}{12}ML^2\text{ kg m}^2. The value of xx is ______.

Figure

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

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Answer: 17

  1. Vertical rod, axis through its end P: ML23\frac{ML^2}{3}.
  2. Horizontal rod's centre is LL below P: ML212+ML2\frac{ML^2}{12} + ML^2.
  3. I=4+1+1212ML2=1712ML2I = \frac{4+1+12}{12}ML^2 = \frac{17}{12}ML^2, so x=17x = 17.
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