Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Moment of Inertia & Radius of Gyration›Question 127

JEE Main 2021 (31 Aug, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2021 (31 Aug, Shift 2)

Q127JEE Main 2021 · 31 AUG · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the ends of a light rod. The distance between the centres of the two spheres is 5 m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint?

Show answer and solution

Answer: (c)

  1. Each sphere: 25mr2=25(1.5)(0.5)2=0.15\frac{2}{5}mr^2 = \frac{2}{5}(1.5)(0.5)^2 = 0.15 kg m2^2.
  2. Each centre is 2.52.5 m from the axis: md2=1.5×6.25=9.375md^2 = 1.5\times6.25 = 9.375 kg m2^2.
  3. I=2(0.15+9.375)=19.05I = 2(0.15 + 9.375) = 19.05 kg m2^2, so the answer is (c).
Report a mistake

More questions on Moment of Inertia & Radius of Gyration