JEE Main 2020 · 4 September · evening · Single correct · +4 / −1

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :

JEE Main PYQ figure
  1. (a) 1/2
  2. (b) 1/4
  3. (c) 1/8
  4. (d) 2/3
Answer and solution

Answer (b) 1/4

Solution

Physics sir · 3 steps

Shift the axis from the centre to the corner: the distance you shift by is half the diagonal.

  1. About the centre O. For a rectangular sheet of sides a and b, the moment of inertia about the perpendicular axis through the centre is I_O = M(a²+b²)/12. Here a = 0.6 m and b = 0.8 m, so a² + b² = 0.36 + 0.64 = 1 m² and I_O = M/12.
  2. Distance to corner. The corner O' is half a diagonal away from O. The full diagonal is √(a²+b²) = 1 m, so d = 0.5 m and d² = (a²+b²)/4 = 0.25 m².

    I_(O') = I_O + Md² = M/12 + M/4 = M/3

  3. So. I_O/I_(O') = (M/12)/(M/3) = 1/4, which is option (b). The sizes cancel, so this ratio is 1/4 for every rectangular sheet.

The idea to keep. The parallel axis theorem works only from the centre-of-mass axis. For any corner of a rectangle, d² = (a²+b²)/4, so I_(corner) = 4I_(centre).

Answer from the NTA final answer key. Solution by PrepWiser.

Study Rotational Motion

Asked about 180 times in JEE Main since 2020. This question is from 2020.