Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centres at the corners of a square of side 4 cm . The moment of inertia of the system about the diagonal of the square is N × 10 - 4 kg m 2 , then N is

Step-by-step solution

r = d 2 = 5 2 cm = 5 2 × 10 - 2   m m = 0.5   kg a = 4   cm = 4 × 10 - 2   m I XX = I 1 + I 2 + I 3 + I 4 = 2 5 mr 2 + m a 2 2 + 2 5 mr 2 + 2 5 mr 2 + m a 2 2 + 2 5 mr 2 Substituting the values, we get, I XX = 9 × 10 - 4   kg   m 2 Therefore, N = 9