Four solid spheres each of diameter 5 ~cm and mass 0.5 ~kg ar 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

aligned r & = d 2 = 5 2 ~cm & = 5 2 10⁻² ~m m & =0.5 ~kg a & =4 ~cm & =4 10⁻² ~m I_ X X & =I₁+I₂+I₃+I₄ aligned aligned = & [ 2 5 m r^2+m ( a 2 )^2 ]+ 2 5 m r^2 & + [ 2 5 m r^2+m ( a 2 )^2 ]+ 2 5 m r^2 aligned Substituting the values, we get aligned & I_ X X & =9 10⁻⁴ Kgm ⁻² & N & =9 aligned Answer is 9. Analysis of Question (i) Question is simple. (ii) Only theorem of parallel axes is to be used properly.