Four identical spheres, each of mass m , are placed at the vertices of a square of side a . When released from rest, they interact only through their mutual gravitational forces and collide at the centre of the square after a time T = 16 seconds. If the side of the square is increased to 3a and the mass of each sphere is changed to x m , they are found to collide at the centre after a time T = 24 seconds. The value o

Step-by-step solution

By dimensional analysis, the time T taken for the spheres to collide under mutual gravitational force depends on the mass m , the side length a , and the universal gravitational constant G . Let T a^x m^y G^z . Substituting the dimensions: [T] = [L]^x [M]^y [M⁻¹ L^3 T⁻²]^z [M^0 L^0 T^1] = [M^ y-z L^ x+3z T^ -2z ] Equating the powers: -2z = 1 z = - 1 2 y - z = 0 y = - 1 2 x + 3z = 0 x = 3 2 Thus, T a^3 m . Given T₁ = 16 s for side a and mass m . For side 3a and mass x m , the time is T₂ = 24 s. T₂ T₁ = (3a)^3 / (x m