Two objects of masses $m_1$ and $m_2$ having the same size are dropped simultaneously from heights $h_1$ and $h_2$ respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if $(i)$ one of the objects is hollow and the other one is solid and $(ii)$ both of them are hollow,size remaining the same in each case. Give reason.

(N/A) For an object dropped from height $h$,the time $t$ taken to reach the ground is given by the equation of motion $h = \frac{1}{2} g t^2$,where $g$ is the acceleration due to gravity.
For the first object: $h_1 = \frac{1}{2} g t_1^2 \implies t_1 = \sqrt{\frac{2h_1}{g}}$.
For the second object: $h_2 = \frac{1}{2} g t_2^2 \implies t_2 = \sqrt{\frac{2h_2}{g}}$.
The ratio of time taken is $\frac{t_1}{t_2} = \frac{\sqrt{2h_1/g}}{\sqrt{2h_2/g}} = \sqrt{\frac{h_1}{h_2}}$.
$(i)$ If one object is hollow and the other is solid,the ratio remains the same because the acceleration due to gravity $g$ is independent of the mass and density of the object.
(ii) If both are hollow,the ratio still remains the same for the same reason. In free-fall,the acceleration of an object is independent of its mass,shape,or size.