As the two masses are dropped,`u=0,a=g` (which does not depend upon mass)
From `s =ut+ (1)/(2) at^(2)`
`h_(1) = 0 +(1)/(2) g t_(1)^(2)` and `h_(2) = 0 +(1)/(2) g t_(2)^(2)`
`:. (h_(1))/(h_(2))=(t_(1)/t_(2))^(2) or (t_(1))/(t_(2))=sqrt(h_(1)/(h_(2)))`
This ratio of time shall remain the same if
(i)one of the objects is hollow and the other one is solid-as the ratio is independent of masses. (ii) both the objects are hollow, as acc. due is gravity (g) does not depent upon mass and also whether the body is hollow/solid.
As size remains the same in each case, there is no effect of resistance due to air in the two cases.