A) 1 :\[\sqrt{2}\]
B) \[\sqrt{2}\]:1
C) 2 : 1
D) 1 : 2
Correct Answer: A
Solution :
For first ball, \[h=0\times {{t}_{1}}+\frac{1}{2}gt_{1}^{2}\] \[\Rightarrow \] \[{{t}_{1}}=\sqrt{\frac{2h}{g}}\] ?(i) For second ball, \[2h=0\times {{t}_{2}}-\frac{1}{2}gt_{2}^{2}\] \[\Rightarrow \] \[{{t}_{2}}=\sqrt{\frac{4h}{g}}\] ?(ii) From equation (i) and (ii) \[{{t}_{1}}:{{t}_{2}}=1:\sqrt{2}\]
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