A) \[{{t}_{1}}>{{t}_{2}}\]
B) \[{{t}_{2}}>{{t}_{1}}\]
C) \[{{t}_{1}}={{t}_{2}}\]
D) \[{{t}_{1}}>>{{t}_{2}}\]
Correct Answer: A
Solution :
Time taken by coin to reach the floor is given by \[h=\frac{1}{2}g{{t}^{2}}\] \[(\because u=0)\] \[\Rightarrow \] \[t=\sqrt{\frac{2h}{g}}\] In stationary lift, \[{{t}_{1}}=\sqrt{\frac{2h}{g}}\] In upward moving lift with constant acceleration a, \[g'=g+a\] \[\therefore \] \[{{t}_{2}}=\sqrt{\frac{2h}{g+a}}\] Clearly, \[g'>g\] Thus, \[{{t}_{2}}<{{t}_{1}}\]
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