A) \[{{t}_{1}}>{{t}_{2}}\]
B) \[{{t}_{2}}>{{t}_{1}}\]
C) \[{{t}_{1}}={{t}_{2}}\]
D) \[{{t}_{1}}>>{{t}_{2}}\]
Correct Answer: A
Solution :
From equation of motion, we have \[S=ut+\frac{1}{2}g{{t}^{2}}\] where \[u\] is initial velocity, \[t\] is time, \[g\] is acceleration due to gravity. At the time of dropping\[u=0\]. \[\because \] \[S=\frac{1}{2}gt_{1}^{2}\] \[\Rightarrow \] \[t_{1}^{2}=\frac{2\,\,s}{g}\] When lift moves up \[g=g+a\] \[\therefore \] \[t_{2}^{2}=\frac{2\,\,s}{g+a}\] \[\Rightarrow \] \[t_{2}^{2}<t_{1}^{2}\] \[i.e.,\] \[{{t}_{2}}<{{t}_{1}}\] or \[{{t}_{1}}>{{t}_{2}}\]You need to login to perform this action.
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