A) t1 > t2
B) t2 > t1
C) t1 = t2
D) t1 >> t2
Correct Answer: A
Solution :
Key Idea: When lift is moving up, resultant acceleration increases. From equation of motion, we have\[s=ut+\frac{1}{2}g{{t}^{2}}\]where u is initial velocity, t is time and g is acceleration due to gravity. At the time of dropping u = 0. \[\because \] \[s=\frac{1}{2}gt_{1}^{2}\] \[\Rightarrow \] \[t_{1}^{2}=\frac{2s}{g}\] When lift moves up \[g'=g+a\] \[\therefore \] \[t_{2}^{2}=\frac{2s}{g+a}\] \[\Rightarrow \] \[t_{2}^{2}<t_{1}^{2}\] i.e., \[{{t}_{2}}<{{t}_{1}}\]or \[{{t}_{1}}<{{t}_{2}}\]You need to login to perform this action.
You will be redirected in
3 sec