question_answer

A person standing on the floor of an elevator drop a coin. The coin reaches the floor of the elevator in time t, if elevator is moving uniformly & in time \[t{{ & }_{2}}\] if is accelerating up, then

A) \[t{{ & }_{1}}>{{t}_{2}}\]                              

B) \[t{{ & }_{1}}={{t}_{2}}\]

C) \[t{{ & }_{1}}<{{t}_{2}}\]

D)  Information is insufficient

Correct Answer: A

Solution :

When the elevator is moving uniformly \[{{\text{v}}_{\text{coin,}\,\text{elevator}}}\text{=a}{{\,}_{\text{coin,}}}_{\text{ground}}\text{=g(downward)}\] \[{{\text{v}}_{\text{coin,}\,\text{elevator}}}\text{=0}\] \[{{s}_{\text{coin,}\,\text{elevator}}}\text{=h(downward)}\] from \[s=ut+\frac{1}{2}a{{t}^{2}}\]with elevator For \[h=0+\frac{1}{2}g_{1}^{2}\Rightarrow {{t}_{1}}=\sqrt{\frac{2h}{g}}\] When elevator is accelerating up with an acceleration a \[{{a}_{\text{coin,}\,\text{elevator}}}\text{=a+g(downward)}\] \[{{u}_{\text{coin,}\,\text{elevator}}}\text{=0}\] \[{{s}_{\text{coin,}\,\text{elevator}}}\text{=h(downward)}\] so,\[h=0+\frac{1}{2}(a+g)t_{2}^{2}\Rightarrow {{t}_{2}}=\sqrt{\frac{2h}{a+g}}\] so,\[{{t}_{1}}>{{t}_{2}}\]