question_answer

A elevator car whose floor to ceiling distance is 2.7 m starts ascending with a constant acceleration of \[1.2\,\,m/{{s}^{2}}\], 2 s after the start a bolt is begin to fall from the ceiling of the car. The free fall time of the bolt is \[\left( g=9.8\,\,m/{{s}^{2}} \right):\]

A)  \[\sqrt{\frac{2.7}{9.8}s}\]                           

B)  \[\sqrt{\frac{5.4}{9.8}s}\]

C)  \[\sqrt{\frac{5.4}{8.6}s}\]                           

D)  \[\sqrt{\frac{5.4}{11}s}\]

Correct Answer: D

Solution :

Key idea: Net acceleration on the elevator car is more than acceleration due to gravity. Since elevator car is ascending upwards, from Newton?s second law, the net force is acting upwards, hence resultant acceleration is \[{{a}_{r}}=\left( g+a \right)\] \[=\left( 9.8+1.2 \right)=11\,\,m/{{s}^{2}}\] Relative velocity of observer to elevator is \[{{u}_{r}}=0\] From the equation \[s={{u}_{r}}t+\frac{1}{2}{{a}_{r}}{{t}^{2}}\] \[2.7=0+\frac{1}{2}\times 11\times {{t}^{2}}\] \[\Rightarrow \]               \[{{t}^{2}}=\frac{5.4}{11}\] \[\Rightarrow \]               \[t=\sqrt{\frac{5.4}{11}s}\]