A) \[\sqrt{\frac{2}{5}T}\]
B) \[\sqrt{\frac{5}{2}T}\]
C) \[\frac{\sqrt{5}\,T}{2}\]
D) \[\frac{2T}{\sqrt{5}}\]
Correct Answer: D
Solution :
Key Idea: when the lift moves upwards effective acceleration increases. When the lift is moving upwards, then from Newton?s second law of motion net force on him is \[F=ma\] (upwards) Or \[F=-ma\] (downwards) Therefore, \[-ma=mg-R\] \[\Rightarrow \] \[R=mg+ma\] \[\therefore \] \[R=m\left( g+a \right)\] Time Period, \[T=2\pi \sqrt{\frac{l}{g}}\,is\] \[T=2\pi \sqrt{\frac{l}{g+a}}\,\] ?(1) When lift is stationary time period is \[T=2\pi \sqrt{\frac{l}{g}}\,\] ?(2) Also, \[g'=g+a=g+\frac{g}{4}=\frac{5g}{4}\] Dividing Eq. (2) by (1), we get \[\frac{T}{T'}=\sqrt{\frac{5/4g}{g}}=\frac{\sqrt{5}}{2}\] \[T'=\frac{2}{\sqrt{5}}T\]You need to login to perform this action.
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