A) \[1.53\times {{10}^{7}}m/s\]
B) \[A{{m}^{2}}\]
C) \[A{{m}^{2}}\]
D) \[A{{m}^{2}}\]
Correct Answer: C
Solution :
Here: Time period of simple pendulum in stationary lift is \[{{T}_{1}}=T\] Acceleration of the lift \[a=g/3\] The acceleration of the lift in upward direction is \[g=g+a=g+\frac{g}{3}=\frac{4g}{3}\] Now the time period of simple pendulum is \[T=2\pi \frac{\sqrt{l}}{g}\] \[\Rightarrow \] \[T\propto \frac{\sqrt{l}}{g}\] Hence, \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{\sqrt{{{g}_{1}}}}{{{g}_{2}}}=\frac{\frac{\sqrt{4}}{3}g}{g}=\frac{2}{\sqrt{3}}\] So \[{{T}_{2}}=\frac{\sqrt{3}}{2}T\]
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