A) \[\frac{6}{5}\]
B) \[\frac{5}{6}\]
C) 1
D) 4
Correct Answer: A
Solution :
According to the relation, \[y=k{{t}^{2}}\Rightarrow \frac{dy}{dt}=2kt\] \[\therefore \] \[a=\frac{{{d}^{2}}y}{d{{t}^{2}}}=2k=2\times 1=2\] \[[\because k=1]\] The point of suspension is moving upward with acceleration a, then effective acceleration due to gravity on pendulum \[g'=g+a=10+2=12\,m/{{s}^{2}}\] \[\therefore \] \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g}}\] And \[{{T}_{2}}=2\pi \sqrt{\frac{l}{g+a}}\] \[\therefore \] \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{g+a}{g}=\frac{12}{10}=\frac{6}{5}\]You need to login to perform this action.
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