A) \[\sqrt{\frac{40}{8}}\]
B) \[\sqrt{\frac{72}{40}}\]
C) \[\sqrt{\frac{40}{12}}\]
D) \[\frac{8}{4}\]
Correct Answer: C
Solution :
The relation for time period is given by \[t\propto \frac{1}{\sqrt{g}}\] \[\because \] \[T=2\pi \sqrt{\frac{l}{g}}\] Hence \[\frac{{{t}_{1}}}{t}=\frac{1}{\sqrt{g+a}}\times \frac{\sqrt{g}}{1}=\sqrt{\frac{g}{g+a}}=\sqrt{\frac{10}{12}}\] hence \[{{t}_{1}}=t\sqrt{\frac{10}{12}}=2\sqrt{\frac{10}{12}}=\sqrt{\frac{40}{12}}\]
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