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  • A seconds' pendulum is taken to a place where acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?

A seconds' pendulum is taken to a place where acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?

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best_answer

The time period of a pendulum is inversely proportional to the square root of acceleration due to gravity.
i.e T \ \alpha \ \frac{1}{\sqrt{g}}

And Time period of a seconds' pendulum = 2s

let original time period = T= 2sec
final time period = T'

As g' = g/4

SO

\begin{array}{l} \frac{T^{\prime}}{T}=\sqrt{\frac{g}{g^{\prime}}} \\ T^{\prime}=T \sqrt{\frac{g}{g^{\prime}}} \\ T^{\prime}=T \sqrt{\frac{g}{g / 4}} \\ T^{\prime}=T \sqrt{4}=2 T =2*2=4 \ sec\end{array}

 

Posted by

avinash.dongre

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