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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Solution

Time period of a pendulum is inversely proportional to the square root of acceleration due to gravity.

Given that gravity falls to one-fourth.
g' = g/4
let original time period = T
final time period = T'

$\frac{T'}{T} = \sqrt{ \frac{g}{g'} } \\ \\ T'=T \sqrt{ \frac{g}{g'} } \\ \\ T'=T \sqrt{ \frac{g}{g/4} } \\ \\ T'=T \sqrt{4} =2T$

So the time period becomes 2 times the original time period.

Time period of a seconds' pendulum = 2s
New Time period = 2×2s = 4s

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