Question

A seconds pendulum is taken to a place where acceleration due to gravity becomes 8 times its initial value.What will be its new time period?​

Answer 1

Explanation:

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

T α 1/√g

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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