Question

a seconds pendulum is taken to a deep mine where the acceleration due to gravity becomes one fourth what is time period​

Answer 1

Answer:

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'

\begin{gathered} \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\end{gathered}

T

T

′

=

g

′

g

T

′

=T

g

′

g

T

′

=T

g/4

g

T

′

=T

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