What effect occurs on the frequency of a pendulum if it is taken from the earth surface to deep into a mine

Options

A. Increase
B. decrease
C. First increase and decrease
D. Remains unchanged

Answer

A

Explanation

Time period of a pendulum, `T=2pisqrt((l)/(g))`
`therefore` Frequency, `upsilon=(1)/(T)=(1)/(2pi)sqrt((g)/(l))orupsilonpropsqrt(g)`
Also is deep mine, `g.=g(1-(d)/(R))` i.e., g. decreases with increase in depth in mine, so `upsilon` decreases as pendulum is moved into the mine.