Question

If T1 and T2 are the time periods of oscillation of a simple pendulum on the surface of the earth (of radius R) and at a depth d, then d is​

Answer 1

We know that the time period of a simple pendulum

T=2π√(l/g)

⇒T ∝ l/√g

Here in a given condition,

T₁ =k/√g    …(i)

and T₂= k/[√g(1−d/R)]     ...(ii)

From Eqs. (i) and (ii), we get

T₁/T₂=[k/√3]/[k/√g(1−d/R)]

or, T₁/T₂ = √(1-d/R)

or, T₁^2/T₂^2 = 1−d/R

⇒d = [1 − T₁^2/T₂^2] R

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Answer 2

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