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`         A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.`
3 years ago

Kevin Nash
332 Points
```										Sol. Length of the pendulum = 40cm = 0.4m.
Let acceleration due to gravity be g at the depth of 1600km.
∴ gd = g(1-d/R) = 9.8 (1- 1600/6400) = 9.8 (1- 1/4) = 9.8 × 3/4 = 7.35m/s2
∴ Time period T’ = 2π √(l/gδ)
= 2π √(0.4/7.34) = 2π √0.054 = 2π × 0.23 = 2 × 3.14 × 0.23 = 1.465 ≈ 1.47 sec

```
3 years ago
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