AP EAMCET · PHYSICS · Oscillations
The time period of a simple pendulum on the surface of the earth is T. The height above the surface of the earth at which the time period of the pendulum becomes \(2 \mathrm{~T}\) is (Radius of the earth \(=6400 \mathrm{~km}\) )
- A \(3200 \mathrm{~km}\)
- B \(6400 \mathrm{~km}\)
- C \(1600 \mathrm{~km}\)
- D \(800 \mathrm{~km}\)
Answer & Solution
Correct Answer
(A) \(3200 \mathrm{~km}\)
Step-by-step Solution
Detailed explanation
Time period of a simple pendulum is given by, \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\) \(2 \mathrm{~T}=\frac{\mathrm{T}}{1+\frac{\mathrm{h}}{\mathrm{R}}} \Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=\frac{1}{2}\)…
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