AP EAMCET · PHYSICS · Oscillations
Time period of a simple pendulum is \(4 \mathrm{~s}\) at a place on the earth where the acceleration due to gravity is \(\pi^2 \mathrm{~ms}^{-2}\). Then the length of the pendulum in meters is
- A \(4\)
- B \(2\)
- C \(\pi\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
We have \(\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \Rightarrow 4=2 \pi \sqrt{\frac{\ell}{\pi^2}} \Rightarrow 16=4 \pi^2 \cdot \frac{\ell}{\pi^2}\) \(\Rightarrow \ell=4 \mathrm{sec}\)
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