TS EAMCET · Physics · Oscillations
If a body dropped freely from a height of 20 m reaches the surface of a planet with a velocity of \(31.4 \mathrm{~ms}^{-1}\), then the length of a simple pendulum that ticks seconds on the planet is
- A 1 m
- B 0.625 m
- C 2.5 m
- D 2 m
Answer & Solution
Correct Answer
(C) 2.5 m
Step-by-step Solution
Detailed explanation
\(\mathrm{h}=20 \mathrm{~m}, \mathrm{v}=31.4 \mathrm{~ms}^{-1}\) By conservation of energy \(\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2\) \(\Rightarrow \mathrm{g}=\frac{\mathrm{v}^2}{2 \mathrm{~h}}=\frac{31.4 \times 31.4}{2 \times 20}=\frac{5 \pi^2}{2} \mathrm{~ms}^{-2}\) For simple…
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