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JEE Mains · Physics · STD 11 - 13. oscillations
In an experiment for determining the gravitational acceleration \(g\) of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure. What is the value of \(g\) at the place? ...... \(m/s^2\)
- A \(9.81\)
- B \(9.87\)
- C \(9.91\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(9.87\)
Step-by-step Solution
Detailed explanation
From graph it is clear that wher \(\mathrm{L}=1 \mathrm{m}, \mathrm{T}^{2}=4 \mathrm{s}^{2}\) As we know, \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\) \(\Rightarrow \mathrm{g}=\frac{4 \pi^{2} \mathrm{L}}{\mathrm{T}^{2}}\)…
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