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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2014
  • Medium
  • MCQ

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? ...... \(m/s^2\)

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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,

\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\)

\(\Rightarrow \mathrm{g}=\frac{4 \pi^{2} \mathrm{L}}{\mathrm{T}^{2}}\)

\(=4 \times\left(\frac{22}{7}\right)^{2} \times \frac{1}{4}=\left(\frac{22}{7}\right)^{2}\)

\(\therefore \mathrm{g}=\frac{484}{49}=9.87 \mathrm{m} / \mathrm{s}^{2}\)

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