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  • 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 ?

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?

 

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\\\text{From graph it is clear that when} \\ L=1 m, T^{2}=4 s^{2} \\ \text{As we know,} \\ T=2 \pi \sqrt{\frac{L}{g}} \\

\large \begin{array}{l} \Rightarrow g=\frac{4 \pi^{2} L}{T^{2}} \\\\ =4 \times\left(\frac{22}{7}\right)^{2} \times \frac{1}{4}=\left(\frac{22}{7}\right)^{2} \\\\ \therefore g=\frac{484}{49}=9.87 \mathrm{m} / \mathrm{s}^{2} \end{array}

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