# 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?

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

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

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-