Question

In a 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?

• A. $9.81m/s^2$
• B. $9.87m/s^2$
• C. $9.91m/s^2$
• D. $10.0m/s^2$

Answer 1

The correct option is B $9.87m/s^2$

$\text{Solution}:$

From graph it is clear that when

$L=1m,T^2=4s^2$

As we know,

$T=2\pi \sqrt{\frac{L}{g}}$

$\Rightarrow g=\frac{4{\pi }^{2}L}{T^2}$

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

$\therefore g=\frac{484}{49}=9.87m/s^2$

$\text{Hence B is the correct option}$