Question

The slope of $T^2$ vs $l$ graph for a simple pendulum is___.

• A. $\frac{2{\pi }^4}{g}$
• B. $\frac{4\pi }{g}$
• C. $\frac{4{\pi }^2}{g}$
• D. $\frac{4\pi }{g^2}$

Answer 1

The correct option is C

$\frac{4{\pi }^2}{g}$

The slope of $T^2$ vs $l$ graph is found to be constant at any place and is equal to $\frac{4{\pi }^2}{g}$ where g is the acceleration due to gravity at that place.

Explanation: In a $T^2$ vs $l$ graph, with $T^2$ in the Y-axis and $l$ in th X-axis, slope = $\frac{T^2}{l}$. We have $T^2=4{\pi }^2\frac{l}{g}$ $\Rightarrow \frac{T^2}{l}=\frac{4{\pi }^2}{g}$