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(D) Given the equation for the frequency of a simple pendulum: $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$.Squaring both sides,we get: $f^2 = \frac{1}

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$f^2$ on the ordinate and $1/\ell$ on the abscissa

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(D) Given the equation for the frequency of a simple pendulum: $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$.Squaring both sides,we get: $f^2 = \frac{1}{4 \pi^2} \cdot \frac{g}{\ell}$.This can be rewritten as: $f^2 = \left( \frac{g}{4 \pi^2} \right) \cdot \left( \frac{1}{\ell} \right)$.Comparing this with the equation of a straight line $y = mx + c$,where $y = f^2$,$x = 1/\ell$,$m = \frac{g}{4 \pi^2}$,and $c = 0$.Since the equation represents a linear relationship between $f^2$ and $1/\ell$,the graph between $f^2$ (on the ordinate) and $1/\ell$ (on the abscissa) will be a straight line passing through the origin.

Frequency $f$ of a simple pendulum depends on its length $\ell$ and acceleration $g$ due to gravity according to the following equation $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$. $A$ graph between which of the following quantities is a straight line?

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