The figure shows the variation of capacitive reactance ($X_C$) of two ideal capacitors of capacitances $C_1$ & $C_2$ with the reciprocal of angular frequency ($1/\omega$) of ac source. The value of $C_1/C_2$ is
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Always double-check the precise axis labels on physics graphs. A steeper slope here physically means a smaller capacitance because the slope represents $1/C$, not $C$. Intuition can easily be flipped if you don't check the math.
Concept: • In any alternating current (AC) circuit, a capacitor actively opposes the continuous flow of charge. This specific opposition is formally quantified as capacitive reactance, denoted as $X_C$.
• The explicit mathematical formula firmly defining capacitive reactance is $X_C = \frac{1}{\omega C}$, where $\omega$ is the driving angular frequency and $C$ is the physical capacitance.
• If we intentionally plot a mathematical graph with $X_C$ safely on the y-axis and the reciprocal factor $(1/\omega)$ strictly on the x-axis, the resulting relation $y = \left(\frac{1}{C}\right) x$ vividly forms a straight line passing smoothly through the origin.
• The steepness, or mathematical slope $m$, of this straight line is definitively equal to $\frac{1}{C}$. Therefore, the slope is strictly inversely proportional to the actual capacitance. Step 1: Extract slopes from the given graph
From the provided image, we meticulously observe two distinctly sloped lines corresponding to the two capacitors.
The plotted line representing capacitor $C_1$ creates an angle of exactly $\theta_1 = 45^\circ$ directly with the positive x-axis.
The mathematical slope $m_1$ for this first line is computed using the tangent function:
\[ m_1 = \tan(45^\circ) = 1 \]
The plotted line representing capacitor $C_2$ creates an angle of exactly $\theta_2 = 30^\circ$ directly with the positive x-axis.
The mathematical slope $m_2$ for this second line is similarly computed:
\[ m_2 = \tan(30^\circ) = \frac{1}{\sqrt{3}} \] Step 2: Relate the slopes back to capacitance
As robustly established in the concept section, the slope $m$ of this specific type of graph is mathematically identical to $\frac{1}{C}$.
Therefore, for the first capacitor:
\[ \frac{1}{C_1} = m_1 = 1 \implies C_1 = 1 \]
And for the second capacitor:
\[ \frac{1}{C_2} = m_2 = \frac{1}{\sqrt{3}} \implies C_2 = \sqrt{3} \] Step 3: Calculate the required final ratio
We are strictly asked to find the exact numerical value of the specific fractional ratio $C_1 / C_2$.
We meticulously substitute the derived capacitance values into this fraction:
\[ \frac{C_1}{C_2} = \frac{1}{\sqrt{3}} \] Step 4: Conclusion
The rigorously derived ratio of the two capacitances is exactly $1 / \sqrt{3}$. This flawlessly and undeniably matches option (D).
