Question:
The figure shows the variation of capacitive reactance $X_C$ with $1/\omega$ gives two straight lines for capacitors $C_1$ and $C_2$. The ratio $C_1/C_2$ is
The figure shows the variation of capacitive reactance $X_C$ with $1/\omega$ gives two straight lines for capacitors $C_1$ and $C_2$. The ratio $C_1/C_2$ is
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In graph-based physics questions, slope interpretation is the key step.
Updated On: Jul 2, 2026
- $\frac{1}{2}$
- $2$
- $\sqrt{3}$
- $\frac{1}{\sqrt{3}}$
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The Correct Option is D
Solution and Explanation
Concept:
Capacitive reactance:
\[
X_C = \frac{1}{\omega C}
\]
Let $x = \frac{1}{\omega}$, then:
\[
X_C = \frac{x}{C}
\]
Thus slope of graph:
\[
\text{slope} = \frac{1}{C}
\]
Step 1: Use given angles Slope $m = \tan\theta$ For $C_1$: \[ m_1 = \tan 45^\circ = 1 \] For $C_2$: \[ m_2 = \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Step 2: Ratio of slopes \[ \frac{m_1}{m_2} = \frac{1}{1/\sqrt{3}} = \sqrt{3} \] Since: \[ m \propto \frac{1}{C} \Rightarrow \frac{m_1}{m_2} = \frac{C_2}{C_1} \] \[ \frac{C_1}{C_2} = \frac{1}{\sqrt{3}} \] Final Answer: \[ \boxed{(D)} \]
Step 1: Use given angles Slope $m = \tan\theta$ For $C_1$: \[ m_1 = \tan 45^\circ = 1 \] For $C_2$: \[ m_2 = \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Step 2: Ratio of slopes \[ \frac{m_1}{m_2} = \frac{1}{1/\sqrt{3}} = \sqrt{3} \] Since: \[ m \propto \frac{1}{C} \Rightarrow \frac{m_1}{m_2} = \frac{C_2}{C_1} \] \[ \frac{C_1}{C_2} = \frac{1}{\sqrt{3}} \] Final Answer: \[ \boxed{(D)} \]
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