Question:
Some smoke is trapped in a small glass container and is viewed through a microscope. A number of very small smoke particles are seen in continuous random motion as a result of bombardment by air molecules. If the mass of the smoke particle is about \(10^{12}\) times higher than that of an air molecule, the average speed of a smoke particle is:
Some smoke is trapped in a small glass container and is viewed through a microscope. A number of very small smoke particles are seen in continuous random motion as a result of bombardment by air molecules. If the mass of the smoke particle is about \(10^{12}\) times higher than that of an air molecule, the average speed of a smoke particle is:
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Speed is inversely proportional to square root of mass in kinetic theory.
Updated On: May 1, 2026
- \(10^6\) times the average speed of an air molecule
- \(10^{-12}\) times the average speed of an air molecule
- \(10^{12}\) times the average speed of an air molecule
- \(10^{-6}\) times the average speed of an air molecule
- \(10^{-10}\) times the average speed of an air molecule
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The Correct Option is D
Solution and Explanation
Concept:
From kinetic theory: \[ v \propto \frac{1}{\sqrt{m}} \]
Step 1: Given relation
\[ m_{\text{smoke}} = 10^{12} m_{\text{air}} \]
Step 2: Speed ratio
\[ \frac{v_{\text{smoke}}}{v_{\text{air}}} = \sqrt{\frac{m_{\text{air}}}{m_{\text{smoke}}}} \] \[ = \sqrt{\frac{1}{10^{12}}} \]
Step 3: Simplify
\[ = 10^{-6} \] \[ \boxed{v_{\text{smoke}} = 10^{-6} \, v_{\text{air}}} \]
From kinetic theory: \[ v \propto \frac{1}{\sqrt{m}} \]
Step 1: Given relation
\[ m_{\text{smoke}} = 10^{12} m_{\text{air}} \]
Step 2: Speed ratio
\[ \frac{v_{\text{smoke}}}{v_{\text{air}}} = \sqrt{\frac{m_{\text{air}}}{m_{\text{smoke}}}} \] \[ = \sqrt{\frac{1}{10^{12}}} \]
Step 3: Simplify
\[ = 10^{-6} \] \[ \boxed{v_{\text{smoke}} = 10^{-6} \, v_{\text{air}}} \]
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