Question:
Two capillary tubes of the same diameter are kept vertically in two different liquids whose density are in the ratio \(4:3\). Rise of liquid in both the capillaries is \(h_{1}\) and \(h_{2}\) respectively. Their surface tensions are in the ratio \(6:5\), the ratio of the heights \(\left(\frac{h_{1}}{h_{2}}\right)\) is (Assume that their angles of contact are same)
Two capillary tubes of the same diameter are kept vertically in two different liquids whose density are in the ratio \(4:3\). Rise of liquid in both the capillaries is \(h_{1}\) and \(h_{2}\) respectively. Their surface tensions are in the ratio \(6:5\), the ratio of the heights \(\left(\frac{h_{1}}{h_{2}}\right)\) is (Assume that their angles of contact are same)
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Capillary rise is proportional to surface tension and inversely proportional to density for the same tube and contact angle.
Updated On: Nov 12, 2026
- \(0.4\)
- \(0.5\)
- \(0.8\)
- \(0.9\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Capillary rise is \(h=\dfrac{2T\cos\theta}{r\rho g}\). For the same radius and the same angle of contact, \(h\propto\dfrac{T}{\rho}\).
Step 2: Ratio
\(\dfrac{h_{1}}{h_{2}}=\dfrac{T_{1}}{T_{2}}\times\dfrac{\rho_{2}}{\rho_{1}}\)
Step 3: Calculation
\(\dfrac{h_{1}}{h_{2}}=\dfrac{6}{5}\times\dfrac{3}{4}=\dfrac{18}{20}=0.9\)
Final Answer:
The ratio of the heights is 0.9. \[ \boxed{\text{(D) }0.9} \]
Capillary rise is \(h=\dfrac{2T\cos\theta}{r\rho g}\). For the same radius and the same angle of contact, \(h\propto\dfrac{T}{\rho}\).
Step 2: Ratio
\(\dfrac{h_{1}}{h_{2}}=\dfrac{T_{1}}{T_{2}}\times\dfrac{\rho_{2}}{\rho_{1}}\)
Step 3: Calculation
\(\dfrac{h_{1}}{h_{2}}=\dfrac{6}{5}\times\dfrac{3}{4}=\dfrac{18}{20}=0.9\)
Final Answer:
The ratio of the heights is 0.9. \[ \boxed{\text{(D) }0.9} \]
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