JEE Mains · Physics · STD 11 - 9.2 surface tension
Surface tension of two liquids (having same densities), \(T_1\) and \(T_2\), are measured using capillary rise method utilizing two tubes with inner radii of \(r_1\) and \(r_2\) where \(r_1>r_2\). The measured liquid heights in these tubes are \(h_1\) and \(h_2\) respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights \(h _1\) and \(h _2\) and surface tensions \(T _1\) and \(T _2\) satisfy the relation :
- A \(h_{1} < h_{2} \) and \( T_{1}=T_{2} \)
- B \( h_{1}=h_{2} \) and \( T_{1}=T_{2} \)
- C \( h_{1}>h_{2} \) and \( T_{1}=T_{2} \)
- D \( h_{1}>h_{2} \) and \( T_{1} < T_{2} \)
Answer & Solution
Correct Answer
(A) \(h_{1} < h_{2} \) and \( T_{1}=T_{2} \)
Step-by-step Solution
Detailed explanation
\( h=\frac{2T}{\rho gr} \) \( h\propto\frac{1}{r} \) if \( r_{1}>r_{2}\Rightarrow h_{2}>h_{1} \)
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