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JEE Mains · Physics · STD 11 - Surface tension
- JEE Mains 2020
- Medium
- Numerical
When a long glass capillary tube of radius \(0.015\;cm\) is dipped in a liquid, the liquid rises to a height of \(15\, cm\) within it. If the contact angle between the liquid and glass to clase to \(0^{\circ},\) the surface tension of the liquid, in \(milli\,Newton \;m ^{-1}\) is \(.....\) \(\left[\rho_{\text {(liquid) }}=900\; kg\,m ^{-3}, g =10\, ms ^{-2}\right]\) (Give answer in closest integer)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(101\)
Step-by-step Solution
Detailed explanation
Capillary rise
\(h =\frac{2 S \cos \theta}{\rho gr }\)
\(S=\frac{\rho grh}{2 \cos \theta}\)
\(=\frac{(900)(10)\left(15 \times 10^{-5}\right)\left(15 \times 10^{-2}\right)}{2}\)
\(S=1012.5 \times 10^{-4}\)
\(S=101.25 \times 10^{-3}=101.25 mN / m\)
In question closest integer is asked
so closest integer \(=101.00\)
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