RHDG}{2\COS \THETA }\Rightarrow \]Pressure DIFFERENCE \[=hdg=\frac{2S}{R}\cos \theta \]

"> RHDG}{2\COS \THETA }\Rightarrow \]Pressure DIFFERENCE \[=hdg=\frac{2S}{R}\cos \theta \]

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A capillary tube of radius r is dipped in a liquid of density r and surface tension S. If the angle of contact is q, the pressure difference between the two surfaces in the beaker and the capillary

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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\[S=\frac{RHDG}{2\COS \THETA }\Rightarrow \]Pressure DIFFERENCE \[=hdg=\frac{2S}{R}\cos \theta \]

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Posted on 02 Sep 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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