question_answer 30)

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U shaped tube open at both ends. If the U tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is \[=2\cdot 026\times {{10}^{4}}\] Take the angle contact to be zero, and density of water to be \[1\cdot 03\times {{10}^{3}}g{{m}^{-3}}?\]

Answer:

Here, \[=45\cdot 8\times {{10}^{-11}}p{{a}^{-1}}.\]\[=1\cdot 013\times {{10}^{5}}pa.\] \[p=80\cdot 0\times 1\cdot 013\times {{10}^{5}}pa;\] For narrow tube \[\frac{1}{B}=45\cdot 8\times {{10}^{-11}}P{{a}^{-1}}\] or \[p=1\cdot 03\times {{10}^{3}}kg{{m}^{-3}}\] For wider tube, \[\text{V=}\frac{\text{M}}{\text{P}}\text{ and V }\!\!'\!\!\text{ =}\frac{\text{M}}{\text{P }\!\!'\!\!\text{ }}\] or \[\therefore \] Let \[\vartriangle V=V-V'=M\left( \frac{1}{P}-\frac{1}{P'} \right)\] be the heights to which water rises in narrow tube and wider tube respectively Then, \[\therefore \] and \[\frac{\vartriangle V}{V}=M\left( \frac{1}{P}-\frac{1}{P'} \right)\] \[\times \frac{P}{M}=1-\frac{P}{P'}\]Difference in levels of water in two limbs of tube is, \[\frac{\vartriangle V}{V}=1-\frac{1\cdot 03\times {{10}^{3}}}{P'}\] \[B=\frac{pV}{\vartriangle V}\]\[\frac{\vartriangle V}{V}=\frac{P}{B}\]\[\frac{\vartriangle V}{V}=\left( 80\cdot 0\times 1\cdot 013\times {{10}^{5}} \right)\times 45\cdot 8\times {{10}^{-11}}\]