question_answer 13)

Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is \[\text{4 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}{{\text{m}}^{\text{2}}}\], what is the pressure difference between the two ends of the tube? (density of glycerine \[=1.3\times {{10}^{3}}\,kg{{s}^{-3}}\] and viscosity of glycerine \[{{\text{l}}_{\text{1}}}\text{=4 }\!\!\times\!\!\text{ 7m;}\]).

Answer:

Here, \[\vartriangle {{l}_{1}}=\vartriangle l;\] \[{{F}_{1}}=F\] Mass of glycerine flowing per second, \[{{\text{a}}_{\text{2}}}\text{=4 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}{{\text{m}}^{\text{2}}}\text{,}\] Volume of glycerine flowing per second, \[{{l}_{2}}=3\cdot 5m,\] If \[\vartriangle {{l}_{2}}=\vartriangle l;\]is the difference of pressure between two ends of the tube, then using poiseuilles formula, we have \[{{F}_{2}}=F\]or \[{{Y}_{1}},{{Y}_{2}}\] \[\therefore \]