","comment":{"@type":"Comment","text":"The terminal velocity of a ball decreases as its radius or viscosity coefficient decreases, or vice versa."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{V}_{1}}>{{V}_{2}}>{{V}_{3}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $${{V}_{3}}>{{V}_{2}}>{{V}_{1}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $${{V}_{1}}>{{V}_{3}}>{{V}_{2}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{V}_{2}}>{{V}_{1}}>{{V}_{3}}$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{Terminal velocity of ball in a liquid is given by-}\\\\ v=\\dfrac{2{{R}^{2}}g({{\\rho }_{s}}-{{\\rho }_{l}})}{9\\eta }\\\\ \\text{So for ball one terminal velocity will be-}\\\\ {{V}_{1}}=\\dfrac{2{{R}^{2}}g(\\rho -{}^{\\rho }/{}_{3})}{9\\eta }\\\\ \\Rightarrow{{V}_{1}} =\\dfrac{4{{R}^{2}}g\\rho }{27\\eta }\\\\ \\text{For ball two-}\\\\ {{V}_{2}}=\\dfrac{2{{(2R)}^{2}}g(2\\rho -\\rho )}{9\\times 2\\eta }\\\\ \\Rightarrow {{V}_{2}} =\\dfrac{8{{R}^{2}}g\\rho }{18\\eta }\\\\ \\Rightarrow {{V}_{2}} =\\dfrac{4{{R}^{2}}g\\rho }{9\\eta }\\\\ \\text{For ball three-}\\\\ {{V}_{3}}=\\dfrac{2{{\\left( {}^{R}/{}_{2} \\right)}^{2}}g(3\\rho -\\rho )}{9\\times 3\\eta }\\\\ \\therefore{{V}_{3}}=\\dfrac{{{R}^{2}}g\\rho }{27\\eta }\\\\ \\text{From the above values we can find the relation between three terminal velocities.}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Laminar and turbulent flow Quiz 1","text":" A fluid of viscosity coefficient $$\\eta =0.5\\,{Ns}/{{{m}^{2}}}$$ and relative density $$\\rho =1300kg/{{m}^{2}}$$ flowing through a narrow tube of diameter 0.8 m and velocity 3 m/s. The type of flow of fluid will be- ","comment":{"@type":"Comment","text":"If the value of Reynolds number is greater than 2000 then flow is turbulent else it’s laminar. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Laminar","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" Can’t say","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" both (a) and (b)","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Turbulent","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{Reynolds number is given by,}\\\\ \\Rightarrow {{R}_{e}}=\\dfrac{\\rho vd}{\\eta }\\\\ \\Rightarrow {{R}_{e}}=\\dfrac{1300\\times 3\\times 0.8}{0.5}\\\\ \\therefore {{R}_{e}}=6240\\\\ \\text{The value is greater than 2000 so the flow is turbulent.}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Laminar and turbulent flow Quiz 1","text":" A man is running for an hour, due to which his blood flow increases to 5 times as initial rate.The blood’s viscosity is reduced to 80 % of initial value and the pressure difference across the arteries increased by 70 %. The change in radius of artery will be- ","comment":{"@type":"Comment","text":"By increasing in pressure difference across pipe or decreasing viscosity coefficient rate of flow will increase or vice versa. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 1.8 times","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 2.0 times","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" remains same","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 1.2 times","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{Rate of flow is given by,}\\\\ Q=\\dfrac{\\Delta \\operatorname{P}\\pi {{r}^{4}}}{8\\eta l}\\\\ \\text{New rate flow will be-}\\\\ 5Q=\\dfrac{1.7\\Delta \\operatorname{P}\\pi r{{'}^{4}}}{8\\times 0.8\\eta \\times l}\\\\ \\text{By dividing both equation we have-}\\\\ \\Rightarrow r{{'}^{4}}=2.352{{r}^{4}}\\\\ \\therefore r'=1.2r$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}Laminar and turbulent flow Quiz 1