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JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    Two tubes of radii \[{{r}_{1}}\] and \[{{r}_{2}},\] and lengths \[{{I}_{1}}\] and \[{{I}_{2}},\] respectively, are connected in series and a liquid flows through each of them in stream line conditions. \[{{P}_{1}}\] and \[{{P}_{2}}\] are pressure differences across the two tubes. If \[{{P}_{2}}\] is \[4{{P}_{1}}\] and \[{{I}_{2}}\] is \[\frac{{{I}_{1}}}{4},\] then the radius \[{{r}_{2}}\] will be equal to -                        [JEE Online 09-04-2017]

    A)  \[4{{r}_{1}}\]                                    

    B)  \[{{r}_{1}}\]

    C)  \[2{{r}_{1}}\]                                    

    D)  \[\frac{{{r}_{1}}}{2}\]

    Correct Answer: D

    Solution :

                    \[\frac{\phi v}{dt}=\frac{\pi }{8}\,\frac{p{{r}^{4}}}{nL}\] \[\frac{{{p}_{1}}r_{1}^{4}}{{{L}_{1}}}\,=\frac{{{p}_{2}}r_{2}^{4}}{{{L}_{2}}}\] \[\frac{{{p}_{1}}r_{1}^{4}}{{{I}_{2}}}\,=\frac{4{{p}_{1}}r_{2}^{4}}{{{I}_{1/4}}}\,=r_{2}^{4}\,=\frac{r_{1}^{4}}{16}\] \[{{r}_{2}}=\frac{{{r}_{1}}}{2}\]


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