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JEE MAIN 2022
26-06-2022 S1
Question
An ideal fluid of density 800 $\mathrm{kgm}^{-3}$, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2}$. . The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt{x}}{6} \mathrm{~ms}^{-1}$ for $\mathrm{x}=$ $\_\_\_\_$ (Given $\mathrm{g}=10 \mathrm{~m}^{-2}$ )
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Solution
From continuity equation $$ \begin{aligned} & a v_1=\frac{a}{2} v_2 \\ & v_2=2 v_1 \end{aligned} $$ From Bernoulli's theorem, $$ \begin{aligned} & P_1+\rho g h_1+\frac{1}{2} \rho v_1^2=P_2+\rho g h_2+\frac{1}{2} \rho v_2^2 \\ & P_1+P_2=\rho\left[\left(\frac{v_2^2-v_1^2}{2}\right)+g\left(h_2-h_1\right)\right] \\ & 4100=800\left[\left(\frac{4 v_1^2-v_1^2}{2}\right)+10 \times(0-1)\right] \\ & \frac{41}{8}+10=\frac{3 v_1^2}{2} \\ & \frac{121}{8} \times \frac{2}{3}=v_1^2 \\ & v_1=\sqrt{\frac{121}{4 \times 3} \times \frac{3}{3}} \\ & v_1=\frac{\sqrt{363}}{6} \mathrm{~m} / \mathrm{s} \\ & X=363 \end{aligned} $$
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