JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 27)
An ideal fluid of density 800 kgm$$-$$3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $${a \over 2}$$. The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $${{\sqrt x } \over 6}$$ ms$$-$$1 for x = ___________. (Given g = 10 ms$$-$$2)
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Explanation
From Bernoulli's equation
$${P_1} + {1 \over 2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + {1 \over 2}\rho {v_2}^2 + \rho g{h_2}$$
$${P_1} - {P_2} + \rho g({h_1} - {h_2}) = {1 \over 2}\rho ({v_2}^2 - {v_1}^2)$$ ...... (i)
Also, from equation of continuity
$${A_1}{v_1} = {A_2}{v_2}$$
$$A{v_1} = {A \over 2}{v_2}$$
$${v_2} = 2{v_1}$$ ...... (ii)
put equation (ii) in (i),
$$4100 \times 800 \times 10 \times 1 = {1 \over 2} \times 800 \times (4{v_1}^2 - {v_1}^2)$$
$$4100 + 8000 = 400 \times 3{v_1}^2$$
$${v_1}^2 = {{12100} \over {3 \times 400}} = {{121} \over {12}}$$
$${v_1} = \sqrt {{{121} \over {12}}} $$
Now, $${{\sqrt x } \over 6} = \sqrt {{{121} \over {12}}} $$
$${x \over {36}} = {{121} \over {12}}$$
$$x = 121 \times 3 = 363$$
$$\therefore$$ $$x = 363$$
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