JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 26)
A liquid of density 750 kgm$$-$$3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A1 = 1.2 $$\times$$ 10$$-$$2 m2 to A2 = $${{{A_1}} \over 2}$$. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is ___________ $$\times$$ 10$$-$$3 m3s$$-$$1.
Answer
24
Explanation
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Using Bernoulli's equation
$${P_1} + {1 \over 2}\rho {v^2} = {P_2} + {1 \over 2}\rho 4{v^2}$$
$${3 \over 2}\rho {v^2} = {P_1} - {P_2}$$
$$ \Rightarrow v = \sqrt {{{2({P_1} - {P_2})} \over {3\rho }}} $$
$$ = \sqrt {{{2 \times 4500} \over {3 \times 750}}} = 2$$ m/sec
So $$Q = {A_1}v = 24 \times {10^{ - 3}}$$ m3/sec
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