QJEEM22Ph612d: JEE Main 2022 Physics - Class 11 JEEM22Ph612d Solution: An ideal fluid of density 800 kgm 3 , flows smoothly through a…

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, ${{\sqrt x } \over 6} = \sqrt {{{121} \over {12}}}$

${x \over {36}} = {{121} \over {12}}$

$x = 121 \times 3 = 363$

$\therefore$ $x = 363$