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If cross sectional area of limb I and that of limb II is A2, then the velocity of the liquid in the tube will be,


 

Option: 1

\sqrt{2 g(x-y)}


Option: 2

\frac{A_1}{A_2} \sqrt{2 g(x-y)}


Option: 3

\frac{A_2}{A_1} \sqrt{2 g(x-y)}


Option: 4

None of the above


Answers (1)

best_answer

P_2=P_3 \,\,\,and\,\,\, P_1>P_2

\rho_1 =\rho_0+\rho g x \\

\rho_2 =\rho_0+\rho g y \\

\therefore \Delta \rho =\rho_1-\rho_2=\rho g(x-y)

Between 1 and 2,

Difference in pressure energy = difference in kinetic energy. 

\rho g (x-y)=\frac{1}{2} \rho v^2 \text { or } v=\sqrt{2 g(x-y)}
 

Posted by

avinash.dongre

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