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Find out the value of Sin \theta in the given figure where a zig-zag tube open at N having liquids of densities \rho_1, \rho_2, \rho_3  and  is placed in a vertical plane as shown in figure. (The pressure at  M is equal to atmospheric pressure.)

 

                                                

Option: 1

\sin \theta=\frac{\rho_{2} h}{\left(\rho_{3}-\rho_{1}\right) l}


Option: 2

Sin \theta=\frac{\rho_{3} h}{\left(\rho_{3}-\rho_{1}\right) l}


Option: 3

\sin \theta=\frac{\rho_{1} h}{\left(\rho_{3}-\rho_{2}\right) l}


Option: 4

\sin \theta=\frac{\rho_{2} h}{\left(\rho_{3}-\rho_{2}\right) l}


Answers (1)

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           \text { Pressure at } A \text { is } P_{A}=P_{\text {atm }}-\rho_{1} g l \sin \theta

           \begin{array}{l}{\text { Pressure at } B \text { is } P_{B}=P_{\text {atm}}+\rho_{2} g h} \\ {\text { But } P_{\mathrm{B}} \text { is also given by } P_{B}=P_{A}+\rho_{3} g l \sin \theta} \\ {\text { Hence, } P_{\text {atm }}+\rho_{2} g h=P_{A}+\rho_{3} g l \sin \theta}\end{array}

          \begin{array}{l}{P_{\text {atm }}+\rho_{2} g h=P_{\text {atm }}-\rho_1 g l \sin \theta+\rho_{3} g l \sin \theta} \\ \\ {\sin \theta=\frac{\rho_{2} h}{\left(\rho_{3}-\rho_{1}\right) l}}\end{array}

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SANGALDEEP SINGH

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