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The cross-section area of the pipe shown in Fig. is 50 cm^(2)at the wider portions and 20 cm^(2)at the constriction. The rate of flow of water through the pipe is 4000 cm^(3)//s. Find , (i) the velocities at the wide and the narrow portions (ii) the pressure difference between these portions (iii) the difference in height between the mercury columns in the U-tube. |
Answer» <html><body><p><br/></p>Solution :` V_1 = ( 4000 //50)cm s ^(-1) = 0.8 ms ^(-1), v_2 = ( 4000 // <a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> )cm s^(-1)= 2 ms ^(-1) ` <br/>` P_1 - P_2 = rho ( v_2^(2) -v_1^(2))//2 = 1680 <a href="https://interviewquestions.tuteehub.com/tag/pa-1145246" style="font-weight:bold;" target="_blank" title="Click to know more about PA">PA</a>, h= P_1 - P_2 //rhog =1680//13.6 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> ^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)xx 9.8 = 0.013 m.`</body></html> | |