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The diameter of one end of a tube is 2 cm and that of another end is 3 cm . Velocity and pressure of water at narrow end are 2ms^(-1)and1.5xx10^(5)Nm^(9-2) respectively . If the height differnce between narrow and broad ends is 2.5 m , find the velocity and pressure of water at the broad end .(Density of water is 1xx10^(3)kgm^(-3)). The narrow end is higher .

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The <a href="https://interviewquestions.tuteehub.com/tag/narrow-1110745" style="font-weight:bold;" target="_blank" title="Click to know more about NARROW">NARROW</a> end of the flow tube :<br/>`d_(1)=2cm`<br/>`thereforer_(1)=1cm=<a href="https://interviewquestions.tuteehub.com/tag/1xx10-1804645" style="font-weight:bold;" target="_blank" title="Click to know more about 1XX10">1XX10</a>^(-2)m`<br/>`v_(1)=<a href="https://interviewquestions.tuteehub.com/tag/2ms-1837988" style="font-weight:bold;" target="_blank" title="Click to know more about 2MS">2MS</a>^(-1)`<br/>`P_(1)=1.5xx10^(5)Nm^(-2)`<br/>The broad end of the flow tube :<br/>`d_(2)=3cm`<br/>`thereforer_(2)=1.5cm=1.5xx10^(-2)m`<br/>`v_(2)=?,P_(2)=?`<br/>`A_(1)v_(1)=A_(2)v_(2)`<br/>`thereforev_(2)=(A_(1))/(A_(2))v_(1)=(pir_(1)^(2))/(pir_(2)^(2))*v_(1)`<br/>`thereforev_(2)=(r_(1)^(2))/(r_(2)^(2))*v_(1)=((1xx10^(-2))^(2))/((1.5xx10^(-2))^(2))xx2`<br/>`thereforev_(2)=089m//s`<br/>According to Bernoulli.s equation,<br/>`P_(1)+(1)/(2)rhov_(1)^(2)+rhogy_(1)=P_(2)+(1)/(2)rhov_(1)^(2)+rhogy_(2)`<br/>`thereforeP_(2)=P_(1)+(1)/(2)rho(v_(1)^(2)-v_(2)^(2))+rhog(y_(1)-y_(2))`<br/>`=(1.5xx10^(5))+(1)/(2)xx1xx10^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)[(2)^(2)-(0.89)^(2)]+1xx10^(3)xx9.8xx2.5`<br/> `thereforeP_(2)=1.76xx10^(5)N//m^(2)`</body></html>


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