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At two points on a horizontal tube of varying cross section carrying water, the radii are 1cm and 0.4cm. The pressure difference between these points is 4.9cm of water. How much liquid flows through the tube per second?

Answer» <html><body><p>100c.c, per sec<br/>80 c.c per sec<br/>50 c.c per sec<br/>70 c.c per sec</p>Solution :Using Bernoulli.s equation, `P_(1) + (1)/(2) rho v_(1)^(2) = P_(2) + (1)/(2) rho v_(2)^(2)` …(i) <br/>where `rho` is the density of liquid, <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> its velocity, P its pressure and <a href="https://interviewquestions.tuteehub.com/tag/subscripts-1231334" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSCRIPTS">SUBSCRIPTS</a> 1 and 2 refer to two points. Also `A_(1)v_(1) = A_(2) v_(2)` by equation of continuity ...(<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) <br/> `P_(1) - P_(2) = rho g xx 4.9` ...(<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) <br/> From (i) and (iii), we get <br/>`v_(2)^(2) - v_(1)^(2) = (2 (P_(1) -P_(2)))/(rho) = (2rho g xx 4.9)/(rho)= (2g) xx 4.9 = 2 xx 980 xx 4.9` <br/> or `v_(2)^(2) - v_(1)^(2) = 98^(2) <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>^(2)//sec^(2)` ...(iv) <br/> Using (ii), `(v_(1))/(v_(2)) = (A_(2))/(A_(1)) = (pi xx 0.4^(2))/(pi xx 1^(2)) = 0.16` <br/> Substituting `v_(1) = 0.16 xx v_(2)` in (iv), we get <br/> `v_(2)^(2) [1-0.16^(2)] = 98^(2) or v_(2) = sqrt((98^(2))/(0.9744))` <br/>Quantity of water flowing <br/> `=A_(1)v_(1) = A_(2) v_(2) = pi xx 0.4^(2) xx sqrt((98^(2))/(0.9744)) = 50cc`. per sec</body></html>


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