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A cylindrical vessel has three identical horizontal tubes of length 40 cm, each coming out at heights 0, 5 and 9 cm, respectively. What will be the length of a single overflow tube of the same radius as that of identical tubes which can replace the three tubes if placed at the bottom of the vessel if the level of liquid in the cylindrical vessel is kept constant at 20 cm. |
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Answer» Solution :Let the three tubes be P, Q and R. Height of tube P=0 cm from bottom Pressure head for tube, `h_p= 20 cm` of water column Height of tube Q = 5 cm from bottom Pressure head for tube Q. `h_Q= 20 -5= 15` cm of water column Height of tube R =9 cm from bottom Pressure head for tube R, `h_r=` 20 - 9 =11 cm of water column Let r cm be the radius of each tube. Given, length of each tube, l= 40 cm Rate of flow of liquid through DIFFERENT tubes will be `Q_(p)=(pi p_(1) r_(1)^(4))/(8 eta n)=(pi x h_(p) pg xx r^(4))/(8 eta l)` `=(pi xx 20 xx pg xx r^(4))/(8 eta xx 40)` Similarlry, `Q_(Q)=(pi x h_(Q) pg xx r^(4))/(8 eta l)=(pi xx 15 xx pg xx r^(4))/(8 eta xx 40)` `Q_(R)=(pi xx h_(R) pg xx r^(4))/(8 eta l)=(pi xx 1 xx pg xx r^(4))/(8 eta xx 40)` Total VOLUME of liquid FLOWING through the three tubes P,Q and R is `Q=Q_(P)+Q_(Q)+Q_(R)` `=(pi xx pg r^(4))/(8 eta xx 40) xx 46` If l. is the length of the required single tubem then `Q.=(pi h pg r^(4))/(8 eta l.)=(pi xx 20 xx pg r^(4))/(8 eta l.)` Since Q.=Q `RARR (pi xx 20 xx pg r^(4))/(8 eta l.)=(pi xx pg r^(4) xx 46)/(8 xx 40 eta)` `rArr (20)/(l.)=(46)/(40)` `rArr l.=(20 xx 40)/(46) =17.39cm` |
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