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?

At two points on a horizontal tube of varying cross section carrying w

1.	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» 100c.c, per sec 80 c.c per sec 50 c.c per sec 70 c.c per sec Solution :Using Bernoulli.s equation, `P_(1) + (1)/(2) rho v_(1)^(2) = P_(2) + (1)/(2) rho v_(2)^(2)` …(i) where `rho` is the density of liquid, V its velocity, P its pressure and SUBSCRIPTS 1 and 2 refer to two points. Also `A_(1)v_(1) = A_(2) v_(2)` by equation of continuity ...(II) `P_(1) - P_(2) = rho g xx 4.9` ...(III) From (i) and (iii), we get `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` or `v_(2)^(2) - v_(1)^(2) = 98^(2) CM^(2)//sec^(2)` ...(iv) Using (ii), `(v_(1))/(v_(2)) = (A_(2))/(A_(1)) = (pi xx 0.4^(2))/(pi xx 1^(2)) = 0.16` Substituting `v_(1) = 0.16 xx v_(2)` in (iv), we get `v_(2)^(2) [1-0.16^(2)] = 98^(2) or v_(2) = sqrt((98^(2))/(0.9744))` Quantity of water flowing `=A_(1)v_(1) = A_(2) v_(2) = pi xx 0.4^(2) xx sqrt((98^(2))/(0.9744)) = 50cc`. per sec