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?

Options

A. 100c.c, per sec
B. 80 c.c per sec
C. 50 c.c per sec
D. 70 c.c per sec

Answer

C

Explanation

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