Question

Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. the separation between the cross sections at A and B is 15/16 cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m/s².

Answer 1

Given:
Separation between the cross sections at A and B = 15/16 cm
Speed of water at A, v_A = 25 cm/s
Speed of water at B, v_B = 50 cm/s
(c) By Bernoulli's equation, we get:

${\displaystyle \frac{1}{2}p{v}_A^2+\rho g{h}_A+P_A=\frac{1}{2}p{v}_B^2+\rho g{h}_B+P_B}$

${\displaystyle \Rightarrow P_A-P_B=\left(\frac{1}{2}\right)p(v_B^2-v_A^2)+pg(h_B-h_A) [h_A-h_B=\left(\frac{15}{16}\right)cm]}$

${\displaystyle \Rightarrow P_A-P_B=\frac{1}{2}(2500-625)-1000\times \frac{15}{16}}$
= 0