Question

Water is flowing steadily through a horizontal tube of non-uniform cross-section.If the pressure of water is $4\times {10}^{4}N/m{m}^2$ at a point when cross section is $0.02m^2$ velocity of flow is 2m/s what is pressure at a point where cross section reduces to $0.01m^2$.

• A. $1.4\times {10}^{4}N{m}^2$
• B. $3.4\times {10}^{4}N{m}^2$
• C. $2.4\times {10}^{-4}N{m}^2$
• D. none of these

Answer 1

The correct option is B $3.4\times {10}^{4}N{m}^2$

As we know

$A_1{V}_1=A_2{V}_2$

Where $A_1=0.02m^2$

$V_1=2m/s$

$A_2=0.01m^2$

$⟹0.02\times 2=0.01\times V_2$

$⟹V_2=4m/s$

Now using Bernoulli's theorem

$\frac{P_1}{\rho }+\frac{1}{2}{V}_1^2=\frac{P_2}{\rho }+\frac{1}{2}{v}_2^2$

$⟹P_2=P_1+\frac{{\rho }_1}{2}[V_1^2-V_1^2]$

$⟹P_2=4\times {10}^{10}+\frac{1000}{2}[2^2-4^2]-4\times {10}^{10}=3.4\times {10}^{4}N/m^2$