Question

Water flows in a horizontal tube (see figure). The pressure of water changes by $700\, Nm^{-2}$ between $A$ and $B$ where the area of cross section are $40\, cm^2$ and $20\, cm^2$, respectively. Find the rate of flow of water through the tube. (density of water = $1000\, kgm^{-3}$)

• A. $3020\, cm^3/s$
• B. $2420\, cm^3/s$
• C. $2720\, cm^3/s$
• D. $1810\,cm^3/s$

Answer 1

The Correct Option is C

Rate of flow of water $= A_AV_A = A_BV_B$
$(40)V_A = (20)V_B$
$V_B=2V_A\,......(1)$
Using Bernoulli's theorem
$P+\frac{1}{2}\rho V^{2}=P_{B}+\frac{1}{2}\rho V^{2}_{B}$
$P_{A}-P_{B}=\frac{1}{2}\rho\left(V^{2}_{B}-V^{2}_{A}\right)$
$700=\frac{1}{2}\times1000\left(4V^{2}_{A}-V^{2}_{A}\right)$
$V_{A}=0.68\,m/s = 68\, cm/s$
Rate of flow $= A_{A}V_{A}$
$-\left(40\right)\left(68\right)=2720\,cm^{3}/s$