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Question

Physics Question on mechanical properties of fluid

Water flows in a horizontal tube (see figure). The pressure of water changes by 700Nm2700\, Nm^{-2} between AA and BB where the area of cross section are 40cm240\, cm^2 and 20cm220\, cm^2, respectively. Find the rate of flow of water through the tube. (density of water = 1000kgm31000\, kgm^{-3})

A

3020cm3/s3020\, cm^3/s

B

2420cm3/s2420\, cm^3/s

C

2720cm3/s2720\, cm^3/s

D

1810cm3/s1810\,cm^3/s

Answer

2720cm3/s2720\, cm^3/s

Explanation

Solution

Rate of flow of water =AAVA=ABVB= A_AV_A = A_BV_B
(40)VA=(20)VB(40)V_A = (20)V_B
VB=2VA......(1)V_B=2V_A\,......(1)
Using Bernoulli's theorem
P+12ρV2=PB+12ρVB2P+\frac{1}{2}\rho V^{2}=P_{B}+\frac{1}{2}\rho V^{2}_{B}
PAPB=12ρ(VB2VA2)P_{A}-P_{B}=\frac{1}{2}\rho\left(V^{2}_{B}-V^{2}_{A}\right)
700=12×1000(4VA2VA2)700=\frac{1}{2}\times1000\left(4V^{2}_{A}-V^{2}_{A}\right)
VA=0.68m/s=68cm/sV_{A}=0.68\,m/s = 68\, cm/s
Rate of flow =AAVA= A_{A}V_{A}
(40)(68)=2720cm3/s-\left(40\right)\left(68\right)=2720\,cm^{3}/s