Suppose the tube in the previous problem is kept vertical with B upward. Water enters through B at the rate of 1 $c m^{3} s^{- 1}$ . Repeat parts (a), (b) and (c). Note that the speed decreases as the water falls down.

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Solution

(a) ${\to V}_{A}$ = 25 cm/sec

(b) ${\to V}_{B}$ = 50 cm/sec

(c) $\frac{1}{2} p v_{B}^{2} + ρ g h_{B} + p_{B}$

= $\frac{1}{2} p v_{A}^{2} + ρ g h_{A} + p_{A}$

$\Rightarrow P_{A} - P_{B} \frac{1}{2} \times 1 \times 1875 + \frac{1}{2} \times 1000 \times \frac{15}{16}$

= $1875 \frac{Dyne}{c m^{2}} = 188 N / m^{2}$

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Suppose the tube in the previous problem is kept vertical with B upward. Water enters through B at the rate of 1 cm3s−1. Repeat parts (a), (b) and (c). Note that the speed decreases as the water falls down.

(a) →VA = 25 cm/sec (b) →VB = 50 cm/sec (c) 12pv2B+ρghB+pB = 12pv2A+ρghA+pA ⇒PA−PB12×1×1875+12×1000×1516 = 1875 Dynecm2=188N/m2

Suppose the tube in the previous problem is kept vertical with B upward. Water enters through B at the rate of 1 $c m^{3} s^{- 1}$ . Repeat parts (a), (b) and (c). Note that the speed decreases as the water falls down.