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 $\frac{15}{16}$ cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m $s^{- 2}$ .

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Solution

(a) $v_{A}$ = 25 cm/sec

(b) $v_{B}$ = 50 cm/sec

(c) From Bernoullis equation,

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

$\Rightarrow P_{A} - P_{B} = (\frac{1}{2}) p (v_{B}^{2} - v_{A}^{2}) + ρ g (h_{B} - h_{A})$

[Given $h_{A} - h_{B} = (\frac{15}{16}) c m]$

$\Rightarrow P_{A} - P_{B} = \frac{1}{2} (2500 - 625) - 1000 \times \frac{15}{16}$

= 0

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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 1516 cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 ms−2.

(a) vA = 25 cm/sec (b) vB = 50 cm/sec (c) From Bernoullis equation, 12pv2A+ρghA+pA=12pv2B+ρghB+pB ⇒PA−PB=(12)p(v2B−v2A)+ρg(hB−hA) [Given hA−hB=(1516)cm] ⇒PA−PB=12(2500−625)−1000×1516 = 0

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 $\frac{15}{16}$ cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m $s^{- 2}$ .