Water (ideal fluid) is flowing through a horizontal tube having cross sectional areas of its two ends being A and A' such that the ratio A/A' is 5.If the pressure difference of water between the two ends is 3×105Nm−2, the velocity of water with which it enters the tube will be (neglect gravity effects)
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a
5ms−1
b
10ms−1
c
25ms−1
d
5010 ms−1
answer is A.
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Detailed Solution
According to Bernouli’s theoremP1+12ρv12=P2+12ρv22 From equation ,P1−P2=3×105,A1A2=5 According to equation of continuity A1v1=A2v2 Or, A1A2=v2v1=5⇒v2=5v1From equation (i)P1−P2=12ρ(v22−v12) Or 3×105=12×1000(25v12−v12)⇒600=6v1×4v1 ⇒v12=25 ∴v1=5m/s