The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross sectional areas at A is 1.5 cm2, and B is 25 mm2, if the speed of liquid at B is 60 cm/s, then (PA – PB) is: (Given, PA and PB are liquid pressures at A and B points, Density of liquid= 1000 kg m-3, A and B are on the axis of tube)

Apply Bernoulli's theorem PA+21ρVA2=PB+21ρvB2By any A1V1=A2V2PA−PB=12ρVB21−VA2VB2=12ρVB21−AB2AA2=12.×103(60×10−2)21−251502=175Pa

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross sectional areas at A is 1.5 cm2, and B is 25 mm2, if the speed of liquid at B is 60 cm/s, then (PA – PB) is: (Given, PA and PB are liquid pressures at A and B points, Density of liquid= 1000 kg m-3, A and B are on the axis of tube)

Apply Bernoulli's theorem PA+21ρVA2=PB+21ρvB2By any A1V1=A2V2PA−PB=12ρVB21−VA2VB2=12ρVB21−AB2AA2=12.×103(60×10−2)21−251502=175Pa

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The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross sectional areas at A is 1.5 cm², and B is 25 mm², if the speed of liquid at B is 60 cm/s, then (P_A – P_B) is: (Given, P_A and P_B are liquid pressures at A and B points, Density of liquid= 1000 kg m^-3,A and B are on the axis of tube)

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175 Pa

27 Pa

135 Pa

36 Pa

answer is C.

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Detailed Solution

Apply Bernoulli's theorem $P_{A} +_{2}^{1} {ρV}_{A}^{2} = P_{B} +_{2}^{1} {ρv}_{B}^{2}$
${By any A}_{1} V_{1} = A_{2} V_{2}$
$P_{A} - P_{B} = \frac{1}{2} {ρV}_{B}^{2} (1 - \frac{V_{A}^{2}}{V_{B}^{2}})$

$= \frac{1}{2} {ρV}_{B}^{2} (1 - \frac{A_{B}^{2}}{A_{A}^{2}})$

$= \frac{1}{2} . \times 10^{3} {(60 \times 10^{- 2})}^{2} (1 - {(\frac{25}{150})}^{2})$

$= 175 Pa$

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