A non-viscous liquid of constant density 1000 kg m -3 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in figure. The areas of cross-section of the tube at P and Q at heights of 2 m and 5 m are respectively 4 × 10-3 m 2 and 8 × 10-3 m 2. The velocity of liquid at point is 1 m s -1. The velocity of liquid at Q is

Question

A non-viscous liquid of constant density 1000 kg m -3 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in figure. The areas of cross-section of the tube at P and Q at heights of 2 m and 5 m are respectively 4 × 10-3 m 2 and 8 × 10-3 m 2. The velocity of liquid at point is 1 m s -1. The velocity of liquid at Q is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8afb1549ba2747ecf88ece439697f564-.png" /> (A) 0.5 m s -1 (B) 1 m s -1 (C) 1.5 m s -1 (D) 2 m s -1

Tardigrade Admin · Accepted Answer

AP vP=AQ vQ ⇒ vQ=(AP vP/AO) =((4 × 10-3 m 2)(1 m s -1)/(8 × 10-3 m 2)) =0.5 m s -1

$0.5 m s^{- 1}$

$1 m s^{- 1}$

$1.5 m s^{- 1}$

$2 m s^{- 1}$

$A_{P} v_{P} = A_{Q} v_{Q}$
$\Rightarrow v_{Q} = \frac{A _{P} v _{P}}{A _{O}}$
$= \frac{( 4 \times 1 0 ^{- 3} m ^{2} ) ( 1 m s ^{- 1} )}{( 8 \times 1 0 ^{- 3} m ^{2} )}$
$= 0.5 m s^{- 1}$

0.5ms−1

1ms−1

1.5ms−1

2ms−1

AP​vP​=AQ​vQ​ ⇒vQ​=AO​AP​vP​​ =(8×10−3m2)(4×10−3m2)(1ms−1)​ =0.5ms−1

$0.5 m s^{- 1}$

$1 m s^{- 1}$

$1.5 m s^{- 1}$

$2 m s^{- 1}$

$A_{P} v_{P} = A_{Q} v_{Q}$
$\Rightarrow v_{Q} = \frac{A _{P} v _{P}}{A _{O}}$
$= \frac{( 4 \times 1 0 ^{- 3} m ^{2} ) ( 1 m s ^{- 1} )}{( 8 \times 1 0 ^{- 3} m ^{2} )}$
$= 0.5 m s^{- 1}$