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Standard XII

Physics

Bernoulli's Principle

A horizontal ...

Question

A horizontal pipeline carries water in a streamline flow. At a point along the tube where the cross sectional area is $10^{- 2} m^{2}$ , the water velocity is 2m/s and the pressure is 8000 Pa. The pressure of water at another point wher cross sectional area is $0.5 \times 10^{- 2} m^{2}$ is

4000Pa

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1000Pa

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2000Pa

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3000Pa

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Solution

The correct option is C 2000Pa
Area at other point is half. So speed will be double.
Now,
$p_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + \frac{1}{2} ρ v_{2}^{2}$
$∴$ $p_{2} = p_{1} + \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2})$
$= (8000) + \frac{1}{2} \times 1000 (4 - 16)$
$= 2000$ Pa

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A horizontal pipeline carries water in a streamline flow. At a point along the tube where the cross sectional area is 10−2m2 , the water velocity is 2m/s and the pressure is 8000 Pa. The pressure of water at another point wher cross sectional area is 0.5×10−2m2 is

The correct option is C 2000PaArea at other point is half. So speed will be double.Now,p1+12ρv21=p2+12ρv22∴ p2=p1+12ρ(v21−v22) =(8000)+12×1000(4−16) =2000 Pa

A horizontal pipeline carries water in a streamline flow. At a point along the tube where the cross sectional area is $10^{- 2} m^{2}$ , the water velocity is 2m/s and the pressure is 8000 Pa. The pressure of water at another point wher cross sectional area is $0.5 \times 10^{- 2} m^{2}$ is

The correct option is C 2000Pa
Area at other point is half. So speed will be double.
Now,
$p_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + \frac{1}{2} ρ v_{2}^{2}$
$∴$ $p_{2} = p_{1} + \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2})$
$= (8000) + \frac{1}{2} \times 1000 (4 - 16)$
$= 2000$ Pa