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Standard XII

Physics

Floating

An incompress...

Question

An incompressible fluid is flowing in a cylindrical pipe of internal diameter $4 c m$ with a velocity of $5 m / s$ . If this tube is joined with another tube of internal diameter $2 c m$ , then the velocity of flow of fluid in the smaller tube will be

10 m / s

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20 m / s

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40 m / s

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5 m / s

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Solution

The correct option is B $20 m / s$
According to equation of continuity for incompressible fluid
$A_{1} V_{1} = A_{2} V_{2}$
Thus ${R_{1}}^{2} V_{1} = {R_{2}}^{2} V_{2}$ $(A r e a = π R^{2})$
$\frac{2^{2}}{100^{2}} \times 5 = \frac{1^{2}}{100^{2}} \times V_{2}$
$V_{2} = 20 m / s$

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An incompressible fluid is flowing in a cylindrical pipe of internal diameter 4 cm with a velocity of 5 m/s. If this tube is joined with another tube of internal diameter 2 cm, then the velocity of flow of fluid in the smaller tube will be

The correct option is B 20 m/sAccording to equation of continuity for incompressible fluid A1V1=A2V2 Thus R12V1=R22V2 (Area=πR2) 221002×5=121002×V2 V2=20 m/s

An incompressible fluid is flowing in a cylindrical pipe of internal diameter $4 c m$ with a velocity of $5 m / s$ . If this tube is joined with another tube of internal diameter $2 c m$ , then the velocity of flow of fluid in the smaller tube will be

The correct option is B $20 m / s$
According to equation of continuity for incompressible fluid
$A_{1} V_{1} = A_{2} V_{2}$
Thus ${R_{1}}^{2} V_{1} = {R_{2}}^{2} V_{2}$ $(A r e a = π R^{2})$
$\frac{2^{2}}{100^{2}} \times 5 = \frac{1^{2}}{100^{2}} \times V_{2}$
$V_{2} = 20 m / s$