Question #8ae4b

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Answer 1 · 2017-03-25T11:30:09

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Given that an ideal fluid flows in the tube as shown figure.

The pressure of the fluid at the bottom $(P_2)$ is equal to the pressure of the fluid at the top $(P_1)$ i.e. $(P_1=P_2)$ )

The difference in height of the top and bottom $h_1-h_2=3m$

The velocity of the fluid at the top $v_1=2m"/"s$

The velocity of the fluid at the bottom $v_2="unknown"$

The area of cross-section of the pipe at the top $A_1$

The area of cross-section of the pipe at the bottom $A_2$

We are to find out the ratio of $A_1:A_2$

Now applying Bernoulli principle

$"Energy per unit volume at the top"= "Energy per unit volume at the bottom"$

we have the following equation

$P_1+1/2rhov_1^2+rhogh_1=P_2+1/2rhov_2^2+rhogh_2$

where

$rho= "density of the fluid"$

$and g = "acceleration due to pop gravity"=10m"/"s^2$

Inserting given values in the above equation we get

$cancelP_2+1/2cancelrho2^2+cancelrho10h_1=cancelP_2+1/2cancelrhov_2^2+cancelrho10h_2$

$=>2+10h_1=1/2v_2^2+10h_2$

$=>2+10h_1-10h_2=1/2v_2^2$

$=>2+10(h_1-h_2)=1/2v_2^2$

$=>2+10xx3=1/2v_2^2$

$=>v_2^2=64$

$=>v_2=sqrt64=8m"/"s$

Now applying principle of continuity of flow of ideal fluid we can write

$A_1/A_2=v_2/v_1=(8m"/"s)/(2m"/"s)=4/1$ which is option (2)