Question

Question #8ae4b

Answer 1

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)