91Ó°ÊÓ

Radii for input and output sections are shown in the figure.

This problem is based on the work-energy theorem. The input & output section of the pipe is placed at the same level, so work done is due to a change in velocity only.

Formula:

The net work done by the body due to change in energy, $W=∆PE+∆KE$ (i)

From the equation of continuity through a pipe, $A_i{v}_i=A_0{v}_0$ (ii)

According to work-energy theorem that is from equation (i), we can get the net work as follows:

$$\begin{gathered} W=∆K.E. \\ =\left(\frac{1}{2}m{v}_0^2\right)-\left(\frac{1}{2}m{v}_i^2\right).................\left(a\right) \end{gathered}$$

Now, using the velocity equation from equation (ii), we get that

$v_0=A_i{v}_i/A_0$

Now, substituting the above value in equation (a), we get the net work equation as follows:

role="math" localid="1657250271274" $$\begin{gathered} W=\left(\frac{A_i}{A_0}\right)\left(\frac{1}{2}m{v}_i^2\right)-\left(\frac{1}{2}m{v}_i^2\right) \\ =K.E{.}_i\left(\frac{r_i^4}{r_0^4}\right).........................\left(b\right)(âˆ\mu A={\mathrm{Ï€°ù}}^2) \end{gathered}$$

For cases1and4, input area is the same as the output area.

Thus, velocities would be the same considering equation (b), that is$W_1=W_4=0J.$

Therefore, the net work done for cases 1 AND 4 is zero.

For case 2:$A_i>A_0$

Thus, from equation (b), we can see that the value of work done is positive.

Therefore, work done would be positive for situation 2.

For case 3: $A_I<A_0$

Thus, from equation (b), we can see that the value of work done is negative.

Therefore, work done would be negative for situation 3.