Answer:

The shear stress (\(\tau\)) in laminar flow through a pipe can be calculated using the formula:

\[ \tau = \mu \frac{du}{dr} \]

where:

- \(\mu\) is the dynamic viscosity of the fluid (given as 2 centipoise),

- \(du/dr\) is the velocity gradient with respect to the radial distance.

Given the velocity distribution \(u = 5 \left(1 - \frac{r^2}{25}\right)\), we can find \(du/dr\) by taking the derivative of \(u\) with respect to \(r\).

\[ \frac{du}{dr} = -\frac{10r}{25} \]

Now, substitute the values into the shear stress formula:

\[ \tau = (2 \, \text{centipoise}) \times \left(-\frac{10r}{25}\right) \]

\[ \tau = -\frac{4}{5} \, r \]

Now, for the resisting force per unit length (\(F\)) in the pipe, you can integrate the shear stress over the cross-sectional area:

\[ F = \int_{0}^{R} \tau \, dA \]

where \(R\) is the radius of the pipe.

Substitute the expression for \(\tau\) and the area \(A = \pi r^2\) into the integral:

\[ F = \int_{0}^{R} -\frac{4}{5} \, r \, \pi r^2 \, dr \]

After integrating, you'll get the resisting force per unit length. Multiply it by 1000 to convert it to per kilometer length.

Please note that the actual calculations involve integration, and the result will depend on the limits of integration and the final expression after integration. If you have a specific range for \(r\), you can substitute those limits accordingly.

Explanation:

please brainlist to me bro