Answer :

Correct answer is : 8

$K.E={\int }_o^R\frac{1}{2}\left(\rho v2\pi r.dr\right)v^2={\int }_o^R\frac{1}{2}\rho v^{3}2\pi r.dr$

$K.E={\int }_o^R\frac{1}{2}\rho 2\pi r.{\left[-C\left(r^2-\frac{D^2}{4}\right)\right]}^{3}dr$

$\because v=-C\left(r^2-\frac{D^2}{4}\right)=-C\left(r^2-R^2\right)=C\left(R^2-r^2\right)$

$\therefore K.E={\int }_o^R\rho \pi r{\left[C\left(R^2-r^2\right)\right]}^{3}dr$

$\therefore K.E={\int }_o^R\rho \pi C^3{R}^6{\left[\left(1-\frac{r^2}{R^2}\right)\right]}^{3}rdr$

Now,

∵ (a - b)³ = a³ – b³ – 3ab (a - b)

$\therefore K.E=\rho \pi C^3{R}^6{\int }_o^R{\left(1-\frac{r^2}{R^2}\right)}^{3}rdr$

$K.E=\rho C^3\pi R^6\underset{0}{\overset{R}{\int }}\left[1-\frac{r^6}{R^6}-\frac{3r^2}{R^2}\left(1-\frac{r^2}{R^2}\right)\right]rdr$

$K.E=\rho C^3\pi R^6\underset{0}{\overset{R}{\int }}\left[r-\frac{r^7}{R^6}-\frac{3r^3}{R^2}\left(1-\frac{r^2}{R^2}\right)\right]dr$

$K.E=\rho C^3\pi R^6{\left[\frac{r^2}{2}-\frac{r^8}{8R^6}-\frac{3}{R^2}\cdot \frac{r^4}{4}+\frac{3}{R^4}\cdot \frac{r^6}{6}\right]}_0^R$

$K.E=\rho C^3\pi R^6\left[\frac{R^2}{2}-\frac{R^2}{8}-\frac{3}{4}{R}^2+\frac{1}{2}{R}^2-0\right]$

$K.E=\rho C^3\pi R^6\left[\frac{4R^2-R^2-6R^2+4R^2}{8}\right]=\rho C^3\pi R^6\cdot \frac{R^2}{8}=\frac{\rho C^3\pi }{8}{R}^8$

$K.E=\frac{\rho C^3\pi }{8}\frac{D^8}{256}$