**Solution :**

Given :

Velocity of plate, $\quad u=5 \mathrm{~m} / \mathrm{s}$

Length of plate, $\quad L=0.6 \mathrm{~m}$

Width of plate, $\quad b=0.5 \mathrm{~m}$

Density of air, $\quad p=1.24 \mathrm{~kg} / \mathrm{m}^{3}$

Kinematic viscosity, $\quad v=0.15$ stokes

$=0.15 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}$

Reynold number,

$\begin{aligned}
\quad R_{e}&=\frac{u L}{{v}}=\frac{5 \times 0.6}{0.15 \times 10^{-4}}\\
&=200000
\end{aligned}$.

As $R_e$ is less than $5 \times 10^{5}$, hence boundary layer is laminar over the entire length of the plate.

**(i) Thickness of boundary layer at the end of the plate by Blasius's solution is**

$\begin{aligned}
\delta =\frac{4.91 x}{\sqrt{R_{e_{x}}}} &=\frac{4.91 L}{\sqrt{R_{e_{1}}}}\\
&=\frac{4.91 \times 0.6}{\sqrt{200000}}\\
&=0.00658 {~m}\\
&=6.58 {~mm}
\end{aligned}$

**(ii) Drag force on one side of the plate is given by equation as**

$
\begin{aligned}
C_{D} &=\frac{F_{D}}{\frac{1}{2} \rho A u^{2}} \\
F_{D} &=\frac{1}{2} \rho A u^{2} \times C_{D}
\end{aligned}
$

where $C_{D}$ from Blasius's solution,

$\begin {aligned}C_{D}=\frac{1.328}{\sqrt{R_{e}}}&=\frac{1.328}{\sqrt{200000}}\\
&=0.002969\\
&=0.00297 \end{aligned}$

$ \begin{aligned} F_{D} &=\frac{1}{2} \times 1.24 \times 0.6 \times 0.5 \times 5^{2} \times 0.00297 \\
&=0.01373
\end{aligned}$

$\quad\{\because A=L \times b=0.6 \times 0.5\}$