Let $A B$ be the street light post and $C D$ be the height of man i.e., $C D=2 \mathrm{~m}$.

Let $B C=x \mathrm{~m}, C E=y \mathrm{~m}$ and $\frac{d x}{d t}=\frac{-5}{3} \mathrm{~m} / \mathrm{s}$

From $\triangle A B E$ and $\triangle D C E$, we see that

$\triangle A B E \sim \triangle D C E\quad$ [by AAA similarity]

$$\begin{array}{ll} \therefore & \frac{A B}{D C}=\frac{B E}{C E} \Rightarrow \frac{\frac{16}{3}}{2}=\frac{x+y}{y} \\ \Rightarrow & \frac{16}{6}=\frac{x+y}{y} \\ \Rightarrow & 16 y=6 x+6 y \Rightarrow 10 y=6 x \\ \Rightarrow & y=\frac{3}{5} x \end{array}$$

On differentiating both sides w.r.t. $t$, we get

$$\frac{d y}{d t}=\frac{3}{5} \cdot \frac{d x}{d t}=\frac{3}{5} \cdot\left(-1 \frac{2}{3}\right)\quad$$ [since, man is moving towards the light post]

$$=\frac{3}{5} \cdot\left(\frac{-5}{3}\right)=-1 \mathrm{~m} / \mathrm{s}$$

Let $$z=x+y$$

Now, differentiating both sides w.r.t. $t$, we get

$$\begin{aligned} \frac{d z}{d t} & =\frac{d x}{d t}+\frac{d y}{d t}=-\left(\frac{5}{3}+1\right) \\ & =-\frac{8}{3}=-2 \frac{2}{3} \mathrm{~m} / \mathrm{s} \end{aligned}$$

Hence, the tip of shadow is moving at the rate of $2 \frac{2}{3} \mathrm{~m} / \mathrm{s}$ towards the light source and length of the shadow is decreasing at the rate of $1 \mathrm{~m} / \mathrm{s}$.

Let $L$ represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain, then

$$L=200(10-t)^2$$

$\therefore$ Rate at which the water is running out $=-\frac{d L}{d t}$

$$\begin{aligned} \frac{d L}{d t} & =-200 \cdot 2(10-t) \cdot(-1) \\ & =400(10-t) \end{aligned}$$

$$\begin{aligned} &\text { Rate at which the water is running out at the end of } 5 \mathrm{~s}\\ &\begin{aligned} & =400(10-5) \\ & =2000 \mathrm{~L} / \mathrm{s}=\text { Final rate } \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Since, } \quad \text { initial rate }=-\left(\frac{d L}{d t}\right)_{t=0}=4000 \mathrm{~L} / \mathrm{s}\\ &\begin{aligned} \therefore \quad \text { Average rate during } 5 \mathrm{~s} & =\frac{\text { Initial rate }+ \text { Final rate }}{2} \\ & =\frac{4000+2000}{2} \\ & =3000 \mathrm{~L} / \mathrm{s} \end{aligned} \end{aligned}$$

Let the side of a cube be $x$ unit.

$$\therefore \quad \text { Volume of cube }(V)=x^3$$

On differentiating both side w.r.t. $t$, we get

$$\frac{d V}{d t}=3 x^2 \frac{d x}{d t}=k\quad$$ [constant]

$\Rightarrow \quad \frac{d x}{d t}=\frac{k}{3 x^2}\quad\text{.... (i)}$

Also, surface area of cube, $$S=6 x^2$$

On differentiating w.r.t. $t$, we get

$$\begin{array}{ll} & \frac{d S}{d t}=12 x \cdot \frac{d x}{d t} \\ \Rightarrow & \frac{d S}{d t}=12 x \cdot \frac{k}{3 x^2} \quad\text{[using Eq. (i)]}\\ \Rightarrow & \frac{d S}{d t}=\frac{12 k}{3 x}=4\left(\frac{k}{x}\right) \\ \Rightarrow & \frac{d S}{d t} \propto \frac{1}{x} \end{array}$$

Hence, the surface area of the cube varies inversely as the length of the side.

Since, $x$ and $y$ are the sides of two squares such that $y=x-x^2$.

$\therefore \quad$ Area of the first square $\left(A_1\right)=x^2$

and area of the second square $\left(A_2\right)=y^2=\left(x-x^2\right)^2$

$$\therefore \quad \frac{d A_2}{d t}=\frac{d}{d t}\left(x-x^2\right)^2=2\left(x-x^2\right)\left(\frac{d x}{d t}-2 x \cdot \frac{d x}{d t}\right)$$

$=\frac{d x}{d t}(1-2 x) 2\left(x-x^2\right)$

and $$\frac{d A_1}{d t}=\frac{d}{d t} x^2=2x \cdot \frac{d x}{d t}$$

$\therefore\quad \frac{d A_2}{d A_1}=\frac{d A_2 / d t}{d A_1 / d t}=\frac{\frac{d x}{d t} \cdot(1-2 x)\left(2 x-2 x^2\right)}{2 x \cdot \frac{d x}{d t}}$

$$\begin{aligned} & =\frac{(1-2 x) 2 x(1-x)}{2 x} \\ & =(1-2 x)(1-x) \\ & =1-x-2 x+2 x^2 \\ & =2 x^2-3 x+1 \end{aligned}$$

$$\begin{array}{rlrl} & \text { Given, equation of curves are } & 2 x =y^2 \quad\text{.... (i)}\\ & \text { and } & 2 x y =k \quad\text{.... (ii)}\\ & \Rightarrow & y =\frac{k}{2 x}\quad\text{[from Eq. (ii)]} \end{array}$$

$$\begin{array}{lrl} \text { From Eq. (i), } & 2 x & =\left(\frac{k}{2 x}\right)^2 \\ \Rightarrow & 8 x^3 & =k^2 \\ \Rightarrow & x^3 & =\frac{1}{8} k^2 \\ \Rightarrow & x & =\frac{1}{2} k^{2 / 3} \\ \therefore & y & =\frac{k}{2 x}=\frac{k}{2 \cdot \frac{1}{2} k^{2 / 3}}=k^{1 / 3} \end{array}$$

Thus, we get point of intersection of curves which is $\left(\frac{1}{2} k^{2 / 3}, k^{1 / 3}\right)$.

From Eqs. (i) and (ii),

$$2=2 y \frac{d y}{d x}$$

and $$2\left[x \cdot \frac{d y}{d x}+y \cdot 1\right]=0$$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}=\frac{1}{y} \\ \text { and } & \left(\frac{d y}{d x}\right)=\frac{-2 y}{2 x}=-\frac{y}{x} \end{array}$$

$\Rightarrow\quad\left(\frac{d y}{d x}\right)_{\left(\frac{1}{2} k^{2 / 3}, k^{1 / 3}\right)}=\frac{1}{k^{1 / 3}}\quad \text { [say } m_1 \text { ] }$

$\text{and}\quad\left(\frac{d y}{d x}\right)_{\left(\frac{1}{2} k^{2 / 3}, k^{1 / 3}\right)}=\frac{-k^{1 / 3}}{\frac{1}{2} k^{2 / 3}}=-2 k^{-1 / 3}\quad \text { [say } m_2 \text { ] }$

Since, the curves intersect orthogonally.

$$\begin{array}{lrl} \text { i.e., } & m_1 \cdot m_2 & =-1 \\ \Rightarrow & \frac{1}{k^{1 / 3}} \cdot\left(-2 k^{-1 / 3}\right) & =-1 \end{array}$$

$$\begin{array}{lr} \Rightarrow & -2 k^{-2 / 3}=-1 \\ \Rightarrow & \frac{2}{k^{2 / 3}}=1 \\ \Rightarrow & k^{2 / 3}=2 \\ \therefore & k^2=8 \end{array}$$

which is the required condition.