$$\begin{aligned} & \text { Let internal radius }=r \text { and external radius }=R \\ & \therefore \text { Volume of hollow spherical shell, } V=\frac{4}{3} \pi\left(R^3-r^3\right) \\ & \Rightarrow \quad V=\frac{4}{3} \pi\left[(3.0005)^3-(3)^3\right]\quad\text{.... (i)} \end{aligned}$$

Now, we shall use differentiation to get approximate value of $(3.0005)^3$.

Let $$\quad(3.0005)^3=y+\Delta y$$

$$\begin{aligned} \text{and }\quad & x=3, \Delta x=0.0005 \\ \text{Also, let}\quad & y=x^3 \end{aligned}$$

$$\begin{aligned} &\text { On differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d y}{d x} & =3 x^2 \\ \therefore\quad \Delta y & =\frac{d y}{d x} \times \Delta x=3 x^2 \times 0.0005 \\ & =3 \times 3^2 \times 0.0005 \\ & =27 \times 0.0005=0.0135 \end{aligned} \end{aligned}$$

$$ \begin{aligned} &\begin{aligned} \text{Also,}\quad(3.0005)^3 & =y+\Delta y \\ & =3^3+0.0135=27.0135 \\ \therefore\quad V & =\frac{4}{3} \pi[27.0135-27.000] \quad \text { [using Eq. (i)] }\\ & =\frac{4}{3} \pi[0.0135]=4 \pi \times(0.0045) \\ & =0.0180 \pi \mathrm{~cm}^3 \end{aligned}\\ \end{aligned}$$

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

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

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