Given that, the radius of the sphere is $r$.
Since the cricket ball fits exactly inside the cylinder tin, the height of the cylinder ($h$) will be equal to the diameter of the ball .
$\Rightarrow h=2r$
Ratio of their surface areas
$=\frac{Surfaceareaofball}{Surfaceareaofcylindricaltin}$
$=\frac{4\pi r^2}{2\pi r(r+h)}$
$=\frac{2r}{r+h}$
$=\frac{2r}{r+2r}$
$=\frac{2}{3}$ 
$=2:3$
Hence, required ratio is 2:3

Let the radius of the old cylinder be $R$ and that of the new cylinder be $r$.
$\text{Then,}r=\frac{R}{2}$
$\text{Volume of old cylinder}=\pi R^{2}h$
$\text{Volume of new cylinde}r=\pi r^{2}h$
$=\pi \times (\frac{R}{2}{)}^2\times h$
$=\frac{\pi \times R^2\times h}{4}$
Hence, reduction in volume
$=\text{Volume of old cylinder - volume of new cylinder}$
$=\pi R^{2}h-\frac{\pi \times R^2\times h}{4}=\frac{3}{4}\pi R^{2}h$
$\therefore$ Percentage reduction
$=\frac{\text{Reduction in volume}}{\text{Volume of old cylinder}}\times 100$
$=\frac{\frac{3}{4}\pi R^{2}h}{\pi R^{2}h}\times 100$
$=75\%$

Volume of the big iron ball will be equal to the volume of  27 small iron balls.
Let the radius of the big sphere be R cm.
Let the radius of the small sphere be
r cm = 5 cm.

$\Rightarrow \frac{4}{3}\times \pi R^3=27\times (\frac{4}{3}\times \pi r^3)$
$\Rightarrow R^3=27r^3$

$R=3r=3\times 5=15$ cm.

The area of aluminium sheet = Curved surface area of flask
$\Rightarrow 550m^2=\pi rl$
$\Rightarrow 550=\frac{22}{7}\times 7\times l$ 
$\Rightarrow 550=22l$ 
$\Rightarrow l=25m$ 

Now, height of flask
$\Rightarrow h=\sqrt{(l^2-r^2)}$
$\Rightarrow h=\sqrt{{25}^2-7^2}$
$\Rightarrow h=24m$
 

Hence, the volume of conical flask
$=\frac{1}{3}$ $\pi$$r^{2}h$
$=\frac{1}{3}\times \frac{22}{7}\times 7^2\times 24$
$=1232m^3$ 

We know that 1 m ${}^3=1000litres$ 

So, volume in litres $=1232000litres$

Given that,
Radius of the tube = 3.5 $cm$
Length of the tube = 1 m = 100 $cm$
Curved surface area of cylindrical tube
$=2\pi rh$
$=2\times \frac{22}{7}\times 3.5\times 100$    
$=2\times 22\times 0.5\times 100$
 $=2200c{m}^2$ 
Hence, the required area of the aluminium sheet is 2200 $c{m}^2$.

Water that will flow out will be equal to the volume of spherical ball dropped into the vessel.
So,

$\frac{4}{3}$ $\pi$  $r^3$ = $\pi$  $R^2$ H

$\frac{4}{3}$ $\times$ $\frac{22}{7}$ $\times$ ${\left(0.7\right)}^3$ = $\frac{22}{7}$ $\times$ ${\left(7\right)}^2$ $\times H$

$\Rightarrow H=0.933333cm$
$1m={10}^{-6}\mu m$

$\therefore H=9333.33\mu m$

$\frac{r}{h}$ = $\frac{4}{3}$

Base area = $\pi$$r^2=154$

$\Rightarrow r=7cm$  so $h=\frac{3}{4}\times 7=\frac{21}{4}$
$\Rightarrow l=\sqrt{(h^2+r^2)}=\frac{35}{4}$

$\Rightarrow$ So, curved surface area = $\pi$$rl$

$\Rightarrow$  $\frac{22}{7}\times \left(7\right)\times \frac{35}{4}=192.5c{m}^2$

The sides will be $x,2x,3x$

Total surface area $=2(x.2x+2x.3x+x.3x)$
$\Rightarrow 2(2x^2+6x^2+3x^2)=88$

        $\Rightarrow 22x^2=88$

        $\Rightarrow x=2$

So sides are 2 cm, 4 cm and 6 cm

So volume $=2\times 4\times 6=48m^3$

Given,
Radius $\left(r\right)$ = 7 $\text{cm}$
Slant height $\left(l\right)$ = 10 $\text{cm}$

Area of the aluminium sheet required
=Total surface area of the cone
= $\pi r(l+r)$
= $\frac{22}{7}\times 7(10+7)$
= 374 ${\text{cm}}^2$
Hence, the area of the aluminium sheet required is 374 ${\text{cm}}^2.$