Volume of iron
$=(440\times 260\times 100)c{m}^3$.
Internal radius of the pipe
$=30cm$.
External radius of the pipe
$=(30+5)cm=35cm$.
Let the length of the pipe be $hcm$.
Volume of iron in the pipe = (External volume) - (Internal volume)
$=\left[\pi \right(35{)}^2\times h-\pi (30{)}^2\times h]c{m}^3$  

$=\left(\pi h\right)\left((35{)}^2-(30{)}^2\right)c{m}^3$ 

$=(65\times 5)\pi hc{m}^3$ 
Volume of iron in the pipe = Volume of iron block

$\Rightarrow \left(325\pi h\right)c{m}^3=440\times 260\times 100$

$h=\frac{440\times 260\times 100}{325}\times \frac{7}{22}cm$
$h=11200cm$

$h=\frac{11200}{100}=112m$

Hence, the length of the pipe is 112 m.

Height of Cylinder $=h=AO=2.4\text{cm}$

Diameter of Cylinder $=1.4\text{cm}$

Radius of Cone = Radius of Cylinder
$OB=r=\frac{1.4}{2}=0.7\text{cm}$

Lets find Slant Height $l$ of cone, using pythagoras theorem on $△AOB$ , we get

$AB=l=\sqrt{h^2+r^2}=\sqrt{(2.4{)}^2+(0.7{)}^2}$
$l=\sqrt{5.76+0.49}=\sqrt{6.25}=2.5\text{cm}$

Surface Area of remaining Solid
= Surface Area of the Cylinder $+$ Inner surface Area of the hollow Cone 
          $=(2\pi rh+\pi r^2)+\pi rl=\pi r(2h+r+l)=\frac{22}{7}\times 0.7(2\times 2.4+0.7+2.5)=2.2(4.8+0.7+2.5)=2.2\left(8\right)=17.60{\text{cm}}^2$ 

$\therefore$ Total surface area of the remaining solid is $17.60{\text{cm}}^2$ 

Let the depth of the salt in the cellar be ${}^{\prime }{x}^{\prime }$
Volume of   =   volume of cylinder +
salt cellar          volume of hemisphere
                            
                    $=\pi r^{2}h+\frac{2}{3}\pi r^3=\pi \times 3^2\times 10+\frac{2}{3}\pi \times 3^3=\pi [90+18]=108\pi {\text{cm}}^3$
So height $x$ will come on the cylinder.
Half of total volume $=54\pi {\text{cm}}^3$
                         $\pi r^{2}x=54\pi \pi \times 3^2\times x=54\pi 9x=54\Rightarrow x=6$
$\therefore$ The salt will be 6 cm deep.

Surface area = base circle + curved surface area of the cylinder + curved surface area of the cone.
$=\pi r^2+2\pi rh+\pi rl$

$=\pi (5{)}^2+2\pi (5\left)\right(20)+\pi (5\left)\right(11)$
$=\pi \times 5(5+40+11)$
$=\frac{22}{7}\times 5\times 56$

$=880c{m}^2$.

Radius of the conical vessel, $R=AC=6cm$

Height of the conical vessel, $h=OC=8cm$

Radius of the sphere, $PD=PC=r$

$\therefore PC=PD=rAC=AD=6\text{cm}$

[Since, lengths of two tangents from an external point to a circle are equal] 
$△OCA\&△OPD$ are right triangle.
[$\because$ Tangent and radius  are perpendicular to each other] 
$OA=\sqrt{O{C}^2+A{C}^2}=\sqrt{8^2+6^2}=\sqrt{100}=10\text{cm}O{P}^2=O{D}^2+P{D}^2$
$OD=OA-AD=10-6=4\text{cm}OP=OC-PC=8-r(8-r{)}^2=4^2+r^{2}64-16r+r^2=16+r^{2}16r=48\Rightarrow r=3\text{cm}$. 
Volume of water overflown =  Volume
                                                 of sphere
                                 $=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times (3{)}^3=36\pi {\text{cm}}^3$
Original volume of water = volume of
                                               cone 
                                       $=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 6^2\times 8=96\pi {\text{cm}}^3$ 

$\therefore$  Fraction of water overflown $=\frac{\text{Volume of water overflown}}{\text{Original volume of water}}=\frac{36\pi }{96\pi }=\frac{3}{8}=0.375$ 
$\therefore$  Fraction of water overflown is $0.375$

Let $R$ be the radius of the given cone, $r$ the radius of the small cone, $h$ be the height of the frustum and $h_1$ be the height of the small cone.

In the figure,$△ONC\sim △OMA$
$\therefore \frac{ON}{OM}=\frac{NC}{MA}$    [Sides of similar triangles are proportional]
$\Rightarrow \frac{h_1}{40}=\frac{r}{R}$ 
$\Rightarrow h_1=\left(\frac{r}{R}\right)\times 40$ .......(i)

We are given that $\frac{Volumeofsmallcone}{volumeofgivencone}=\frac{1}{64}$ 

$\Rightarrow \frac{\frac{1}{3}\pi r^2\times h_1}{\frac{1}{3}\pi R^2\times 40}=\frac{1}{64}$

$\Rightarrow \frac{r^2}{R^2}\times \frac{1}{40}\times \left[\left(\frac{r}{R}\right)40\right]=\frac{1}{64}$       [ By (i) ] 

$\Rightarrow {\left(\frac{r}{R}\right)}^3=\frac{1}{64}={\left(\frac{1}{4}\right)}^3$
$\Rightarrow$  $\frac{r}{R}=\frac{1}{4}$    ...... (ii)

From (i) and (ii) $h_1=\frac{1}{4}\times 40=10cm$

$\therefore ,h=40-h_1=(40-10)cm$

$\Rightarrow h=30cm$

Hence, the section is made at a height of 30 cm above the base of the cone  .

S.A. = 1372 $c{m}^2$ (Given)
$\Rightarrow 2(l\times b+b\times h+l\times h)=1372$
$[\because$ Surface area of a cuboid of dimensions $l\times b\times h$ is given by $2(lb+bh+lh).]$
Also, it is given that $l:b:h=4:2:1$.
Let $l=4k,b=2k$ and $h=k$.
$\text{Then,}2(4k\times 2k+2k\times k+4k\times k)=1372$
$⟹28k^2=1372$
$⟹k^2=49$
$⟹k=7$
$\therefore l=4k=4\times 7=28\text{cm}$

CSA of a cylinder $=2\pi rh$

Given,
CSA $=88{\text{cm}}^2$ and $h=14\text{cm}$
$2\pi rh=882r=\frac{88}{\pi h}=88\times \frac{7}{22}\times \frac{1}{14}=4\times 7\times \frac{1}{14}2r=2\text{cm}$

Therefore, diameter = 2r = 2 cm.

Surface area of sphere = 
$4\pi r^2=616\Rightarrow r^2=616\times \frac{7}{22}\times \frac{1}{4}\Rightarrow r^2=49\Rightarrow r=\sqrt{49}\Rightarrow r=7\text{cm}$

$\therefore$ The diameter, $D=2r=2\times 7=14\text{cm}$

Radius of first Sphere = $r_1=6cm$

Radius of second Sphere = $r_2=8cm$

Radius of third Sphere = $r_3=10cm$
Let radius of resulting sphere = r cm 
According to given condition, we have 
Volume of 1st sphere + Volume of 2nd sphere + Volume of 3rd sphere = Volume of resulting sphere  
$\Rightarrow \frac{4}{3}.\pi .(r_1{)}^3+\frac{4}{3}.\pi .(r_2{)}^3+\frac{4}{3}.\pi .(r_3{)}^3=\frac{4}{3}.\pi .r^3$ 

$\Rightarrow \frac{4}{3}\left(\right(r_1{)}^3+(r_2{)}^3+(r_3{)}^3)=\frac{4}{3}.r^3$ 

$\Rightarrow$ $(6^3+8^3+(10{)}^3)=r^3$ 

$\Rightarrow$ ( 216 + 512 + 1000 ) = $r^3$ 

$\Rightarrow$ $r^3$ = 1728

$\Rightarrow$ r = 12 cm