Cube here will be inscribed in a cone as a square is in isosceles triangle.
Let the height of the cone be $h$
Radius=$\sqrt2 h$
Volume of cone=$\dfrac{1}{3}\pi r^2h$
                           =$\dfrac{2\sqrt 2}{3}\pi h^3$
Let the side of the cube be x,the top of the cone above it has the sign $(h-x)$ and radius $\dfrac{x}{2}$
Using properties of similar triangle $\dfrac { \dfrac { x }{ 2 }  }{ h-x } =\dfrac{\sqrt2 h}{h}$
                                                            $=\sqrt 2 x$
                                                             $=\dfrac { 2\sqrt { 2 } h }{ 2\sqrt { 2 } +1 } $
Volume of the cube=$\dfrac { 2\sqrt { 2 } h }{ 2\sqrt { 2 } +1 } $
Ratio of the volume of the cone to volume of the cube=$\dfrac { \dfrac { 2\sqrt { 2 }  }{ 3 } \pi h^{ 3 } }{ (\dfrac { 2\sqrt { 2 } h }{ 2\sqrt { 2 } +1 } )^ 3 } $
                                            $=\dfrac{\pi(2\sqrt { 2 } +1  )^ 3)}{24}$
                                            $=2.35\pi$

$S = 6a^{2}; V = a^{3}$
Then $S^{3} = 216a^{6} = 216(a^{3})^{2}$ or $S^{3} = 216 V^{2}$

Since the box is $\dfrac {1}{4}$ filled with  the given cubes , we need to multiply the total volume of the given cubes by $4$ to get the total volume of thr box.

Let $V _1$ and $V _2$ be two volume of cubes.

Let $l$be the side of cube.

Let $a$ be the initial edge of the cube.

It is given that, $A _,A _2,A _3$ be the areas of $3$ adjacent faces of cuboid

Given,length $=4m$
breadth $=50 cm=0.5 m$
height $20 cm=0.2 m$
Volume of beam $=l \times b \times h=0.4 m^3$
Weight of $1$ $m^3$ wood is $25$ kg
Weight of $0.4$  $m^3$ cubiodal beam$=25\times 0.4=10 $kg

circular cylinder's radius=r
height=h
keeping the height same,  radius is halved
radius of new circular cylinder=r/2
So, ratio of volume = volume of reduced cylinder/volume of original cylinder
$Ratio=\Pi { r /2}^{ 2 }h/\Pi { (r) }^{ 2 }h$
$Ratio=1/4$

Let R be the radius and H be the height of the cylinder and let rand h be the radius and height of the cone respectively. Then,
$3r=2R$
and $H:h=4:3$ ......(i)
$\Rightarrow \dfrac {H}{h}=\dfrac {4}{3}$
$\Rightarrow 3H=4h$ .......(ii)
Let n be the required number of cones which can be made from the materials of the cylinder. Then, the volume of the cylinder will be equal to the sum of the volumes of n cones. Hence, we have
$\pi R^2H=\dfrac {n}{3}\pi r^2h$
$\Rightarrow 3R^2H=nr^2h$
$\Rightarrow n=\dfrac {3R^2H}{r^2H}=\dfrac {3\times \dfrac {9r^2}{4}\times \dfrac {4h}{3}}{r^2h}$ [$\because$ From (i) and (ii), $R=\dfrac {3r}{2}$ and $H=\dfrac {4h}{3}$]
$\Rightarrow n=\dfrac {3\times 9\times 4}{3\times 4}$
$\Rightarrow n=9$
Hence, the required number of cones is 9.

Total surface area of the cube after hemispherical depression
$=$ T.S.A. of cube -Base area of hemisphere + C.S.A of hemisphere
$=6(edge)^2-\pi r^2+2\pi r^2$

Since the thickness of wall and brick is same, we only need to figure out the area of the wall, which on dividing with the area of brick will give us the number of bricks.

Radius of the bowl $ = \dfrac {36}{2} = 18  cm $

Volume of the bowl $ = \dfrac { 2 }{ 3 }

\pi { r }^{ 3 } = \dfrac {2}{3} \times \pi \times 18 \times 18 \times

18 {cm}^{3} $

Volume

of a Cylinder of Radius "R" and height "h" $ = \pi { R }^{

2 }h $





Hence, Volume of one cylindrical bottle, $ = \pi \times 3 \times 3 \times  6 $

Hence, number of bottled required $

= \dfrac {Volume  of  bowl} {Volume  of  each  bottle} = \dfrac{\dfrac {2}{3} \times \pi \times 18 \times 18 \times

18}{ \pi \times 3 \times 3 \times 

6} = 72 $