Mensuration of Solids

Given, perimeter $=4a=20$cm
$\therefore a=5$ cm
Area $=a^2=25  cm^2$
Volume $=$ Area $\times$ Height
Volume $=25 \times 30$
Volume $=750  cm^3$

Given,

The radius of the sphere $=r$

The radius of the cylinder $=r$

The height of the cylinder $=2r$

We know that the volume of the sphere

${{V} _{1}}=\dfrac{4}{3}\pi {{r}^{3}}$

We know that the volume of the cylinder

$ {{V} _{2}}=\pi {{r}^{2}}h $

$ {{V} _{2}}=\pi {{r}^{2}}\left( 2r \right) $

$ {{V} _{2}}=2\pi {{r}^{3}} $

Therefore, the required ratio

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{\dfrac{4}{3}\pi {{r}^{3}}}{2\pi {{r}^{3}}} $

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{2\pi {{r}^{3}}}{3\pi {{r}^{3}}} $

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{2}{3} $

$ {{V} _{1}}:{{V} _{2}}=2:3 $

Hence, this is the answer.

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 } $
                                            $=\dfrac{\pi(2\sqrt { 2 } +1  )^ 3)}{24}$
                                            $=2.35\pi$

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.

Volume of cone $=$ Volume of cylinder

Diameter of cylindrical vessel  $d = 1.06 \ m$
Height of cylindrical vessel  $h = 7.2 \ m$
Volume   $V = \dfrac{\pi d^2 h}{4}$
$\implies \ V = \dfrac{\pi\times (1.06)^2\times 7.2}{4} = 6.354 \ m^3$
Since the number $7.2$ has two significant figures and $1.06$ has three significant figures.
So the final answer must have 2 significant figures.
Thus volume of cylinder   $V = 6.3 \ m^3$

Vol. of cylinder =Vol. of cone 
$\displaystyle \Rightarrow \pi \times 10^{2}\times h=\dfrac{1}{3}\pi \times 14^{2}\times 36$
$\displaystyle \Rightarrow h=23.52cm$

Volume of cylinder = Volume of sphere
$\displaystyle \Rightarrow \pi \left ( \dfrac{d}{2} \right )^{2}h=\dfrac{4}{3}\pi \left ( \dfrac{d}{2} \right )^{3}$
$\displaystyle \Rightarrow h=\dfrac{2}{3}d\Rightarrow \dfrac{h}{2}=\dfrac{d}{3}$

Given $V= \pi \left ( R^{2}-r^{2} \right )l$

The length of the rectangle become the circumference of the base of the cylinder 

Let the base radius of cone and circular cylinder is r and height of circular cylinder is h and height of cone 2h

Given