Mathematics · Quantitative Aptitude

Mensuration of Solids

209 Questions

Mensuration of solids focuses on calculating the volume and surface area of three dimensional shapes like cylinders and cones. These geometry problems require applying standard mathematical formulas. They frequently appear in state public service and engineering exams.

Cylinder volume and areaCone slant heightSurface area ratiosHollow pipe calculationsDimensional optimization

Mensuration of Solids Questions

Multiple choice maths area of complex plane figures 2d and 3d figures

The ratio of the slant height of two right cones of equal base is 3 : 2 then the ratio of their volumes is

  1. $1:4$
  2. $9:4$
  3. $3:2$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the radius of both the cones be r, slant height of first cone $ = 3a $; slant height of second cone $ = 2a $

For a cone, l $ = \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $ where h is the height. 

Hence, for cone 1,  $ 3a = \sqrt { { h 1 }^{ 2 }+  {r}^{ 2 } } $ 

$ 9{a}^{2} = { h 1 }^{ 2 } + {r}^{ 2 }$

$ { h 1 }^{ 2 } = 9{a}^{2} - {r}^{ 2 } $

$ h 1 = \sqrt {9{a}^{2} - {r}^{ 2 }} $

For cone 1,  $ 2a = \sqrt { { h 2 }^{ 2 }+  {r}^{ 2 } } $ 

$ 4{a}^{2} = { h 2 }^{ 2 } + {r}^{ 2 }$

$ { h 2 }^{ 2 } =4{a}^{2} - {r}^{ 2 } $

$ h 2 = \sqrt {4{a}^{2} - {r}^{ 2 }} $
Now, ratio of their volumes is calculated as:
$V _{ 1 } : V _{ 2 }$
$ \frac { 1 }{ 3 }\pi {r }^{ 2 }h 1 = \frac { 1 }{ 3 } \pi { r }^{2 }h 2 $
$ h 1 : h 2 $
$ \sqrt {9{a}^{2} - {r}^{ 2 }} : \sqrt {4{a}^{2} - {r}^{ 2 }}  $
$9{a}^{2} - {r}^{ 2 }: 4{a}^{2} - {r}^{ 2 } $
$ 9{a}^{2} : 4{a}^{2} $
$ 9:4 $

Multiple choice maths measures and the circle surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

The base of a right prism is a square of perimeter 20 cm and its height is 30 cm. The volume of the prism is

  1. $700 cm^3$
  2. $750 cm^3$
  3. $800 cm^3$
  4. $850 cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice maths measures and the circle surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

The base of a right prism is an equilateral triangle of edge $12$m. If the volume of the prism is $288\sqrt 3m^3$, then its height is:

  1. $6$m
  2. $8$m
  3. $10$m
  4. $12$m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

length of Equilateral triangle $= 12 m$
Area of equilateral triangle = $\displaystyle \frac{\sqrt{3}}{4}a^2$ = $\displaystyle \frac{\sqrt{3}}{4}(12)^2$ = $36\sqrt{3}$
Volume of prism = $288\sqrt{3} m^3$ = Area of triangle X height
$288\sqrt{3} m^3$ = $36 \sqrt{3} \times$ height
$\therefore $ Height $= 8 m$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

The radius of a sphere is r and radius of base of a cylinder is r and height is 2r. The ratio of their volumes will be-

  1. $2:3$
  2. $3:4$
  3. $4:3$
  4. $3:2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

Multiple choice maths how big? how heavy? measuring volume volume of solids volume of cube and cuboid

The radius of a cone is $\sqrt2$ times the height of the cone. A cube of maximum possible volume is cut from the same cone. What is the ratio of the volume of the cone to the volume of the cube?

  1. $3.18\pi$
  2. $2.25\pi$
  3. $2.35$

  4. Can't be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice maths how big? how heavy? measuring volume volume of solids volume of cube and cuboid

If the radius of the base of a right circular cylinder is halved, keeping the  height same, what is the ratio of the volume of the reduced cylinder to that of the original.

  1. $1:3$
  2. $1:5$
  3. $1:4$
  4. $1:7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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
$Ratio=\Pi { r /2}^{ 2 }h/\Pi { (r) }^{ 2 }h$
$Ratio=1/4$

Multiple choice maths how big? how heavy? measuring volume volume of solids volume of cube and cuboid

By melting a solid cylindrical metal, a few conical materials are to be made. If three times the radius of the cone is equal to twice the radius of the cylinder and the ratio of the height of the cylinder and the height of the cone is 4: 3, find the number of cones which can be made

  1. 4

  2. 3

  3. 9

  4. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Multiple choice maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

A regular triangular pyramid has an altitude of $9\ m$ and a volume of $187.06\ cu.\ m$. What is the base edge in meters?

  1. $12$
  2. $13$
  3. $14$
  4. $15$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given : Altitude$(height\quad (h))=9\ m$, volume $=187.06\ cu.\ m$

We know that, Volume $=\dfrac{1}{3}Bh$, where $B=x^2 \sin \theta$
$\implies 187.06=\dfrac{1}{3} \left(\dfrac{1}{2}x^2 (\sin \theta)\right) (9)$,  where $x$ is base edge
$\implies 187.06=\dfrac{1}{3} \left(\dfrac{1}{2}x^2 \sin60\right)9$
$\implies x^2=143.9988=144$
$\therefore\ x=12\ m$

Multiple choice maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

The frustum of a regular triangular pyramid has equilateral triangles for its bases. The lower and upper base edges are $9\ m$ and $3\ m$, respectively. If the volume is $118.2\ cu.\ m$, how far apart (m) are the base?

  1. $9$
  2. $8$
  3. $7$
  4. $10$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given : Volume $=118.2\ cu. m$

Upper base edge $=9\ m$, lower base edge $=3\ m$
We know that, 
Volume $=\dfrac{h}{3}(A _{1}+A _{2}+\sqrt{A _{1} A _{2}})$ ....... $(1)$, where $A _{1}, A _{2}$ are area of upper and lower bases.
$A _{1}=\dfrac{\sqrt{3}}{4}\times 9^2=35.074$
$A _{2}=\dfrac{\sqrt{3}}{4}\times 3^2=3.897$
From $(1)$ we get,
$118.2 = \dfrac{h}{3}(35.074+3.897+\sqrt{35.074\times 3.897})$
$\implies 118.2=\dfrac{h}{3}(38.971+11.6911)$
$\implies 118.2\times 3=h(50.6621)$
$\implies h=7\ m$
Hence, the bases are $7\ m$ far from each other.

Multiple choice maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

The base of a right pyramid is an equilateral triangle of perimeter $8$ dm and the height of the pyramid is $30$$\sqrt{3}$ cm. The volume of the pyramid is

  1. $1600$ cm$^{3}$
  2. $16000$ cm$^3$
  3. $\displaystyle \frac{16000}{3} cm^3$
  4. $\displaystyle \frac{5}{4} cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Base of pyramid is an equilateral triangle of parameter
$8dm=80cm$
Let the side of the equilateral triangle be $'a'cm$
$\therefore $Parameter of equilateral triangle$=3a$
$\Rightarrow 3a=80\Rightarrow a=\cfrac { 80 }{ 3 } cm$
Height of pyramid$=30\sqrt { 3 } cm$
Volume of pyramid=Area of base $\times $ height
$=\cfrac { \sqrt { 3 }  }{ 4 } { a }^{ 2 }\times 30\sqrt { 3 } $

$=\cfrac { \sqrt { 3 }  }{ 4 } \times \cfrac { 80 }{ 3 } \times \cfrac { 80 }{ 3 } \times 30\sqrt { 3 } $

$=\cfrac { 3 }{ 4 } \times \cfrac { 80 }{ 3 } \times 80 \times 10 $

$=\cfrac { 1 }{ 4 } \times 80 \times 80 \times 10 $

$=20 \times 80 \times 10 $

$=16000cm^3$
Multiple choice maths perimeter, area and volume volume of prism and pyramid surface areas and volumes of solids recall the surface areas and volumes of different solid shapes

The base of a right pyramid is an equilateral triangle of perimeter 8 cm and the height of the pyramid is $30\sqrt 3$ cm. The volume of the pyramid is

  1. $160 cm^3$
  2. $1600 cm^3$
  3. $\dfrac {160}{3} cm^3$
  4. $\dfrac {5}{4} cm^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of right pyramid $=$ $\dfrac { 1 }{ 3 } \times area\quad of\quad base\times height\quad of\quad pyramid$

Now, base is equilateral $\triangle $, therefore,
area $=\dfrac { \sqrt { 3 }  }{ 4 } \times { \left( side \right)  }^{ 2 }$
Perimeter of triangle $=8cm$
$\therefore \quad \quad 3a=8\Rightarrow a=\dfrac { 8 }{ 3 } cm$
$\therefore \quad \quad area=\dfrac { \sqrt { 3 }  }{ 4 } \times \dfrac { 8 }{ 3 } \times \dfrac { 8 }{ 3 } =\dfrac { 16\sqrt { 3 }  }{ 9 } { cm }^{ 2 }$
Now,  Volume $=\dfrac { 1 }{ 3 } \times \dfrac { 16\sqrt { 3 }  }{ 9 } \times 30\sqrt { 3 } $
                        $=\dfrac { 160\times 3 }{ 9 } =\dfrac { 160 }{ 3 } { cm }^{ 3 }$

Multiple choice maths surface areas and volumes volume of a sphere problems involving volume of combined solids application of surface area and volume of solids

A metallic solid cone is melted and cast into a cylinder of the same base as that of the cone. If the height of the cylinder is $7\;cm$, what was the height of the cone?

  1. $20\;cm$
  2. $21\;cm$
  3. $22\;cm$
  4. $12\;cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of cone $=$ Volume of cylinder

$\dfrac { 1 }{ 3 } \pi { r } _{ 1 }^{ 2 }{ h } _{ 1 }=\pi { r } _{ 2 }^{ 2 }{ h } _{ 2 }$
$\Rightarrow \quad \dfrac { 1 }{ 3 } \times { r }^{ 2 }\times { h } _{ 1 }={ r }^{ 2 }\times 7\quad \left[ { r } _{ 1 }={ r } _{ 2 }=r \right] $
                  $\Rightarrow \quad { h } _{ 1 }=7\times 3=21cm$
Therefore, height of cone is $21$ cm.

Multiple choice physics learning how to measure measurement of area and volume measurement of volume measurement of area, volume and density

The volume enclosed by the cylinder of diameter $1.06\ m$  and height 7.2 m to the correct no. of the significant figure is:

  1. $6.350\ m^3$
  2. $6.35\ m^3$
  3. $6.3\ m^3$
  4. $6\ m^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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   $V = 6.3 \ m^3$

Multiple choice maths problems on measurement basic operations with same units operations involving units of length calculations choosing and converting between units conversion between different units conversion between units

A conical cup 36 cm high has diameter of base 28 cm It is full of water. The water was poured into a cylindrical jar of radius of base 10 cm. The height of water in the vessel is

  1. 23.52 cm

  2. 16.92 cm

  3. 11.76 cm

  4. 13.65 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice maths problems on measurement basic operations with same units operations involving units of length calculations choosing and converting between units conversion between different units conversion between units

A right circular cylinder and a sphere are of equal volumes and their radii are also equal If h is the height of the cylinder and d is the diameter of the sphere then

  1. $\displaystyle \frac{h}{3}=\frac{d}{2} $
  2. $\displaystyle \frac{h}{2}=\frac{d}{3} $
  3. $2h = d$
  4. $h = d$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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