Mathematics · Quantitative Aptitude

Mensuration of Solids

194 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 and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

Mark the correct alternative of the following.
If the heights of two cones are in the ratio of $1:4$ and the radii of their bases are in the ratio $4:1$, then the ratio of their volumes is?

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

The base radius of cone is $'r'$ and vertical height $'h'$.

$\Rightarrow$  Volume of cone $=\dfrac{1}{3}\pi r^2 h$
Let the base radius and height of the two cones be $r _1,h _1$ and $r _2,h _2$ respectively.
It is given that the ratio between the heights of the two cones is $1:4$.
Since, only the ratio is given, to use them in our equation we introduce a constant $'k'.$
So,
$h _1=1k$
$h _2=4k$
It is also given that, the ratio between the base radius of the two cones is $4:1.$
Since, only the ratio is given, to use then in our equation we introduce another constant $'p'$
So,
$r _1=4p$
$r _2=1p$
Let $V _1$ and $V _2$ be the volumes of cones.

$\Rightarrow$  $\dfrac{V _1}{V _2}=\dfrac{\pi\times 4p\times 4p\times 1k\times 3}{3\times \pi\times 1p\times 1p\times 4k}$

$\therefore$   $\dfrac{V _1}{V _2}=\dfrac{4}{1}$

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

If the volume of a vessel in the form of a right circular cylinder is 448 $\pi\, cm^{3}$ and its height is 7 cm, then the curved surface area of the cylinder is

  1. $224\, \pi\, cm^{2}$
  2. $212\, \pi\, cm^{2}$
  3. $112\, \pi\, cm^{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume of a Cylinder of Radius $R$ and height $h$ $ = \pi { R }^{ 2 }h $
$\therefore $ volume of the given cylinder $ =\pi \times {R}^{2} \times 7  = 448 \pi  {cm}^{3} $

$ {R}^{2} = 64 $

$ R = 8 cm $  

Curved surface area of a cylinder of radius "$R$" and height "$h$" $ = 2\pi Rh$

$\therefore$ curved surface area of the given cylinder $ = 2\times \pi \times 8\times 7 =  112 \pi   \  cm^2 $

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

In a cylinder, if the radius is halved and height is doubled, the curved surface area will 

  1. remain same

  2. increase

  3. decrease

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of cylinder $\displaystyle \pi { r }^{ 2 }h$
Now, if $\displaystyle r=\frac { r }{ 2 } & h=h2$
New CSA $\displaystyle =2\pi rh$
$\displaystyle =2\pi \left( \frac { r }{ 2 }  \right) \times \left( h\times 2 \right) $
$\displaystyle =2\pi rh$
It will remain same

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

The circumference of base of cylindrical reservoir is $\displaystyle 30\pi cm$ and height is $10$ cm. How many litres of water can it hold?

  1. $\displaystyle 2.1\pi$ litres
  2. $\displaystyle 2.25\pi$ litres
  3. $\displaystyle 225\pi$ litres
  4. $\displaystyle 2250\pi$ litres
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Circumference $\displaystyle 2\pi r=30\pi ,r=\frac { 30\pi  }{ 2\pi  } =15$

Volume of cylinder $\displaystyle =\pi { r }^{ 2 }h$
$\displaystyle =\left( \pi \times 10\times 15\times 15 \right) { cm }^{ 3 }$
$\displaystyle =2250\pi { cm }^{ 3 }$
$\displaystyle =2.25\pi l$

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

The height of a hollow cylinder is $7 cm$ and its radius is $3.5 cm$. Then the surface area is

  1. $231{ cm }^{ 2 }$
  2. $154{ cm }^{ 2 }$
  3. $308{ cm }^{ 2 }$
  4. $115.5{ cm }^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given : radius $r=3.5 cm$ and height $h=7cm$

Surface area of a hollow cylinder $=2\pi r(h+r)$
                                                        $=2\times 3.14\times 3.5(7+3.5)$
                                                        $=230.79cm^2\approx 231$
$\therefore$ Surface area $=231cm^2$.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The height of a cylinder is equal to the radius. If an error of $\alpha$ % is made in the height, then percentage error in its volume is

  1. $\alpha$ %
  2. $2\alpha$ %
  3. $3\alpha$ %

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Volume of cylinder $V= \pi { r }^{ 2 }h$
Since, $h=r$
$V=\pi {h}^{3}$ 
$\displaystyle \dfrac{dV}{dh}=3\pi h^{2}$
Given, percentage error in measuring height $=\alpha$%
$\Rightarrow \displaystyle \dfrac { \Delta h }{ h } =\dfrac { \alpha }{ 100 } $
$\Rightarrow \displaystyle  { \Delta h }=\dfrac {\alpha h }{ 100 } $
Now, approximate error in measuring V$\displaystyle =dV= (\dfrac{dV}{dh}){ \Delta h}$
                                          $\displaystyle = \dfrac{3\alpha }{100} {\pi h^{3}} =3\alpha$% of V
Percentage error in measuring $V =3\alpha$%
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If the ratio of base radius and height of a cone is 1:2 and percentage error in radius is $\lambda$ %, then the error in its volume is

  1. $\lambda$ %
  2. $2\lambda$%
  3. $3\lambda$%

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Volume of cone $V=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h $
Given, $\displaystyle \dfrac{r}{h}=\dfrac{1}{2}$
$\Rightarrow V=\dfrac { 2 }{ 3 } \pi { r }^{ 3 } $
$\displaystyle \dfrac{dV}{dr}=2\pi r^{2}$
Percentage error in measuring r $=\lambda$%
$\Rightarrow \displaystyle \dfrac{\Delta r}{r}=\dfrac{\lambda}{100}$
$\Rightarrow \displaystyle \Delta r =\dfrac{\lambda r}{100}$
Approximate error in V $\displaystyle=dV=(\dfrac{dV}{dr}) \Delta r $
                                      $\displaystyle=\dfrac{\lambda}{100} (2\pi r^{3})=\lambda\%$ of V
Percentage error in $V =\lambda\%$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If errors of $1\%$ each are made in the base radius and height of a cylinder, then the percentage error in its volume is

  1. $1\%$
  2. $2\%$
  3. $3\%$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, percentage error in r is 1%
$\Rightarrow \displaystyle \frac{\Delta r}{r}=\frac{1}{100}$

$\Rightarrow \displaystyle \Delta r=\frac{r}{100}$
Also given, percentage error in h is 1%
$\Rightarrow \displaystyle \frac{\Delta h}{h}=\frac{1}{100}$

$\Rightarrow \displaystyle \Delta h=\frac{h}{100}$
Now, volume of cylinder $V=\pi r^{2}h$
$\Delta V=\pi [r^{2}\Delta h+2rh\Delta r]$
$\displaystyle \Delta V=\pi[r^{2}\frac{h}{100}+2rh\frac{r}{100}]$

$\displaystyle\Delta V=\pi r^{2}h[\frac{3}{100}]$
$\Rightarrow\displaystyle\frac{\Delta V}{V}=\frac{3}{100}$
Percentage error in V is 3%
 

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The lateral surface area (in ${cm}^{2}$) of a cone with height $3$ cm and radius $4$ cm is:

  1. $62\cfrac{6}{7}$
  2. $52\cfrac{6}{7}$
  3. $31\cfrac{3}{7}$
  4. $15\cfrac{5}{7}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Curved surface area of a cone $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.
For a cone,  $l = \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $, where $h$ is the height.

Hence, $ l = \sqrt { { 3 }^{ 2 }+  {4}^{ 2 } } $ 

$ l = \sqrt {25} $

$ l = 5 $ cm

Hence, Curved surface area of this cone, $ =\displaystyle  \frac {22}{7} \times 4 \times 5 = 62 \frac {6}{7}  {cm}^{2} $

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

Find the total surface area of a cone, if its slant height is $9\ m$ and the radius of its base is $12\ m$.

  1. $792\ {m}^{2}$
  2. $452\ {m}^{2}$
  3. $682\ {m}^{2}$
  4. $987\ {m}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total surface area of a cone $ = \pi r (r + l) $  

Hence, TSA of this cone, $ = \cfrac {22}{7} \times 12 \times (12 + 9) = 792  {m}^{2} $

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The diameter of a cone is $14\ cm$ and its slant height is $9\ cm$. Find the area of its curved surface.

  1. $198\ {cm}^{2}$
  2. $108\ {cm}^{2}$
  3. $152\ {cm}^{2}$
  4. $218\ {cm}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Curved surface area of a cone $= \pi rl$  

Radius of the cone $ = \cfrac {Diameter}{2} = \cfrac {14}{2}  =  7  cm $

Hence, CSA of this cone $ = \cfrac {22}{7} \times 7 \times 9 = 198  {cm}^{2} $

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

Slant height of a cone is 13 cm and radius is 7 cm its lateral surface area is

  1. 280 $\displaystyle cm^{2}$
  2. 282 $\displaystyle cm^{2}$
  3. 284 $\displaystyle cm^{2}$
  4. 286 $\displaystyle cm^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle L.S.A\, of\, a\, cone =\pi rl$
$\displaystyle =\dfrac{22}{7}\times 7\times 13=286cm^{2}$

Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The circumference of the base of a 10 m high conical tent is 44 metres Then the length of canvas used in making the tent if width of canvas is 2 m is (Use $\displaystyle \pi =22/7$)

  1. $132.2 m$
  2. $134.2 m$
  3. $130.2 m$
  4. $136.2 m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let r m be the radius of the base h m be the height and l m be the slant height of the cone Then 
Circumference=44 metres
$\displaystyle \Rightarrow 2\pi r=44\Rightarrow 2\times \frac{22}{7}\times r=44\Rightarrow r=7$
metres It is given that h=10 metres
$\displaystyle \therefore l^{2}=r^{2}+h^{2}\Rightarrow l=\sqrt{r^{2}+h^{2}}$
$\displaystyle=\sqrt{49+100}=\sqrt{149}=12.2m$
Now surface area of the tent $\displaystyle =\pi rl$
$\displaystyle \frac{22}{7}\times 7\times 12.2m^{2}=268.4m^{2}$
$\displaystyle \therefore $ Area of the canvas used $\displaystyle =268.4m^{2}$
It is given that the width of the canvas is 2 m
$\displaystyle \therefore $ Length of the canvas used=$\displaystyle =\frac{area}{width}=\frac{268.4}{2}=134.2m$
Multiple choice maths solids surface area of a cone surface area of cone the surface area of a cone

The volume of a right circular cone of height $8 cm$ and radius of base $3 cm$ is

  1. $12 \pi cm^3$
  2. $24 \pi cm^3$
  3. $48 \pi cm^3$
  4. $72 \pi cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that height of right circular cone $h=8cm$

Radius of right circular cone $r=3cm$

Volume of right circular cone$V=\dfrac{1}{3}{{r}^{2}}\pi .h$

$ =\dfrac{1}{3}{{.3}^{2.}}\pi .8c{{m}^{3}} $

$ =24\pi c{{m}^{3}}$

Hence, this is the answer.