Tag: hollow cylinder

Questions Related to hollow cylinder

Mark the correct alternative of the following.
The volume of a cylinder of radius r is $1/4$ of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is r in terms of x?

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

  1. $\dfrac{x^2}{2\pi}$

  2. $\dfrac{x}{2\sqrt{\pi}}$

  3. $\dfrac{\sqrt{2x}}{\pi}$

  4. $\dfrac{\pi}{2\sqrt{x}}$

Show answer
Correct Option: B
Explanation:

Let the height of the cylinder be $h.$

Volume of the cylinder $=\pi r^2 h$
Height of the rectangular box $=h$
Since, base is square with side $x.$
Volume of the box $=x\times x\times h=x^2 h$
According to question,
$\Rightarrow$  $\pi r^2  h=\dfrac{1}{4} x^2 h$

$\Rightarrow$  $r^2=\dfrac{1}{4\pi}x^2$
Taking square root on both sides,
$\Rightarrow$  $r=\dfrac{x}{2\sqrt{\pi}}$

Mark the correct alternative of the following.
Two circular cylinders of equal volume have their heights in the ratio $1:2$. Ratio of their radii is?

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

  1. $1:\sqrt{2}$

  2. $\sqrt{2}:1$

  3. $1:2$

  4. $1:4$

Show answer
Correct Option: A
Explanation:
$\dfrac{h _1}{h _2}=\dfrac{1}{2}$        [ Given ]
Let $V _1$ and $V _2$ are volume of cylinders.

$\therefore$  $V _1=V _2$          [ Given ]

$\therefore$  $\dfrac{V _1}{V _2}=1$

$\Rightarrow$  $\dfrac{\pi r _1^2h _1}{\pi r _2^2 h _2}=1$

$\Rightarrow$  $\left(\dfrac{r _1}{r _2}\right)^2\left(\dfrac{h _1}{h _2}\right)=1$

But it is given that,
$\dfrac{h _1}{h _2}=\dfrac{1}{2}$

$\therefore$  $\left(\dfrac{r _1}{r _2}\right)^2\times\dfrac{1}{2}=1$

$\Rightarrow$  $\left(\dfrac{r _1}{r _2}\right)^2=2$

$\Rightarrow$  $\left(\dfrac{r _1}{r _2}\right)^2=\dfrac{2}{1}$

$\Rightarrow$  $\dfrac{r _1}{r _2}=\dfrac{\sqrt{2}}{1}$

$\therefore$  The ratio of the radii of the two cylinders is $\sqrt{2}:1$

Mark the correct alternative of the following.
The altitude of a right circular cylinder is increased six times and the base area is decreased one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is?

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

  1. $\dfrac{2}{3}$

  2. $\dfrac{1}{2}$

  3. $\dfrac{3}{2}$

  4. $2$

Show answer
Correct Option: D
Explanation:

Curved surface area of cylinder $=2\pi r h$


Height is increased to $6$ times $=6h$


Base area is decreased to $\left(\dfrac{1}{9}th\right)$ 

i.e. $\pi (r{^{\prime}})^2=\dfrac{1}{9}\pi r^2\Rightarrow r^{\prime}=\dfrac13r $

Now,
New curved surface area $=2\pi\times\dfrac{1}{3}r\times 6h$

                                           $=2\times(2\pi r h)$
$\therefore$  Lateral surface area becomes twice.

Mark the correct alternative of the following.
The height h of a cylinder equal the circumference of the cylinder. In terms of h, what is the volume of the cylinder?

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

  1. $\dfrac{h^3}{4\pi}$

  2. $\dfrac{h^2}{2\pi}$

  3. $\dfrac{h^3}{2}$

  4. $\pi h^3$

Show answer
Correct Option: A
Explanation:

Let $h$ be the height of cylinder with radius $r.$

It is given that,
$2\pi r =h$
$\Rightarrow$  $r=\dfrac{h}{2\pi}$
Therefore, the volume of the cylinder is
$V=\pi r^2 h$

$\Rightarrow$  $V=\pi\left(\dfrac{h}{2\pi}\right)^2h$

$\Rightarrow$  $V=\dfrac{h^3}{4\pi}$

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?

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

  1. $1:2$

  2. $2:3$

  3. $3:4$

  4. $4:1$

Show answer
Correct Option: D
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.

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

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

A hollow cylindrical pipe is $21 \ cm$ long. If its outer and inner diameters are $10 \ cm$ and $6 \ cm$ respectively, them the volume of the metal used in making the pipe is $\displaystyle \left(Take\, \pi\, =\, \frac{22}{7}\right)$

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

  1. $1048\, cm^{3}$

  2. $1056\, cm^{3}$

  3. $1060\, cm^{3}$

  4. $1064\, cm^{3}$

Show answer
Correct Option: B
Explanation:

The pipe is in the shape of a hollow cylinder.
Volume of a hollow Cylinder of outer Radius "R", inner Radius ""r" and height "h" $ = \pi ({ R }^{ 2 }-{ r }^{ 2 })h $
Outer Radius $ = \frac {10}{2} = 5  cm $
Inner Radius $ = \frac {6}{2} = 3  cm $
Hence, volume of the pipe $ = \frac { 22 }{ 7 } \times ({ 5 }^{ 2 }-{ 3 }^{ 2 })\times 21 = 1056  {cm}^{3} $

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

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

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

  2. $212\, \pi\, cm^{2}$

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

  4. None of these

Show answer
Correct Option: C
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 $

A hollow iron pipe of $21 cm$ long and its external diameter is $8 cm$. If the thickness of the pipes is $1 cm$ and iron weights $\displaystyle 8g/cm^{2}$, then the weight of the pipe is equal to

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

  1. $3.6 kg$

  2. $3.696 kg$

  3. $36 kg$

  4. $36.9 kg$

Show answer
Correct Option: B
Explanation:
Let external diameter be ${ d } _{ 2 }$ and internal be ${ d } _{ 1 }$
$\Rightarrow { d } _{ 2 }=8cm$ & ${ d } _{ 1 }=8-1-1=6cm$
Therefore, ${ r } _{ 1 }=3cm$ & ${ r } _{ 2 }=4cm$
Volume of hollow cylindrical pipe$=\pi \left( { r } _{ 2 }^{ 2 }-{ r } _{ 1 }^{ 2 } \right) \times h$
$=\cfrac { 22 }{ 7 } \left( { \left( 4 \right)  }^{ 2 }-{ \left( 3 \right)  }^{ 2 } \right) \times 21$
$=\cfrac { 22 }{ 7 } \times 7\times 21$
$=462cm^3$
$\therefore $Weight of the pipe$=8\times 462=3696g=3.696kg$

A rectangular sheet of width $14$ m is rolled along its width and is converted to form a cylinder. Find the radius of cylinder.

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

  1. $\displaystyle \frac { 22 }{ 49 } $

  2. $\displaystyle \frac { 44 }{ 29 } $

  3. $\displaystyle \frac { 49 }{ 22 } $

  4. None

Show answer
Correct Option: C
Explanation:

Curved surface area of cylinder$=100m^2$

$\therefore 2 \pi r h = 100$
Here, width of the rectangle = height of the cylinder.
$\therefore h=14m$

$\therefore 2 \times \dfrac {22}{7} \times r \times 14 = 100$

$ \therefore r = \dfrac {100 \times 7}{2 \times 22 \times 14}$

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

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

  1. remain same

  2. increase

  3. decrease

  4. none of the above

Show answer
Correct Option: A
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