Tag: maths

Questions Related to maths

Mark the correct alternative of the following.
The radius of a wire is decreased to one-third. If volume remains the same, the length will become?

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

  1. $3$ times

  2. $6$ times

  3. $9$ times

  4. $27$ times

Show answer
Correct Option: C
Explanation:

Let $V _1$ and $V _2$ be the volume of the two cylinders with $h _1$ and $h _2$ as their heights.

Let $r _1$ and $r _2$ be their base radius.
It is given that, the radius of a wire is decreased to on-third.
$\therefore$  $r _2=\dfrac{1}{3}r _1$

$\Rightarrow$  $V _1=V _2$             [ Given ]
$\Rightarrow$  $\pi r _1^2 h _1=\pi r _2^2 h _2$

$\Rightarrow$  $r _1^2 h _1=\left(\dfrac{1}{3}r _1\right)^2 h _2$

$\Rightarrow$  $r _1^2 h _1=\dfrac{1}{9} r _1^2 h _2$

$\Rightarrow$  $h _2=9h _1$

$\therefore$  The length will become $9$ times.

Mark the correct alternative of the following.
If the height of a cylinder is doubled and radius remains the same, then volume will be?

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

  1. Doubled

  2. Halved

  3. Same

  4. Four times

Show answer
Correct Option: A
Explanation:

Let $V _1$ be the volume of the cylinder with radis $r _1$ and height $h _1,$ then

$\Rightarrow$  $V _1=\pi r _1^2 h _1$            ---- ( 1 )
Now, let $V _2$ be the volume after changing the dimension, then
$\Rightarrow$  $r _2=r _1,$  $h _2=2h _1$
So,
$\Rightarrow$  $V _2=\pi r _2^2h _2$

$\Rightarrow$  $V _2=\pi\times{r _1}^2\times 2h _1$

$\Rightarrow$  $V _2=2\times \pi r _1^2 h _1$
From ( 1 ),

$\Rightarrow$  $V _2=2V _1$

$\therefore$  If the height of a cylinder is doubled and radius remains the same, then volume will be $Doubled.$

Mark the correct alternative of the following.
In a cylinder, if radius is halved and height is doubled, the volume will be?

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

  1. Same

  2. Doubled

  3. Halved

  4. Four times

Show answer
Correct Option: C
Explanation:

Let $V _1$ be the volume of the cylinder with radis $r _1$ and height $h _1,$ then

$\Rightarrow$  $V _1=\pi r _1^2 h _1$            ---- ( 1 )
Now, let $V _2$ be the volume after changing the dimension, then
$\Rightarrow$  $r _2=\dfrac{1}{2}r _1,$  $h _2=2h _1$
So,
$\Rightarrow$  $V _2=\pi r _2^2h _2$

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

$\Rightarrow$  $V _2=\dfrac{1}{2}\times \pi r _1^2 h _1$
From ( 1 ),

$\Rightarrow$  $V _2=\dfrac{1}{2}V _1$

$\therefore$  In a cylinder, if radius is halved and height is doubled, the volume will be $Halved$

Mark the correct alternative of the following.
If the radius of a cylinder is doubled and the height remains same, the volume will be?

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

  1. Doubled

  2. Halved

  3. Same

  4. Four times

Show answer
Correct Option: D
Explanation:

Let $V _1$ be the volume of the cylinder with radius $r _1$ and height $h _1,$ then

$V _1=\pi r _1^2 h _1$           ---- ( 1 )
Now, let $V _2$ be the volume after changing the dimensions, then
$r _2=2r _1,\,h _2=h _1$
So,
$\Rightarrow$  $V _2=\pi r _2^2 h _2$

$\Rightarrow$  $V _2=\pi\times(2r _1)^2\times h _1$

$\Rightarrow$  $V _2=4\times \pi r _1^2 h _1$
From ( 1 ),
$\therefore$  $V _2=4V _1$
Hence, If the radius of a cylinder is doubled and the height remains same, the volume will be Four times.

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} $