Tag: maths

Questions Related to maths

The curved surface of a circular cylinder of height 'h' and the curved surface area of the cone of slant height 2 'h' having the same circular base are in the ratio of

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

  1. 1 : 2

  2. 2 : 1

  3. 1 : 1

  4. 1 : 3

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Correct Option: C
Explanation:

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

Then Curved surface area of circular cylinder =$2\pi rh$
And curved surface area of cone=$\pi r(2h)=2\pi rh$
So ratio of Curved surface area of circular cylinder : curved surface area of cone :: $2\pi r(h)=2\pi rh$ : $2\pi rh$=1:1

A hollow sphere of internal and external radii 3 cm and 4 cm respectively is malted into a cylinder of diameter 37 cm The height of the cylinder is

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

  1. 2 cm

  2. 2.5 cm

  3. 3 cm

  4. none

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Correct Option: D
Explanation:

Given the external  radius of hollow sphere is 4 cm and internal radius 3 cm 

Then volume of metal used=$\frac{4}{2}\pi (R^{2}-r^{2})=\frac{4}{3}\pi \left ( (4)^{2}-(3)^{2} \right )=\frac{4}{3}\pi (16-9)=\frac{4}{3}7\pi $
The diameter of cylinder 37 cm 
Radius of cylinder is =18.5 cm
Volume of cylinder=$\pi r^{2}h=\pi (18.5)^{2}h$
But cylinder made by metal of hollow sphere
$\pi (18.5\times 18.5)h=\frac{4}{3}7\pi \Rightarrow h=\frac{4\times 7}{3\times 18.5\times 18.5}\Rightarrow \Rightarrow h=0.027 cm$

A cistern $6$ m long and $4$ m wide contains water to a depth of $1.25$ m. What is the area of wetted surface?

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

  1. $40$ sq. m

  2. $45$ sq. m

  3. $49$ sq. m

  4. $73$ sq. m

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Correct Option: D
Explanation:

Given, $l = 6, b = 4$

Also given depth i.e., $ h = 1.25$
Area of the wetted surface $= 2[lb + bh + hl]$
$=2[(6\times 4)+(4\times 1.25)+(1.25\times 6)]$
$=2[24+5+7.5]$
$= 73$ sq. m
Therefore, the area of wetted surface is $73$ sq. m.

The outer and inner diameters of a circular pipe are $6$ cm and $4$ cm respectively. If its length is $10$ cm then what is the total surface area in square centimetres?

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

  1. $55\pi$

  2. $110\pi$

  3. $150\pi$

  4. None of the above

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Correct Option: D
Explanation:

Given, outer and inner diameters of circular pipe are $6$ cm and $4$ cm
Therefore, outer and inner radii of a circular pipe are $3$ cm and $2$ cm.
Thus total surface area would be $ 10\times \pi (3^{2} - 2^{2})$ $= 50\pi $ sq. cm.

What will be the volume of a sphere of diameter $15\ cm$? (Correct up to $2$ decimal place)

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

  1. $1436.76\ {cm}^{3}$

  2. $1767.15\ {cm}^{3}$

  3. $14137.17\ {cm}^{3}$

  4. $4188.79\ {cm}^{3}$

Show answer
Correct Option: B
Explanation:

The Volume of a sphere$=\dfrac{4}{3}\pi \ \text{R}^{3}$
Here Radius $R=7.5 \ cm$
On solving We get Volume$=1767.15\ {cm}^{3}$ 

What will be the radius of a sphere whose surface area is 616 $cm^{2}$? (Use $\pi=\dfrac{22}{7}$)

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

  1. $6$ cm

  2. $7$ cm

  3. $8$ cm

  4. $9$ cm

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Correct Option: B
Explanation:

Radius $=\dfrac{1}{2}\times\sqrt{\dfrac{\text{surface area}}{\pi}}$
on solving we get radius $=7$ cm

If volume of sphere is $850$ $m^{3}$ then its radius and surface area are

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

  1. $6m$, $450$ $m^{2}$

  2. $5m$, $560$ $m^{2}$

  3. $2m$, $780$ $m^{2}$

  4. $5.88m$, $434$ $m^{2}$

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Correct Option: D
Explanation:
Volume of Sphere $=850m^3=\cfrac{4}{3}\pi r^3 \Rightarrow r^3=\cfrac{850\times 3\times 7}{4\times 22}=202.84 \\ \Rightarrow r=\sqrt[3]{202.84}=5.88m$
Surface area $=4\pi r^2=4\times \cfrac{22}{7}\times 5.88\times 5.88 \approx 434m^2$

 If the radius of a sphere is doubled, then what is the ratio of new to the old surface area?

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

  1. $1:2$

  2. $2:1$

  3. $1:4$

  4. $4:1$

Show answer
Correct Option: D
Explanation:

$S _1=4\pi r^2 \ S _2=4\pi (2r)^2=16\pi r^2 \ \cfrac{S _2}{S _1}=\cfrac{16\pi r^2}{4\pi r^2}=\cfrac{4}{1} \Rightarrow 4:1$

A test-tube consists of a hollow cylindrical tube joined to a hemi-spherical bown of the same internal radius. The whole tube holds $350$ cc of water and in the cylindrical portion falls $1$ cm if $19.64$ cc of water is removed. Find the length of the cylindrical portion of the tube. (Take $\pi =$ $22/7$)

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

  1. $12.15$ cm

  2. $16.15 $ cm

  3. $24.15 $ cm

  4. None of these

Show answer
Correct Option: B
Explanation:

Let r cm. be the radius of the hemisphere and h cm be the length of the cylindrical portion.
Volume of water removed $\pi r^2 (1) = 19.64cc$
$\Rightarrow r^2 = 19.64 \times \displaystyle \frac{7}{22}  \Rightarrow r = 2.5 cm$
Volume of the whole tube $= \pi r^2 h + \displaystyle \frac{2}{h} \pi r^3 = 350 c.c.$
$\Rightarrow \pi r^2 \displaystyle \left ( h + \frac{2}{3} r \right ) = 350$
$\Rightarrow \displaystyle \frac{22}{7} \times 2.5^2 \times \left ( h + \frac{2}{3} \times 2.5 \right ) = 350$
$\displaystyle \frac{22}{7} \times 6.25 \times (h+ 1.67) = 350 $
$ \Rightarrow \displaystyle 350 \times \frac{7}{22} \times 6. 25 - 1.67 cm    \Rightarrow h = 16.15 cm$

The height of a hollow cylinder is $14cm$ if external diameter is $16cm$ and total curved surface area of the hollow cylinder is $1320sq.cm$, then its internal diameter is

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

  1. $14cm$

  2. $16cm$

  3. $7cm$

  4. $8cm$

Show answer
Correct Option: C
Explanation:

Given     

external radius $r _2=8$, height of cylinder $h=14$
 we have,


$2\pi h(r _{1}+r _{2})=1320$

$ \implies8+r _1=\displaystyle \frac{1320\times7}{2\times22\times14}$

$\implies r _1=7cm$