Tag: hollow cylinder

Questions Related to hollow cylinder

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?

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

  1. $\displaystyle 2.1\pi$ litres

  2. $\displaystyle 2.25\pi$ litres

  3. $\displaystyle 225\pi$ litres

  4. $\displaystyle 2250\pi$ litres

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Correct Option: B
Explanation:

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

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

The inner diameter of a circular well is $3.5$ m. It is $10$ m deep. Find the cost of plastering this curved surface at the rate of Rs. $40$ per m$^2$.

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

  1. Rs. $4000$

  2. Rs. $4400$

  3. Rs. $4500$

  4. Rs. $4800$

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Correct Option: B
Explanation:

Given diameter of well is $3.5$ m and depth is $10$ m and cost of plastering is Rs. $40$ per sq m.

Then radius of well $=\dfrac{3.5}{2}=1.75$ m
And height of well $=10$ m
Then curved surface area of well $=$ $2\pi rh=2\times \dfrac{22}{7}\times 1.75\times 10$
$=$ $2\times 22\times 0.25\times 10=110 m^{2}$
Then cost of plastering curved surface area $=$ $110\times 40=$ Rs. $4400$.

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

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

  1. $231{ cm }^{ 2 }$

  2. $154{ cm }^{ 2 }$

  3. $308{ cm }^{ 2 }$

  4. $115.5{ cm }^{ 2 }$

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Correct Option: A
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 $=231cm^2$.

A magnet is in the form of a ring with inner diameter $4cm$ and outer diameter $6cm$. If the thickness of the magnet is $2cm$ . What is the cost of fabricating the surface of the magnet if the cost of fabrication per ${cm}^{2}$ is $Rs.10$

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

  1. $Rs.950$

  2. $Rs.945$

  3. $Rs.942$

  4. $Rs.1000$

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Correct Option: C

A flower pot is in the form of a hollow cylinder with a closed base with inner radius $2cm$ and outer radius $4cm$ . The height of the flower pot is $10cm$ . If the pot has to be polished find the cost of polishing if the cost of polishing per ${cm}^{2}$ is $Rs.2$.

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

  1. $Rs.276.32$

  2. $Rs.275$

  3. $Rs.270$

  4. $Rs.278.64$

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Correct Option: A
Explanation:

We have to first find the total surface area of the flower pot.
Outer radius $=4cm$
Inner radius $=2cm$
Height of flower pot $=10cm$
Total surface area $=A=2\pi Rh+2\pi rh+\pi { { R }^{ 2 } }+\pi { { r }^{ 2 } }$
$A=2\pi (R+r)h+\pi ({ R }^{ 2 }+{ r }^{ 2 })\ A=2\pi (4+2)2+\pi (16+4)\ A=24\pi +20\pi =44\pi \ A=40\pi =44\times 3.14=138.16{ cm }^{ 2 }$
Cost of fabrication per ${cm}^{2}$ $=Rs.2$
Total cost of fabrication $==138.16\times 2=Rs.276.32$

What will be the Inner surface area of a spherical shell of inner radius $15\ cm$ and outer radius $16\ cm$? (Correct upto 2 decimal places)

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

  1. $706.86\ {cm^2}$

  2. $804.25\ {cm^2}$

  3. $2827.43\ {cm^2}$

  4. $3216.99\ {cm^2}$

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Correct Option: C

The surface area of a solid sphere is always greater than the surface area of a hemisphere for the same value of radius.

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

  1. True

  2. False

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Correct Option: A
Explanation:
Let radius of solid sphere=radius of hemisphere$=r$
Then, S.A of solid sphere$=4\pi { r }^{ 2 }$
S.A of hemisphere$=2\pi { r }^{ 2 }$
$\therefore $S.A. of solid sphere$>$ S.A of hemisphere
Hence statement is true

A hemispherical bowl has inner radius $5cm$ and outer radius $6cm$. What will be the volume of solid enclosed between the two hemispheres? (Correct upto 2 decimal places)

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

  1. $904.78 \ {cm}^{3}$

  2. $523.60 \ {cm}^{3}$

  3. $381.18 \ {cm}^{3}$

  4. $190.59 \ {cm}^{3}$

Show answer
Correct Option: D
Explanation:

The volume of hemispherical shell$= \dfrac{2}{3}\pi*R^{3}-\dfrac{2}{3}\pi*r^{3}$
where R and r are the outer and inner radius of the hemisphere
On solving the equation we get Volume$= 190.59 \ {cm}^{3}$

The surface area of a solid spherical ball of diameter $10\ cm$ is equal to :

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

  1. $25\pi\ {cm}^2$

  2. $50\pi\ {cm}^2$

  3. $100\pi\ {cm}^2$

  4. $200\pi\ {cm}^2$

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Correct Option: C
Explanation:

Given: Diameter of the sphere $= 10\ cm$
Hence, Radius ($r$) of the sphere will be $5\ cm$

We know that, 
Surface area of the sphere is $4\pi r^2$
Therefore, Area will be $4\pi (5)^2 = 100\pi\ {cm}^2$

What will be the Inner and Outer radius of a spherical shell of inner surface area $452.39\ {cm}^2$ and outer surface area $804.25\ {cm}^2$ ? (Surface areas are accurate upto 2 decimal places)

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

  1. $6 \ cm, 7 \ cm$

  2. $7 \ cm, 8 \ cm$

  3. $6 \ cm, 8 \ cm$

  4. $7 \ cm, 9\ cm$

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Correct Option: C
Explanation:
Inner surface area$=4\pi { \left( inner\quad radius \right)  }^{ 2 }$
$\Rightarrow 4\pi { \left( { r } _{ 1 } \right)  }^{ 2 }=452.39cm^{2}\Rightarrow { r } _{ 1 }^{ 2 }=\cfrac { 452.39\times 7 }{ 4\times 22 } =35.99$
$\Rightarrow { r } _{ 2 }=\sqrt { 35.99 } =5.99\approx 6cm$
Outer surface area$=4\pi { \left( outer\quad radius \right)  }^{ 2 }$
$\Rightarrow 4\pi { \left( { r } _{ 2 } \right)  }^{ 2 }=804.25㎠\Rightarrow { r } _{ 2 }^{ 2 }=\cfrac { 804.25\times 7 }{ 4\times 22 } =63.97$
$\Rightarrow { r } _{ 2 }=\sqrt { 63.97 } =7.99\approx 8cm$