Tag: application of surface area and volume of solids

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Questions Related to application of surface area and volume of solids

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

The inside radius of a spherical metal shell is $25$ cm and the thickness of the shell is $10$ cm. Calculate the volume of the material used in the shell to the nearest unit.

  1. $124,087 \space\ cm$
  2. $144,087 \space\ cm$
  3. $114,087 \space\ cm$
  4. $134,087 \space\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Thickness, $T = R - r$
$10 = R - 25$
$R = 35$ cm
$r  = 25$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
$=$ $\cfrac{4}{3}\pi (35^3-25^3)$
$=$ $\cfrac{4}{3}\pi (42875-15625)$
$=$ $\cfrac{4}{3}\pi (27250)$
$=$ $\cfrac{109000 \pi}{3}$
$=$ $114,087 \space\ cm$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

A spherical shell 5 m thick has an outer radius of 7 m. What is the volume of shell?

  1. $1202.53\space\ m$
  2. $1302.53\space\ m$
  3. $1402.53\space\ m$
  4. $1102.53\space\ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Thickness, $T = R - r$
$5 = 7 - r$
$r = 7 - 5 = 2$ m
$R = 7$ m
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (7^3-2^3)$
= $\cfrac{4}{3}\pi (343-8)$
= $\cfrac{4}{3}\pi (335)$
= $\cfrac{1340 \pi}{3}$
= $1402.53\space\ m$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

A hollow spherical shell has inner diameter $4$ cm and outer diameter $8$ cm. Determine the volume of the shell.

  1. $204.45 \space\ cm^3$
  2. $134.45 \space\ cm^3$
  3. $234.45 \space\ cm^3$
  4. $334.45 \space\ cm^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Outer radius, $R = 4$ cm
Inner radius, $r  = 2$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (4^3-2^3)$
= $\cfrac{4}{3}\pi (64-8)$
= $\cfrac{4}{3}\pi (56)$
= $\cfrac{224 \pi}{3}$
= $74.666\pi $
= $234.45 \space\ cm^3$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

A spherical shell has a outer radius $14$ m and inner radius $7$ m. What's the volume of the sphere?

  1. $\approx 9000 \space\ m^3$
  2. $\approx 8000 \space\ m^3$
  3. $\approx 10000 \space\ m^3$
  4. $\approx 7000 \space\ m^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Outer radius, $R = 14$ cm
Inner radius, $r  = 7$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (14^3-7^3)$
= $\cfrac{4}{3}\pi (2744-343)$
= $\cfrac{4}{3}\pi (2401)$
= $\cfrac{9604 \pi}{3}$
= $3201.33\pi $
$\approx 10000 \space\ m^3$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

What is the volume of material that is needed to form a spherical shell whose outer radius is $5$ ft and whose inner radius is $3$ ft?

  1. $610.293 \space\ ft^3$
  2. $510.293 \space\ ft^3$
  3. $450.293 \space\ ft^3$
  4. $410.293 \space\ ft^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$R = 5$ ft
$r  = 3$ ft
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (5^3-3^3)$
= $\cfrac{4}{3}\pi (125-27)$
= $\cfrac{4}{3}\pi (98)$
= $\cfrac{392 \pi}{3}$
= $130.666\pi $
= $410.293 \space\ ft^3$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

Calculate the volume of the material used in the shell to the nearest unit. The inside radius of a spherical metal shell is $2.5$ cm and the outer radius of the shell is $5$ cm.

  1. $257.91 \space\ cm^3$
  2. $457.91 \space\ cm^3$
  3. $417.91 \space\ cm^3$
  4. $357.91 \space\ cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$R = 5$ cm
$r  = 2.5$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (5^3-2.5^3)$
= $\cfrac{4}{3}\pi (125-15.625)$
= $\cfrac{4}{3}\pi (109.375)$
= $\cfrac{437.5 \pi}{3}$
= $1312.5\pi $
= $457.91 \space\ cm^3$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

Determine the volume of a spherical shell which has an inner radius of $6$ cm and an outer radius of $24$ cm.

  1. $44972 \space\ cm^3$
  2. $56972 \space\ cm^3$
  3. $66972 \space\ cm^3$
  4. $56000 \space\ cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Outer radius, $R = 24$ cm
Inner radius, $r  = 6$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (24^3-6^3)$
= $\cfrac{4}{3}\pi (13824-216)$
= $\cfrac{4}{3}\pi (13608)$
= $\cfrac{54432 \pi}{3}$
= $18144\pi $
= $56972 \space\ cm^3$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

Find the volume of material that is needed to form a spherical shell whose outer radius is $3.0$ inches and whose inner radius is $0.1$ inches.

  1. $103.035 \space\ in^3$
  2. $93.035 \space\ in^3$
  3. $123.035 \space\ in^3$
  4. $113.035 \space\ in^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$R = 3.0$ in
$r  = 0.1$ in
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (3^3-0.1^3)$
= $\cfrac{4}{3}\pi (27-0.001)$
= $\cfrac{4}{3}\pi (26.999)$
= $\cfrac{107.996 \pi}{3}$
= $35.998\pi $
= $113.035 \space\ in^3$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

A spherical shell of lead, whose external diameter is $24$ cm, is melted and recast into a right circular cylinder, whose height is $12$ cm and diameter $16$ cm. Determine the internal diameter of the shell.

  1. $8(18)^{1/3}$ cm
  2. $10$ cm
  3. $12$ cm
  4. $18(18)^{1/3}$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

  

Outer radius of the spherical lead $ = \dfrac {24}{2} = 12 $ cm 
Radius of the cylinder $ = \dfrac {16}{2} = 8 $ cm  
Since the spherical lead is recasted into the cylinder, their volumes are equal. 
Volume of a hollow sphere of outer radius $R$ and inner radius $r$ $ = \dfrac { 4 }{ 3 } \pi ({R}^{3} -{ r }^{ 3 }) $
Volume of a Cylinder of Radius "$R$" and height "$h$" $ = \pi { R }^{ 2 }h $
Hence, $ \dfrac { 4 }{ 3 } \pi ({12}^{3} -{ r }^{ 3 }) = \pi { 8 }^{ 2 } \times 12 $ 
Thus $ 1728 - { r }^{ 3 } = 576 $
$\Rightarrow  { r }^{ 3 } = 1152 $
$\Rightarrow  r = \sqrt [3] {1152} = 4 \sqrt [3] {18}   $ cm 
Inner diameter of the spherical lead $ = 2 \times \ \text{radius }= 2 \times 4 \sqrt [3] {18} $ cm $= 8 \sqrt [3] {18} $ cm

Multiple choice maths surface areas and volumes volume of a sphere problems involving volume of combined solids application of surface area and volume of solids

The diameter of a metallic sphere is $6 cm$. It was melted to make a wire of diameter $4 mm$. Find the length of the wire.

  1. $90mm$
  2. $90cm$
  3. $9cm$
  4. $9m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of metallic sphere = $\dfrac{4}{3} \times \pi \times {r^3}$

                                            = $\dfrac{4}{3} \times \pi \times 6^3$
Volume of cylindrical wire = $ \pi \times r^{2} \times h$

Now,
Volume of metallic sphere = Volume of cylindrical wire
$\therefore \dfrac{4}{3} \times \pi \times 6^3$ = $\pi \times 0.02^2 \times h$
$\therefore h = 900 cm=9 m$