Tag: problems involving volume of combined solids

Questions Related to problems involving volume of combined 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.

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

  1. $124,087 \space\ cm$

  2. $144,087 \space\ cm$

  3. $114,087 \space\ cm$

  4. $134,087 \space\ cm$

Show answer
Correct Option: C
Explanation:

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

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

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

  1. $1202.53\space\ m$

  2. $1302.53\space\ m$

  3. $1402.53\space\ m$

  4. $1102.53\space\ m$

Show answer
Correct Option: C
Explanation:

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

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

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

  1. $204.45 \space\ cm^3$

  2. $134.45 \space\ cm^3$

  3. $234.45 \space\ cm^3$

  4. $334.45 \space\ cm^3$

Show answer
Correct Option: C
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 $
= $234.45 \space\ cm^3$

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

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

  1. $\approx 9000 \space\ m^3$

  2. $\approx 8000 \space\ m^3$

  3. $\approx 10000 \space\ m^3$

  4. $\approx 7000 \space\ m^3$

Show answer
Correct Option: C
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 $
$\approx 10000 \space\ m^3$

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?

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

  1. $610.293 \space\ ft^3$

  2. $510.293 \space\ ft^3$

  3. $450.293 \space\ ft^3$

  4. $410.293 \space\ ft^3$

Show answer
Correct Option: D
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 $
= $410.293 \space\ ft^3$

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.

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

  1. $257.91 \space\ cm^3$

  2. $457.91 \space\ cm^3$

  3. $417.91 \space\ cm^3$

  4. $357.91 \space\ cm^3$

Show answer
Correct Option: B
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 $
= $457.91 \space\ cm^3$

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

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

  1. $44972 \space\ cm^3$

  2. $56972 \space\ cm^3$

  3. $66972 \space\ cm^3$

  4. $56000 \space\ cm^3$

Show answer
Correct Option: B
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 $
= $56972 \space\ cm^3$

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.

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

  1. $103.035 \space\ in^3$

  2. $93.035 \space\ in^3$

  3. $123.035 \space\ in^3$

  4. $113.035 \space\ in^3$

Show answer
Correct Option: D
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 $
= $113.035 \space\ in^3$

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.

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

  1. $8(18)^{1/3}$ cm

  2. $10$ cm

  3. $12$ cm

  4. $18(18)^{1/3}$ cm

Show answer
Correct Option: A
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

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.

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

  1. $90mm$

  2. $90cm$

  3. $9cm$

  4. $9m$

Show answer
Correct Option: D
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
$\therefore \dfrac{4}{3} \times \pi \times 6^3$ = $\pi \times 0.02^2 \times h$
$\therefore h = 900 cm=9 m$