Tag: surface area and volume of sphere

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Questions Related to surface area and volume of sphere

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