Question

a hollow sphere of internal and external diameter is 4cm and 5cm respectively is melted in to a cone of base diameter is 8cm find height of cone

Answer 1

Hi friend,

Here is your answer,

              Volume of sphere                   =        Volume of cone
 4/3 * 22/7 *(2.5*2.5*2.5 - 2*2*2)          =         1/3 * 22/7 * 4 * 4 * h
                   10.16                                 =         16.762 * h
                                             h              =          0.61 cm


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Answer 2

$\huge\sf\pink{Answer}$

☞ Height of the cone is 5 cm

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ External diameter of the hollow sphere = 6 cm

✭ Internal diameter of the hollow sphere = 4 cm

✭ Radius of cone = 2 cm

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ What is the height of the cone?

$\rule{110}1$

$\huge\sf\purple{Steps}$

External diameter of the hollow sphere = 6 cm

≫ $\sf{ Radius = \dfrac{Diameter}{2} = \dfrac{6}{2}}$

≫ $\sf{ External \: radius = 3 \: cm}$

Internal diameter of the Hollow sphere = 4 cm

≫ $\sf{ Internal \: radius = \dfrac{4}{2} }$

≫ $\sf{ Internal \: radius = 2 \: cm}$

Volume of a hollow sphere is given by,

$\underline{\boxed{\sf{Volume = \dfrac{4}{3} \pi (R^3 - r^3) }}}$

Volume of a cone is given by,

$\underline{\boxed{\sf{Volume = \dfrac{1}{3} \pi r^2 h }}}$

Let the height be h cm

$\bullet\underline{\textsf{As Per the Question}}$

➳ $\sf{ \dfrac{4}{{3}} {\pi} (R^3 - r^3) = \dfrac{1}{{3}} {\pi} r^2 h }$

➳ $\sf{ 4(R^3-r^3) = r^2 h }$

Substituting the given values,

➝ $\sf{ 4(3^2-2^2) = 2^2 \times h }$

➝ $\sf{4(9-4) = 4 \times h }$

➝ $\sf{ 4 \times 5 = 4 \times h }$

➝ $\sf{20 = 4h }$

➝ $\sf{ h = \dfrac{20}{4} }$

➝ $\sf{\orange{ h = 5 \: cm }}$

$\rule{170}3$