Question

A hollow aluminium sphere has an internal radius
of 20 cm and an external radius of 30 cm. Given that
the density of aluminium is 2.7 g/cm², find the mass
of the sphere in kg.​

Answer 1

Answer :-

$\\\;\large{\underbrace{\underline{\sf{Question's\;\;Analysis\;\;:-}}}}$

Here the concept of Volume of Sphere has been used. Firstly we need to find out the Internal Volume of Hollow Sphere and then we have to find out External Volume of Sphere. Then we will subtract both to get the Volume Aluminum used in the sphere. After that, we shall apply it in the formula of Density to find the answer.

________________________________________________

★ Formula Used :-

$\\\;\large{\boxed{\sf{Volume\;of\;Sphere\;=\;\bf{\dfrac{4}{3}\:\times\:\pi r^{3}}}}}$

$\\\;\large{\boxed{\sf{Density\;=\;\bf{\dfrac{Mass_{(in\;g)}}{Volume_{(in\;cm^{3})}}}}}}$

$\\\;\large{\boxed{\sf{1\;Kg\;=\;\bf{1000\;\;g}}}}$

________________________________________________

★ Solution :-

Given,

» Internal Radius of Sphere = 20 cm

» External Radius of Sphere = 30 cm

» Density of Aluminum = 2.7 g/cm³

Note :- Density is denoted by g/cm³ since mass is divided by volume.

________________________________________________

~ For the Internal Volume of Sphere :-

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(Internal)}\;=\;\bf{\dfrac{4}{3}\:\times\:\pi r^{3}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(Internal)}\;=\;\bf{\dfrac{4}{3}\:\times\:\dfrac{22}{7}\:\times\:(20)^{3}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(Internal)}\;=\;\bf{\dfrac{4}{3}\:\times\:\dfrac{22}{7}\:\times\:8000}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(Internal)}\;=\;\bf{\dfrac{704000}{21}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(Internal)}\;=\;\underline{\underline{\bf{33523.81\;\;cm^{3}}}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;Internal\;\;Volume\;\;of\;\;Sphere\;=\;\bf{33523.81\;\;cm^{3}}}}}$

________________________________________________

~ For the External Volume of Sphere :-

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(External)}\;=\;\bf{\dfrac{4}{3}\:\times\:\pi r^{3}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(External)}\;=\;\bf{\dfrac{4}{3}\:\times\:\dfrac{22}{7}\:\times\:(30)^{3}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(External)}\;=\;\bf{\dfrac{4}{3}\:\times\:\dfrac{22}{7}\:\times\:27000}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(External)}\;=\;\bf{\dfrac{2376000}{21}}}$

$\\\;\;\;\;\;\sf{:\rightarrow\;\;\;Volume\;of\;Sphere_{(External)}\;=\;\underline{\underline{\bf{113142.86\;\;cm^{3}}}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;External\;\;Volume\;\;of\;\;Sphere\;=\;\bf{113142.86\;\;cm^{3}}}}}$

________________________________________________

~ For Volume of Aluminium used in Sphere :-

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Volume\;of\;Aluminium\;Used\;=\;\bf{External\;Volume\;-\;Internal\;Volume}}$

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Volume\;of\;Aluminium\;Used\;=\;\bf{113142.86\;-\;33523.81}}$

$\\\;\;\;\;\;\sf{:\Longrightarrow\;\;\;Volume\;of\;Aluminium\;Used\;=\;\underline{\underline{\bf{79619.05\;\;cm^{3}}}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;Volume\;\;of\;\;Aluminium\;\;Used\;=\;\bf{79619.05\;\;cm^{3}}}}}$

________________________________________________

~ For Mass of Sphere (in Kg) :-

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Density\;=\;\bf{\dfrac{Mass_{(in\;g)}}{Volume_{(in\;cm^{3})}}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;2.7\;=\;\bf{\dfrac{Mass_{(in\;g)}}{79619.05}}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Mass_{(in\;g)}\;=\;\bf{2.7\;\times\;79619.05}}$

$\\\;\;\;\;\;\sf{:\mapsto\;\;\;Mass_{(in\;g)}\;=\;\underline{\underline{\bf{214971.44\;\;g}}}}$

Now conversion from g to Kg :-

$\\\;\;\;\;\;\sf{:\Rightarrow\;\;\;1\;Kg\;=\;\bf{1000\;\;g}}$

$\\\;\;\;\;\;\sf{:\Rightarrow\;\;\;1\;g\;=\;\bf{\dfrac{1}{1000}\:Kg}}$

$\\\;\;\;\;\;\sf{:\Rightarrow\;\;\;214971.44\;g\;=\;\bf{\dfrac{214971.44}{1000}\:Kg}}$

$\\\;\;\;\;\;\sf{:\Rightarrow\;\;\;214971.44\;g\;=\;\bf{214.971\;\:Kg}}$

$\\\:\large{\underline{\underline{\rm{Thus,\;mass\;of\;aluminium\;used\;in\;Sphere\;is\;\;\boxed{\bf{214.971\;\;Kg}}}}}}$

________________________________________________

★ More to know :-

$\\\;\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\:\pi r^{3}}$

$\\\;\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Cylinder\;=\;\pi r^{2} h}$

$\\\;\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\:\times\:\pi r^{2} h}$

$\\\;\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Cuboid\;=\;Length\:\times\:Breadth\:\times\:Height}$

$\\\;\;\;\;\;\sf{\leadsto\;\;\;Volume\;of\;Cuboid\;=\;(Side)^{3}}$

Answer 2

Answer:

214.971 kg is the answer