Respuesta :
8 solid iron sphere with radius 'a cm' each are melted to form a sphere with radius 'b cm' then the ratio of a:b is 1 : 2
Solution:
Given that 8 solid iron sphere with radius 'a cm' each are melted to form a sphere with radius 'b cm'
Need to find the ratio of a:b
As 8 solid iron sphere with radius 'a cm' each are melted to form a sphere with radius 'b cm'.
For sake of simplicity, let volume of 1 sphere of radius a cm is represented by [tex]V_a[/tex] and volume of 1 sphere of radius b cm is represented by [tex]V_b[/tex]
So volume of 8 solid iron sphere with radius 'a cm' = volume of 1 solid iron sphere with radius 'b cm'
[tex]=>8 \times} \mathrm{V}_{\mathrm{a}}=\mathrm{V}_{\mathrm{b}}[/tex]
[tex]\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{b}}}=\frac{1}{8}[/tex] ---- eqn 1
[tex]\text {Let's calculate } {V}_{a} \text { and } V_{b}[/tex]
Formula for volume of sphere is as follows:
[tex]V=\frac{4}{3} \pi r^{3}[/tex]
Where r is radius of the sphere
Substituting r = a cm in the formula of volume of sphere we get
[tex]V_{a}=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi a^{3}[/tex]
Substituting r = b cm in the formula of volume of sphere we get
[tex]V_{b}=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi b^{3}[/tex]
[tex]\text { Substituting value of } V_{a} \text { and } V_{b} \text { in equation }(1) \text { we get }[/tex]
[tex]\frac{\frac{4}{3} \pi a^{3}}{\frac{4}{3} \pi b^{3}}=\frac{1}{8}[/tex]
[tex]\begin{array}{l}{=>\frac{\frac{4}{3} \pi a^{3}}{\frac{4}{3} \pi b^{3}}=\frac{1}{8}} \\\\ {=>\left(\frac{a}{b}\right)^{3}=\left(\frac{1}{2}\right)^{3}} \\\\ {=>\frac{a}{b}=\frac{1}{2}}\end{array}[/tex]
a : b = 1 : 2
Hence the ratio of a:b is 1 : 2