8 solid iron sphare with radius 'a cm' each are melted to form a sphare with radius 'b cm'. find the ratio of a:b​

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 : 2Solution: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:bAs 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 sphereSubstituting 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 : 2Hence the ratio of a:b is 1 : 2