Question

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​

Answer 1

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 $V_a$ and volume of 1 sphere of radius b cm is represented by $V_b$

So volume of 8 solid iron sphere with radius 'a cm' = volume of  1 solid iron sphere with radius 'b cm'

$=>8 \times} \mathrm{V}_{\mathrm{a}}=\mathrm{V}_{\mathrm{b}}$

$\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{b}}}=\frac{1}{8}$  ---- eqn 1

$\text {Let's calculate } {V}_{a} \text { and } V_{b}$

Formula for volume of sphere is as follows:

$V=\frac{4}{3} \pi r^{3}$

Where r is radius of the sphere

Substituting r = a cm in the formula of volume of sphere we get

$V_{a}=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi a^{3}$

Substituting r = b cm in the formula of volume of sphere we get

$V_{b}=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi b^{3}$

$\text { Substituting value of } V_{a} \text { and } V_{b} \text { in equation }(1) \text { we get }$

$\frac{\frac{4}{3} \pi a^{3}}{\frac{4}{3} \pi b^{3}}=\frac{1}{8}$

$\begin{array}{l}{=>\frac{\frac{4}{3} \pi a^{3}}{\frac{4}{3} \pi b^{3}}=\frac{1}{8}} \\\\ {=>\left(\frac{a}{b}\right)^{3}=\left(\frac{1}{2}\right)^{3}} \\\\ {=>\frac{a}{b}=\frac{1}{2}}\end{array}$

a : b = 1 : 2

Hence the ratio of a:b is 1 : 2