Question

A cube and a sphere made of cast iron (each of volume 1000 $c{m}^3$) were cast under identical conditions. The timer taken for solidifying the cube was 4 s. The solidification time (in s) for the sphere is

• A. 6.157355

Answer 1

The correct option is A 6.157355
$\text{Time of solidification}$

$\left(T\right)=k{\left[\frac{\text{volume}}{\text{surface area}}\right]}^2$

$\left[\frac{T_{\text{sphere}}}{T_{\text{cube}}}\right]={\left[\frac{A_{\text{cube}}}{A_{\text{sphere}}}\right]}^2={\left[\frac{6a^2}{4\pi r^2}\right]}^2-\left(1\right)$

$\because [V_{\text{sphere}}=V_{\text{cube}}]$

$V_{cube}=a^3=1000c{m}^3$

$\Rightarrow a=10cm$

$\text{Similarly,}$

$V_{sphere}=\frac{4}{3}\pi r^3=1000$

$\Rightarrow r=6.2035cm$

$\text{Putting in Eq. (1), we get}$

$\frac{T_{\text{sphere}}}{T_{\text{cube}}}={\left[\frac{6\times {10}^2}{4\pi \times {\left(6.2035\right)}^2}\right]}^2$

$T_{\text{sphere}}=6.157355\sec$