Question

Volume of a cube of side $l$ and volume of a sphere of radius r are equal. Both the cube and the sphere are solid and of the same material. They are being cast. The ratio of the solidification time of the cube to the same of sphere is

• A. ${\left(\frac{4\pi }{6}\right)}^3{\left(\frac{r}{l}\right)}^6$
• B. ${\left(\frac{4\pi }{6}\right)}^2{\left(\frac{r}{l}\right)}^3$
• C. $\left(\frac{4\pi }{6}\right){\left(\frac{r}{l}\right)}^2$
• D. ${\left(\frac{4\pi }{6}\right)}^2{\left(\frac{r}{l}\right)}^4$

Answer 1

The correct option is D ${\left(\frac{4\pi }{6}\right)}^2{\left(\frac{r}{l}\right)}^4$

$\text{Surface area of cube =}6l^2$

$\text{Surface area of sphere =}4\pi r^2$

$\text{Solidification time}\propto {\left(\frac{\text{volume}}{\text{surface area}}\right)}^2$

$\text{As volume of cube and sphere are equal}$

$\frac{t_c}{t_s}={\left(\frac{A_s}{A_c}\right)}^2={\left(\frac{4\pi r^2}{6l^2}\right)}^2$

$\frac{t_c}{t_s}={\left(\frac{4\pi }{6}\right)}^2{\left(\frac{r}{l}\right)}^4$