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 same material. They are being cast. The ratio of the solidification time of the cube to the same of the sphere is

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

Answer & Solution

Correct Answer: Option D

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2 Comments

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Dravida Krishnatreya 5 years ago

Can any one explain this answer please.

Rahul Bhandari 6 years ago

Solidification time T= C (volume/surface area)^n
Where C is a constant depends upon the material.
And n= 2 usually.

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Volume of a cube of side l and volume of a...

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 same material. They are being cast. The ratio of the solidification time of the cube to the same of the sphere is

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