Question

A metal cuboid of dimensions 49 m, 22 m and 14 m is melted and cast into 7 identical cylinders of radius 7 m. These cylinders are again melted and cast into cubes such that the side of each cube is equal to half of the height of each cylinder. The numbers of cubes thus formed is __________.

Answer 1

Let h be the height of each cylinder.

Radius of each cylinder, r = 7 m

It is given that, the metal cuboid of given dimensions is melted and re-cast into 7 identical cylinders.

∴ 7 × Volume of each cylinder = Volume of the cuboid

$\Rightarrow 7\times \mathrm{\pi }{r}^{2}h=\mathrm{Length}\times \mathrm{Breadth}\times \mathrm{Height}$

$\Rightarrow 7\times \frac{22}{7}\times {\left(7\mathrm{m}\right)}^2\times h=49\mathrm{m}\times 22\mathrm{m}\times 14\mathrm{m}$

$\Rightarrow h=\frac{49\times 22\times 14}{22\times 49}=14\mathrm{m}$

So, the height of each cylinder is 14 m.

It is also given that these cylinders are again melted and re-cast into cubes such that the side of each cube is half of the height of each cylinder.

Let the side of each cube be a.

$\therefore a=\frac{h}{2}=\frac{14}{2}=7\mathrm{m}$

Suppose the number of cubes formed be n.

∴ n × Volume of each cube = Volume of 7 identical cylinder = Volume of the cuboid

$\Rightarrow n=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cuboid}}{\mathrm{Volume}\mathrm{of}\mathrm{each}\mathrm{cube}}$

$\Rightarrow n=\frac{49\mathrm{m}\times 22\mathrm{m}\times 14\mathrm{m}}{{\left(7\mathrm{m}\right)}^3}$ [Volume of the cube = (Side)³]

$\Rightarrow n=44$

Thus, the number of cubes formed is 44.

A metal cuboid of dimensions 49 m, 22 m and 14 m is melted and cast into 7 identical cylinders of radius 7 m. These cylinders are again melted and cast into cubes such that the side of each cube is equal to half of the height of each cylinder. The numbers of cubes thus formed is ___44___.