Question

A hollow cube of internal edge $22$ cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that ${\left(\frac{1}{8}\right)}^{th}$ space of the cube remains unfilled. Find the number of marbles that the cube can accommodate?

Answer 1

For Spherical marble,

$d=0.5cm$

$r=\frac{0.5}{2}cm$

Now,

Space unfilled= $\frac{1}{8}\times v$. of cube

Space filled= $(1-\frac{1}{8})\times$ volume of cube

$=\frac{7}{8}\times$ volume of cube

$\therefore$ space filled = no. of marbles $\times$ volume of each marbles.

$\frac{7}{8}\times a^3=N\times \frac{4}{3}\pi r^3$

$\frac{7}{8}\times 22\times 22\times 22=N\times \frac{4}{3}\times \frac{22}{7}\times \frac{0.5}{2}\times \frac{0.5}{2}\times \frac{0.5}{2}$

$N=49\times 484\times 6$

$=142296$

So, $142296$ marbles are accommodated by cube.