If a hollow cube of internal edge 22 cm

Question:

If a hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that – space of the cube remains unfilled.

Then, the number of marbles that the cube can accomodate is

(a) 142244                         

(b) 142344                     

(c) 142444                                  

(d) 142544

Solution:

(a) Given, edge of the cube = 22 cm

$\therefore \quad$ Volume of the cube $=(22)^{3}=10648 \mathrm{~cm}^{3} \quad\left[\because\right.$ volume of cube $\left.=(\text { side })^{3}\right]$

Also, given diameter of marble $=0.5 \mathrm{~cm}$

$\therefore$ Radius of a marble, $r=\frac{0.5}{2}=0.25 \mathrm{~cm}$ $[\because$ diameter $=2 \times$ radius $]$

Volume of one marble $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times(0.25)^{3}$

$\left[\because\right.$ volume of sphere $\left.=\frac{4}{3} \times \pi \times(\text { radius })^{3}\right]$

$=\frac{1.375}{21}=0.0655 \mathrm{~cm}^{3}$

Filled space of cube $=$ Volume of the cube $\frac{1}{8} \times$ Volume of cube

$=10648-10648 \times \frac{1}{8}$

$=10648 \times \frac{7}{8}=9317 \mathrm{~cm}^{3}$

$\therefore$ Required number of marbles $=\frac{\text { Total space filled by marbles in a cube }}{\text { Volume of one marble }}$

$=\frac{9317}{0.0655}=142244$ (approx)

Hence, the number of marbles that the cube can accomodate is $142244 .$

 

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