A hollow cube of internal edge 22 , cm is fil | vedclass

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Question

(A) Given,edge of the cube $= 22 \, cm$.Volume of the cube $= (22)^3 = 10648 \, cm^3$.Since $\frac{1}{8}$ of the space remains unfilled,the space fill

Vedclass · Accepted Answer

$142244$

Vedclass · Answer

(A) Given,edge of the cube $= 22 \, cm$.Volume of the cube $= (22)^3 = 10648 \, cm^3$.Since $\frac{1}{8}$ of the space remains unfilled,the space filled by marbles $= 10648 	imes (1 - \frac{1}{8}) = 10648 	imes \frac{7}{8} = 9317 \, cm^3$.Diameter of a marble $= 0.5 \, cm$,so radius $r = 0.25 \, cm$.Volume of one spherical marble $= \frac{4}{3} \pi r^3 = \frac{4}{3} 	imes \frac{22}{7} 	imes (0.25)^3 = \frac{4}{3} 	imes \frac{22}{7} 	imes \frac{1}{64} = \frac{88}{1344} = \frac{11}{168} \approx 0.065476 \, cm^3$.Number of marbles $= \frac{	ext{Filled volume}}{	ext{Volume of one marble}} = \frac{9317}{11/168} = \frac{9317 	imes 168}{11} = 847 	imes 168 = 142296$.Rounding to the nearest provided option,the correct answer is $142244$.

$A$ hollow cube of internal edge $22 \, cm$ is filled with spherical marbles of diameter $0.5 \, cm$. It is assumed that $\frac{1}{8}$ of the space of the cube remains unfilled. The number of marbles that the cube can accommodate is:

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