A cube of edge 3, cm of iron weighs 12, gm. W | vedclass

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(A) The volume of a cube is given by $V = a^3$,where $a$ is the edge length.For the first cube,$a_1 = 3\, cm$,so $V_1 = 3^3 = 27\, cm^3$.For the secon

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$768$

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(A) The volume of a cube is given by $V = a^3$,where $a$ is the edge length.For the first cube,$a_1 = 3\, cm$,so $V_1 = 3^3 = 27\, cm^3$.For the second cube,$a_2 = 12\, cm$,so $V_2 = 12^3 = 1728\, cm^3$.The ratio of the volumes is $\frac{V_2}{V_1} = \frac{1728}{27} = 64$.Since the material is the same,the weight is directly proportional to the volume.Weight of the second cube = $64 	imes 	ext{Weight of the first cube}$.Weight = $64 	imes 12\, gm = 768\, gm$.

$A$ cube of edge $3\, cm$ of iron weighs $12\, gm$. What is the weight of a similar cube of iron whose edge is $12\, cm$? (in $gm$)

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