What is the density of an iron block which has a volume of $12 c m^{3}$ and a mass of 96g?

8 k g / m^{3}

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80 k g / m^{3}

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8000 k g / m^{3}

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800 k g / m^{3}

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Solution

The correct option is A $8 k g / m^{3}$
Answer is A.
As we know, Density $d = \frac{M a s s, M}{V o l u m e, V}$

In this case, an iron block which has a volume of $12 c m^{3}$ = $0.12 m^{3}$ and a mass of $96 g = 0.096 k g$ .

Therefore, Density $d = \frac{0.096}{0.12} = 8 k g / m^{3}$ .

Hence, the density of the iron block is 8 $k g / m^{3}$ .

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What is the density of an iron block which has a volume of 12cm3 and a mass of 96g?

The correct option is A 8kg/m3Answer is A.As we know, Density d=Mass,MVolume,VIn this case, an iron block which has a volume of 12cm3 = 0.12m3 and a mass of 96g=0.096kg.Therefore, Density d=0.0960.12=8kg/m3.Hence, the density of the iron block is 8 kg/m3.

What is the density of an iron block which has a volume of $12 c m^{3}$ and a mass of 96g?