The mass of $3 m^{3}$ of cement of density $3000 k g / m^{3}$ is

1000 kg

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9000 kg

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10000 kg

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90000 kg

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Solution

The correct option is B 9000 kg
Answer is B.
The Mass of a block is given as volume of the block $\times$ Density of the block.
In this case, the volume of the cement is $3 m^{3}$ and the density is $3000 k g / m^{3}$ .
Therefore, $M = 3 \times 3000 = 9000 k g$ .
The mass of cement is $9000 k g .$

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