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Q.

What is the density of the earth of its mass is $6 \times 10^{24} kg$ and radius is $6.378 \times 10^{6} m$ ?

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a

5.5 \times 10^{3} kg m^{3}

b

5.6 \times 10^{3} kg m^{3}

c

5.9 \times 10^{3} kg m^{3}

d

6.0 \times 10^{3} kg m^{3}

answer is A.

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Detailed Solution

The density of earth of its mass is

6 \times 10^{24} kg

and radius

6.378 \times 10^{6} m

is

5.5 \times 10^{3} kg m^{3}

.
Earth is assumed as a sphere. To calculate density of earth is mass per unit volume, given that mass of earth is

6 \times 10^{24} kg

Volume of earth

V = \frac{4}{3} π r^{3}

= \frac{4}{3} \times π \times {(6.378 \times 10^{6} m)}^{3}

Density of earth is

d = \frac{M}{V}

M =

mass of earth

V =

volume of earth

d = \frac{6 \times 10^{24} kg}{\frac{4}{3} \times π \times {(6.378 \times 10^{6} m)}^{3}}

d = 5.5 \times 10^{3} kg m^{3}

Hence the density of the earth is

5.5 \times 10^{3} kg m^{3}

.

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