The mass and radius of the earth are 5.975x 10²⁴ kg and 6.37x 10⁶ m respectively. Calculate the average earth's density to correct significant figures. Take π=3.142.

The correct option is 5.52 x 10³ kg m⁻³.

To find the average density of the Earth, we use the formula for density:

$$\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}}$$

Assuming the Earth is a perfect sphere, its volume ($V$) is given by the formula:

$$V=\frac{4}{3}\pi R^3$$

where:
Mass ($M$) = 5.975 x 10²⁴ kg
Radius ($R$) = 6.37 x 10⁶ m
$\pi$ = 3.142

First, let's calculate the volume of the Earth:

$$V=\frac{4}{3}\times 3.142\times {(6.37\times {10}^6)}^3$$

Calculating ${6.37}^3$:
${6.37}^3\approx 258.47$

Now, substituting this back into the volume formula:

$$V\approx \frac{4}{3}\times 3.142\times 258.47\times {10}^{18}$$

$$V\approx 1.0827\times {10}^{21}{m}^3$$

Now, we calculate the average density ($\rho$):

$$\rho =\frac{5.975\times {10}^{24}}{1.0827\times {10}^{21}}$$

$$\rho \approx 5.5186\times {10}^3{\mathrm{kg\; m}}^{-3}$$

According to the rules of significant figures, when multiplying or dividing numbers, the final result should be rounded to the same number of significant figures as the measurement with the least number of significant figures.
The given radius $6.37\times {10}^6$ m has 3 significant figures.
Therefore, we round our final answer to 3 significant figures:

$$\rho \approx 5.52\times {10}^3{\mathrm{kg\; m}}^{-3}$$