Class 11 NCERT PHYSICS • Chapter 1: Units and Measurement • Question 17

Loading content…

We need to calculate the average density of the Sun using its mass and radius to determine its physical state.

Concept: Mass density

The average mass density ( $\rho$ ) of a spherical object is defined as the ratio of its total mass ( $M$ ) to its total volume ( $V$ ). For a sphere of radius $R$ , the volume is given by:

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

The density is then:

\rho = \frac{M}{V}

Given:

Mass of the Sun ( $M$ ) = $2.0 \times 10^{30} \text{ kg}$
Radius of the Sun ( $R$ ) = $7.0 \times 10^8 \text{ m}$

Solving:

First, calculate the volume of the Sun:

\begin{aligned} V &= \frac{4}{3} \pi R^3 \\ &= \frac{4}{3} \times \pi \times (7.0 \times 10^8 \text{ m})^3 \\ &= \frac{4}{3} \times \pi \times 3.43 \times 10^{26} \text{ m}^3 \\ &\approx 1.4 \times 10^{27} \text{ m}^3 \end{aligned}

Next, calculate the average density:

\begin{aligned} \rho &= \frac{M}{V} \\ &= \frac{2.0 \times 10^{30} \text{ kg}}{1.4 \times 10^{27} \text{ m}^3} \\ &\approx 1.4 \times 10^3 \text{ kg} \cdotp \text{ m}^{-3} \end{aligned}

Comparison with known densities:

Density of gases (at STP) is approximately $1 \text{ kg} \cdotp \text{ m}^{-3}$ .
Density of liquids (like water) is approximately $1000 \text{ kg} \cdotp \text{ m}^{-3}$ .
Density of solids typically ranges from $1000 \text{ kg} \cdotp \text{ m}^{-3}$ to $20000 \text{ kg} \cdotp \text{ m}^{-3}$ .

The calculated density of $1.4 \times 10^3 \text{ kg} \cdotp \text{ m}^{-3}$ is much higher than that of typical gases and falls within the range of densities for solids and liquids. This high density, despite the high temperature, is due to the intense gravitational attraction pulling the solar plasma toward the center.

Answer

1.4 \times 10^3 \text{ kg} \cdotp \text{ m}^{-3}

This value lies in the range of densities of solids and liquids.

NCERT Solutions