An object is made of two equal parts by volume; one part has density $

An object is made of two equal parts by volume; one part has density $\rho_0$ and the other part has density $2\rho_0$. What is the average density of the object?

$3 ho_0$
$ rac{3}{2} ho_0$
$ ho_0$
$ rac{1}{2} ho_0$
This question was previously asked in
UPSC NDA-2 – 2022
Let V be the total volume of the object. The object is made of two equal parts by volume, so the volume of each part is $V_1 = V_2 = V/2$.
Let $\rho_1$ be the density of the first part and $\rho_2$ be the density of the second part.
Given: $\rho_1 = \rho_0$ and $\rho_2 = 2\rho_0$.
The mass of the first part is $m_1 = \rho_1 \times V_1 = \rho_0 \times (V/2)$.
The mass of the second part is $m_2 = \rho_2 \times V_2 = 2\rho_0 \times (V/2) = \rho_0 V$.
The total mass of the object is $M = m_1 + m_2 = \rho_0 (V/2) + \rho_0 V = \rho_0 V (\frac{1}{2} + 1) = \rho_0 V (\frac{3}{2})$.
The total volume of the object is $V_{\text{total}} = V_1 + V_2 = V/2 + V/2 = V$.
The average density of the object is $\rho_{\text{avg}} = \frac{M}{V_{\text{total}}} = \frac{\rho_0 V (3/2)}{V} = \frac{3}{2}\rho_0$.
When calculating average density for parts of equal volume, the average density is the simple arithmetic mean of the densities. However, in this case, the masses are different. The calculation involves finding the total mass and dividing by the total volume.
If the parts were of equal mass instead of equal volume, the calculation would be different, involving the reciprocal of the average of reciprocals (harmonic mean of densities).