Question:medium

Equal masses of two substances of densities _1 and _2 are mixed. Density of mixture is:

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Equal masses → Harmonic mean Equal volumes → Arithmetic mean
Updated On: Jun 10, 2026
  • _1 + _22
  • 2_1_2_1 + _2
  • _1_2_1 + _2
  • _1 + _22_1_2
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Set up the problem.
Equal masses of two substances are mixed. One has density $\rho_1$ and the other has density $\rho_2$. We want the density of the mixture. Let each mass be $m$.

Step 2: Recall the link between mass, volume and density.
Density is mass divided by volume, so volume is mass divided by density: \[ V = \frac{m}{\rho} \]

Step 3: Find each volume.
The first substance takes volume $V_1 = \dfrac{m}{\rho_1}$ and the second takes $V_2 = \dfrac{m}{\rho_2}$.

Step 4: Add up total mass and total volume.
The total mass is $m + m = 2m$. The total volume is: \[ V = \frac{m}{\rho_1} + \frac{m}{\rho_2} = m\left(\frac{1}{\rho_1} + \frac{1}{\rho_2}\right) \]

Step 5: Write the mixture density.
Divide total mass by total volume: \[ \rho_m = \frac{2m}{m\left(\frac{1}{\rho_1} + \frac{1}{\rho_2}\right)} \] The $m$ cancels neatly.

Step 6: Simplify the expression.
Combine the fractions in the denominator: \[ \rho_m = \frac{2}{\frac{\rho_1 + \rho_2}{\rho_1 \rho_2}} = \frac{2\rho_1 \rho_2}{\rho_1 + \rho_2} \] This is the harmonic style average, matching the second option. \[ \boxed{\dfrac{2\rho_1 \rho_2}{\rho_1 + \rho_2}} \]
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