Question:
Two substances of densities \(\rho_1\) and \(\rho_2\) are mixed in equal volumes, mixture density = 4. When mixed in equal masses, density = 3. Find \(\rho_1, \rho_2\).
Two substances of densities \(\rho_1\) and \(\rho_2\) are mixed in equal volumes, mixture density = 4. When mixed in equal masses, density = 3. Find \(\rho_1, \rho_2\).
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Equal volume $\Rightarrow$ arithmetic mean; equal mass $\Rightarrow$ harmonic mean.
Updated On: Apr 23, 2026
- \(\rho_1 = 6,\ \rho_2 = 2\)
- \(\rho_1 = 3,\ \rho_2 = 5\)
- \(\rho_1 = 12,\ \rho_2 = 4\)
- None of these
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The Correct Option is A
Solution and Explanation
Step 1: Equal volume mixture
\[ \rho = \frac{\rho_1 + \rho_2}{2} = 4 \Rightarrow \rho_1 + \rho_2 = 8 \]
Step 2: Equal mass mixture
\[ \rho = \frac{2\rho_1\rho_2}{\rho_1 + \rho_2} = 3 \] \[ \frac{2\rho_1\rho_2}{8} = 3 \Rightarrow \rho_1 \rho_2 = 12 \]
Step 3: Solve
\[ x^2 - 8x + 12 = 0 \Rightarrow x=6,2 \] Conclusion: \[ {(A)} \]
\[ \rho = \frac{\rho_1 + \rho_2}{2} = 4 \Rightarrow \rho_1 + \rho_2 = 8 \]
Step 2: Equal mass mixture
\[ \rho = \frac{2\rho_1\rho_2}{\rho_1 + \rho_2} = 3 \] \[ \frac{2\rho_1\rho_2}{8} = 3 \Rightarrow \rho_1 \rho_2 = 12 \]
Step 3: Solve
\[ x^2 - 8x + 12 = 0 \Rightarrow x=6,2 \] Conclusion: \[ {(A)} \]
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