9. (a) $V_1$, cm$^3$ of liquid A of relative density 0.8 is mixed with $V_2$, cm$^3$ of liquid B of relative density 0.9 to form a mixture of relative density 0.88 - Leaving Cert Applied Maths - Question 9 - 2013

To find the value of $a$, we start with the equilibrium of forces acting on the floating object:

1. Understanding the Forces: The weight of the object, $W$, is given by:

   $W = B_C + B_D$

   where:

   • $B_C$ and $B_D$ are the buoyant forces from liquids C and D, respectively.

2. Setting Up the Equation: Using the density values:

   $\rho(V_1 + V_2)g = \rho_{C}(V_1)g + \rho_{D}(V_2)g$

   This simplifies to:

   $(\rho - 800)V_1 = (1200 - \rho)V_2$

3. Solving for Fractions: The fraction of volume immersed in liquid D is given by:

   $\frac{V_2}{V_1 + V_2} = \frac{1}{\frac{V_1}{V_2} + 1}$

   Letting:

   $V_2 = \frac{\rho - 800}{1200 - \rho} V_2$

4. Applying the Provided Formula: We know:

   $\frac{V_2}{V_1 + V_2} = \frac{\rho - 800}{400}$

5. Finding the Value of a: Setting the equation equal to:

   $\frac{\rho - 2a}{a} = \frac{\rho - 800}{400}$

   Cross-multiplying leads to:

   $400(\rho - 2a) = a(\rho - 800)$

   This simplifies to:

   $400\rho - 800a = a\rho - 800a$

   Thus:

   $400\rho = 400a \implies a = 400$