When 2 specific substances are mixed, the volume of the mixture is equal to the sum of the original volumes. When equal volumes of these two substances are mixed, the relative density of the mixture is 0.8. When equal masses of the same two substances are mixed the relative density of the mixture is 0.6. Find the relative densities of the two substances. A liquid of density ρ rests on another liquid of density 2ρ without mixing. A solid of density d floats with its surface totally covered by liquid and with part of its volume immersed in the upper liquid. Show that the fraction of the volume of the solid immersed in the upper liquid is $ \frac{2p - d}{\rho} $

Leaving Cert Applied Maths Hydrostatics: When 2 specific substances are m

When 2 specific substances are mixed, the volume of the mixture is equal to the sum of the original volumes - Leaving Cert Applied Maths - Question 9 - 2019

Question 9

When 2 specific substances are mixed, the volume of the mixture is equal to the sum of the original volumes. When equal volumes of these two substances are mixed, th... show full transcript

Worked Solution & Example Answer:When 2 specific substances are mixed, the volume of the mixture is equal to the sum of the original volumes - Leaving Cert Applied Maths - Question 9 - 2019

Step 1

Relative Densities of the Two Substances (part a)

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Answer

To find the relative densities of the two substances, we denote the densities as ( s_1 ) and ( s_2 ).

Equal Volumes: According to the problem, when equal volumes are mixed: $s_1 + s_2 = 1.6$ Since the relative density of the mixture is 0.8, we have: $\frac{s_1 + s_2}{2} = 0.8 \implies s_1 + s_2 = 1.6$
Equal Masses: When equal masses are mixed: The volume relation is given as follows: $\frac{m}{s_1} + \frac{m}{s_2} = V_t$ Thus: $\frac{1}{s_1} + \frac{1}{s_2} = \frac{2}{V_t}$ Given that the relative density of the mixture is 0.6, we find: $s_1 s_2 = 0.48$
Substituting for s_2: Now, substituting ( s_2 = 1.6 - s_1 ):
$s_1 (1.6 - s_1) = 0.48 \implies s_1^2 - 1.6s_1 + 0.48 = 0$ Using the quadratic formula: $s_1 = \frac{-(-1.6) \pm \sqrt{(-1.6)^2 - 4 \cdot 1 \cdot 0.48}}{2 \cdot 1}$ This leads to the solutions ( s_1 = 0.6 ) and ( s_2 = 1.0. )

Step 2

Fraction of Volume Immersed in the Upper Liquid (part b)

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Answer

To show the fraction of the volume of the solid immersed in the upper liquid:

Balance of Forces: Consider the weight of the solid and the buoyant forces acting on it:
- Weight of solid: ( W = dVg )
- Buoyant force in upper liquid: ( B_u = \rho V_{immersed} g )
- Buoyant force in lower liquid: ( B_l = 2\rho V_{total} g )
Equilibrium Condition: Setting the total upward buoyant force equal to the weight: $\rho V_{immersed} g + 2\rho (V_{total} - V_{immersed}) g = dV g$
Expressing Volume Immersed: Solving the above gives: $k \rho g + 2\rho (1 - k)g = dg$ where ( k = \frac{V_{immersed}}{V_{total}} ightarrow 2p - d = k \rho $$
Conclusion: Finally, rearranging this relation yields: $k = \frac{2p - d}{\rho}$ which is the desired fraction of the volume of the solid immersed in the upper liquid.

When 2 specific substances are mixed, the volume of the mixture is equal to the sum of the original volumes - Leaving Cert Applied Maths - Question 9 - 2019

Worked Solution & Example Answer:When 2 specific substances are mixed, the volume of the mixture is equal to the sum of the original volumes - Leaving Cert Applied Maths - Question 9 - 2019

Relative Densities of the Two Substances (part a)

Fraction of Volume Immersed in the Upper Liquid (part b)

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