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9. (a) When equal volumes of two substances are mixed the density of the mixture is 4000 kg m^-3 - Leaving Cert Applied Maths - Question 9 - 2014
Question 9
9. (a) When equal volumes of two substances are mixed the density of the mixture is 4000 kg m^-3. When equal masses of the same two substances are mixed the density ... show full transcript
Worked Solution & Example Answer:9. (a) When equal volumes of two substances are mixed the density of the mixture is 4000 kg m^-3 - Leaving Cert Applied Maths - Question 9 - 2014
Step 1
(a) Find the density of each of the substances.
Answer
Let the densities of the two substances be ( \rho_1 ) and ( \rho_2 ).
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Equal Volumes Mixture: When equal volumes are mixed the density of the mixture is given by:
[ \rho_m = \frac{\rho_1 V + \rho_2 V}{V + V} = \frac{\rho_1 + \rho_2}{2} \quad (1) ]
Given ( \rho_m = 4000 \text{ kg m}^{-3} ), we have:
[ \rho_1 + \rho_2 = 8000 \quad (2) ]
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Equal Masses Mixture: When equal masses are mixed:
[ \rho_m = \frac{m_1}{V_1} + \frac{m_2}{V_2} \quad (3) ]
Let the mass of each substance be ( m ) and their volumes be ( V_1 = \frac{m}{\rho_1} ) and ( V_2 = \frac{m}{\rho_2} ). Hence,
[ \rho_m = \frac{\rho_1 \rho_2}{\rho_1 + \rho_2} = 3840 \quad (4) ]
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Solving Equations: From (2) and (4): Substitute ( \rho_2 = 8000 - \rho_1 ) into (4):
[ \frac{\rho_1 (8000 - \rho_1)}{8000} = 3840 ]
Rearranging gives:
[ \rho_1^2 - 8000 \rho_1 + 15360000 = 0 \quad (5) ]
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Using the Quadratic Formula: Solving equation (5) using the quadratic formula:
[ \rho_1 = \frac{8000 \pm \sqrt{8000^2 - 4 \cdot 15360000}}{2} = 3200 ext{ or } 4800 ]
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Finding Densities: Thus, densities are:
- ( \rho_1 = 3200 \text{ kg m}^{-3} )
- ( \rho_2 = 4800 \text{ kg m}^{-3} ).
Step 2
(b) (i) the value of x
Answer
Let ( x ) be the side length of the cube.
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Calculating Volume: The initial water volume is:
[ V_w = x^2 \times 0.18 \text{ m}^3 ]
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Using Buoyancy Principle: The buoyant force on the solid body when floating equals its weight.
[ B = \rho_{water} \times V_{displaced} \times g ]
- Here, ( g = 9.81 \text{ m/s}^2 ), so( B = 120 ext{ N} ).
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Displacement Calculation: The volume of water displaced when floating:
[ V_{displaced} = x^2 \times (0.24 - 0.18) \text{ m}^3 = x^2 \times 0.06 \text{ m}^3 ]
Thus,
[ 1000 \times x^2 \times 0.06 = 120 ]
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Solving for x:
[ x^2 = \frac{120}{60} = 2 \implies x \approx 1.41 \text{ m} \text{ (or in cm) } ]
Hence, ( x = 45.2 ext{ cm} ).
Step 3
(b) (ii) the tension in the string
Answer
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Total Force Calculation: When submerged, the total buoyant force is:
[ T + W = \rho_{water} \times x^2 \times 0.14 \times g ]
- Substituting known values:
[ T + 120 = 1000 \times (x^2) \times 0.14 ]
- Replacing ( x ) gives:
[ T + 120 = 280 \implies T = 280 - 120 = 160 ext{ N} ].
Step 4