2020 A/L Physics Past Paper · Question 46 (English Medium)

Two joined tubes — real cubical expansivity of a liquid

Two vertical glass tubes T1T_1, T2T_2 are joined at their lower ends by a horizontal capillary tube and filled with a liquid. One tube (T1)(T_1) is immersed in a mixture of ice and water at 0∘C0^\circ\text{C}, and the other tube (T2)(T_2) is immersed in water kept at a constant temperature of 40∘C40^\circ\text{C}. The difference between the heights of the liquid columns in the two tubes is 1.6 cm1.6\ \text{cm}, and the height of the liquid column at 0∘C0^\circ\text{C} is 80 cm80\ \text{cm}. (The figure is not drawn to scale.) The real cubical expansivity of the liquid is

  1. (1) 2.5×10−4 ∘C−12.5 \times 10^{-4}\ {}^\circ\text{C}^{-1}
  2. (2) 5.0×10−4 ∘C−15.0 \times 10^{-4}\ {}^\circ\text{C}^{-1}
  3. (3) 6.0×10−4 ∘C−16.0 \times 10^{-4}\ {}^\circ\text{C}^{-1}
  4. (4) 1.0×10−3 ∘C−11.0 \times 10^{-3}\ {}^\circ\text{C}^{-1}
  5. (5) 1.2×10−3 ∘C−11.2 \times 10^{-3}\ {}^\circ\text{C}^{-1}
Answer and explanation

Answer: (2) 5.0×10−4 ∘C−15.0 \times 10^{-4}\ {}^\circ\text{C}^{-1}

Given information

We know: the 0∘C0^\circ\text{C} column h0=80h_0 = 80 cm; the 40∘C40^\circ\text{C} column h=80+1.6=81.6h = 80 + 1.6 = 81.6 cm; temperature difference θ=40∘C\theta = 40^\circ\text{C}; the two tubes are joined at the bottom.

What the question asks

We must find: the real (true) cubical expansivity γ\gamma of the liquid.

Relevant theory

So it is clear that the density of liquids decreases with increase of temperature.

V2 = V1(1 + γθ)

Hence pA=hρgp_A = h\rho g

The pressure at the point B which is at the same level, is also the same.

Required equations

p0+h0ρ0g=p0+hρθgp_0 + h_0\rho_0 g = p_0 + h\rho_\theta g

ρθ=ρ01+γθ\rho_\theta = \frac{\rho_0}{1 + \gamma\theta}

h=h0(1+γθ)⇒γ=h−h0h0 θh = h_0(1 + \gamma\theta) \Rightarrow \gamma = \frac{h - h_0}{h_0\,\theta}

Applying it to this question

Plan: (1) the pressure is equal on both sides of the join⇒(1)\,the\,pressure\,is\,equal\,on\,both\,sides\,of\,the\,join \Rightarrow h0ρ0=hρθh_0\rho_0 = h\rho_\theta.

(2) substitute ρθ=ρ0/(1+γθ)\rho_\theta = \rho_0/(1 + \gamma\theta) to get h/h0h/h_0.

(3) find γ\gamma.

Atmospheric pressure p0p_0 is the same at both surfaces.

Reasoning and calculation

1.021.02

ρ0ρθ=hh0=81.680\frac{\rho_0}{\rho_\theta} = \frac{h}{h_0} = \frac{81.6}{80}

1+γ×40=1.021 + \gamma \times 40 = 1.02

0.020.02

γθ=0.02\gamma\theta = 0.02

5.0×10−45.0 \times 10^{-4}

γ=1.680×40\gamma = \frac{1.6}{80 \times 40}

Checks

Unit check: cmcm×∘C=∘C−1\dfrac{\text{cm}}{\text{cm} \times {}^\circ\text{C}} = {}^\circ\text{C}^{-1} ✓.

Physical plausibility: 5×10−4 ∘C−15 \times 10^{-4}\ {}^\circ\text{C}^{-1} is a typical value for liquids (the book's own example is 10−310^{-3} K−1K^{-1}).

The hotter column is taller — the less dense liquid needs a taller column for the same pressure ✓.

Since the glass's own expansion does not affect the heights here, this is the real expansivity.

dh/(h0∗th)dh / (h0 * th)

Conclusion

γ=h−h0h0θ=5.0×10−4 ∘C−1\gamma = \frac{h - h_0}{h_0\theta} = 5.0 \times 10^{-4}\ ^\circ\text{C}^{-1}

h0ρ0=hρθh_0\rho_0 = h\rho_\theta, ρθ=ρ0/(1+γθ)\rho_\theta = \rho_0/(1 + \gamma\theta) ⇒ γ=1.680×40=5.0×10−4 ∘C−1\gamma = \dfrac{1.6}{80 \times 40} = 5.0 \times 10^{-4}\ {}^\circ\text{C}^{-1} — the answer is (2).

Expecting both columns to be at the same level.

At different temperatures the densities differ; for equal pressure at the bottom, the hotter (less dense) column must be taller.

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