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U-tube capillaries find liquid density

Short answer

Capillary rise difference Δh = (4σ/(ρg))(1/d₁ – 1/d₂), so ρ = (4σ/(g Δh))(1/d₁ – 1/d₂) = [4×0.075/(9.81×0.002)] × (1/0.0015 – 1/0.002) ≈ 2548 kg/m³.

A U-tube is made up of two capillaries of bores 1.5 mm and 2 mm respectively. The U tube is held vertical and partially filled with liquid whose surface tension σ\sigma = 0.075 N/m. Find out the mass density of the liquid if the difference in two menisci is 2 mm. Assume angle of contact is zero.

Data: d1=1.5d_1=1.5 mm, d2=2d_2=2 mm, σ=0.075\sigma=0.075 N/m, θ=0∘\theta=0^\circ, difference of menisci Δh=2\Delta h=2 mm =0.002=0.002 m.

Capillary rise in each limb: h=4σcos⁡θρgdh=\dfrac{4\sigma\cos\theta}{\rho g d}. The narrower limb rises higher, so

Δh=h1−h2=4σρg(1d1−1d2)\Delta h=h_1-h_2=\frac{4\sigma}{\rho g}\left(\frac{1}{d_1}-\frac{1}{d_2}\right) ρ=4σg Δh(1d1−1d2)=4(0.075)9.81(0.002)(10.0015−10.002)\rho=\frac{4\sigma}{g\,\Delta h}\left(\frac{1}{d_1}-\frac{1}{d_2}\right)=\frac{4(0.075)}{9.81(0.002)}\left(\frac{1}{0.0015}-\frac{1}{0.002}\right) ρ=0.30.01962×166.67=2548 kg/m3\rho=\frac{0.3}{0.01962}\times166.67=2548\ \text{kg/m}^3

Answer: ρ≈2548 kg/m3\rho\approx2548\ \text{kg/m}^3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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