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Worked Examples · Example 2.7

Q.A capillary tube of radius 5 × 10⁻⁴ m is immersed in a beaker filled with mercury. The mercury level inside the tube is found to be 8 × 10⁻³ m below the level of reservoir. Determine the angle of contact between mercury and glass. Surface tension of mercury is 0.465 N/m and its density is 13.6 × 10³ kg/m³. (g = 9.8 m/s²)

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✓ Free question

Solve T=rhρg/(2cos⁡θ)T = rh\rho g / (2\cos\theta) for cos θ with the depressed (negative) h.

With r=5×10−4r=5\times10^{-4} m, h=−8×10−3h=-8\times10^{-3} m, T=0.465T=0.465 N/m, ρ=13.6×103\rho=13.6\times10^3 kg/m³: from T=hrρg2cos⁡θT = \dfrac{hr\rho g}{2\cos\theta}, cos⁡θ=−8×10−3×5×10−4×13.6×103×9.82×0.465⇒−cos⁡θ=0.5732\cos\theta = \frac{-8\times10^{-3}\times5\times10^{-4}\times13.6\times10^3\times9.8}{2\times0.465} \Rightarrow -\cos\theta = 0.5732 cos⁡(π−θ)=0.5732⇒180°−θ=55°2′⇒θ=124°58′\cos(\pi-\theta) = 0.5732 \Rightarrow 180°-\theta = 55°2' \Rightarrow \theta = 124°58'

✓Final answer

cos⁡(π−θ)=0.5732\cos(\pi-\theta) = 0.5732, so 180°−θ=55°2′180°-\theta = 55°2' and θ=124°58′\theta = 124°58'.

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