A tapering glass capillary tube A of length 0.1 m has diameters 10−3 m and 5×10−4m at the ends. When it is just immersed in a liquid at 0C ∘ with larger radius in contact with liquid surface, the liquid rises 8×10−2 m in the tube. In another experiment, in the cylindrical glass capillary tube B, when immersed in the same liquid at 0C ∘ , the liquid rises to 6×10−2m height. The rise of liquid in tube B is only 5.5×10−2m when the liquid is at 50∘C . Find the rate at which the surface tension changes with temperature considering the changes to be linear. The density of liquid is 114×104kg/m3and the angle of contact is zero. Effect of temperature on the density of liquid and glass is negligible.

R2−rh=R2−R1l=tanα ⇒r=R1−(R1−R2)hl⇒r=0.5×10−3−(0.5×10−3−2.5×10−4)8×10−210−1 =0.5×10−3−(0.25×10−3)×8×10−1=0.5×10−3−2×10−4=0.3×10−3m Surface tension of liquid at 00CS0=rhρg2cosθ(θ=0)=0.3×10−3×8×10−2×10414×9.82=0.3×8×9.8×10−12×14=0.6×9.87×10−1=0.084 N/m If r0→ radius of cylindrical tubeS0=r0hρg2(at 00C)S50=r0h50ρg2(at 500C)S50S0=h50h0=5.5×10−26×10−1=1112S50=1112×S0=1112×0.084=0.077N/m Rate of change of surface tension with temperatureS50−S050=0.077−0.08450=−1.4×10−4Nm0C

A tapering glass capillary tube A of length 0.1 m has diameters 10−3 m and 5×10−4m at the ends. When it is just immersed in a liquid at 0C ∘ with larger radius in contact with liquid surface, the liquid rises 8×10−2 m in the tube. In another experiment, in the cylindrical glass capillary tube B, when immersed in the same liquid at 0C ∘ , the liquid rises to 6×10−2m height. The rise of liquid in tube B is only 5.5×10−2m when the liquid is at 50∘C . Find the rate at which the surface tension changes with temperature considering the changes to be linear. The density of liquid is 114×104kg/m3and the angle of contact is zero. Effect of temperature on the density of liquid and glass is negligible.

R2−rh=R2−R1l=tanα ⇒r=R1−(R1−R2)hl⇒r=0.5×10−3−(0.5×10−3−2.5×10−4)8×10−210−1 =0.5×10−3−(0.25×10−3)×8×10−1=0.5×10−3−2×10−4=0.3×10−3m Surface tension of liquid at 00CS0=rhρg2cosθ(θ=0)=0.3×10−3×8×10−2×10414×9.82=0.3×8×9.8×10−12×14=0.6×9.87×10−1=0.084 N/m If r0→ radius of cylindrical tubeS0=r0hρg2(at 00C)S50=r0h50ρg2(at 500C)S50S0=h50h0=5.5×10−26×10−1=1112S50=1112×S0=1112×0.084=0.077N/m Rate of change of surface tension with temperatureS50−S050=0.077−0.08450=−1.4×10−4Nm0C

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A tapering glass capillary tube A of length 0.1 m has diameters $10^{- 3}$ m and $5 \times 10^{- 4} m$ at the ends. When it is just immersed in a liquid at $0 {}^{\circ}C$ with larger radius in contact with liquid surface, the liquid rises $8 \times 10^{- 2}$ m in the tube. In another experiment, in the cylindrical glass capillary tube B, when immersed in the same liquid at $0 {}^{\circ}C$ , the liquid rises to $6 \times 10^{- 2} m$ height. The rise of liquid in tube B is only $5.5 \times 10^{- 2} m$ when the liquid is at $50^{\circ} C$ . Find the rate at which the surface tension changes with temperature considering the changes to be linear. The density of liquid is $\frac{1}{14} \times 10^{4} k g / m^{3}$ and the angle of contact is zero. Effect of temperature on the density of liquid and glass is negligible.

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$- 1.4 \times 10^{- 4} \frac{N}{m^{0} C}$

$- 3.8 \times 10^{- 4} \frac{N}{m^{0} C}$

$- 4.6 \times 10^{- 4} \frac{N}{m^{0} C}$

$- 2.6 \times 10^{- 4} \frac{N}{m^{0} C}$

answer is A.

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Detailed Solution

$\frac{R_{2} - r}{h} = \frac{R_{2} - R_{1}}{l} = \tan α$
$\begin{array}{l} \Rightarrow r = R_{1} - (R_{1} - R_{2}) \frac{h}{l} \\ \Rightarrow r = 0.5 \times 10^{- 3} - (0.5 \times 10^{- 3} - 2.5 \times 10^{- 4}) \frac{8 \times 10^{- 2}}{10^{- 1}} \end{array}$
$\begin{array}{l} = 0.5 \times 10^{- 3} - (0.25 \times 10^{- 3}) \times 8 \times 10^{- 1} \\ = 0.5 \times 10^{- 3} - 2 \times 10^{- 4} \\ = 0.3 \times 10^{- 3} m \end{array}$
Surface tension of liquid at $0^{0} C$
$\begin{array}{l} S_{0} = \frac{r h ρ g}{2 \cos θ} (θ = 0) \\ = \frac{0.3 \times 10^{- 3} \times 8 \times 10^{- 2} \times \frac{10^{4}}{14} \times 9.8}{2} \\ = \frac{0.3 \times 8 \times 9.8 \times 10^{- 1}}{2 \times 14} = \frac{0.6 \times 9.8}{7} \times 10^{- 1} \\ = 0.084 N / m \end{array}$
If $r_{0} \to$ radius of cylindrical tube
$S_{0} = \frac{r_{0} h ρ g}{2} (a t 0^{0} C)$
$S_{50} = \frac{r_{0} h_{50} ρ g}{2} (a t 50^{0} C)$
$\begin{array}{l} \frac{S_{50}}{S_{0}} = \frac{h_{50}}{h_{0}} = \frac{5.5 \times 10^{- 2}}{6 \times 10^{- 1}} = \frac{11}{12} \\ S_{50} = \frac{11}{12} \times S_{0} = \frac{11}{12} \times 0.084 = 0.077 N / m \end{array}$
Rate of change of surface tension with temperature
$\frac{S_{50} - S_{0}}{50} = \frac{0.077 - 0.084}{50} = - 1.4 \times 10^{- 4} \frac{N}{m^{0} C}$