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A glass capillary tube with a radius of r=0.2 \mathrm{~mm} is dipped into a container of liquid mercury. The liquid mercury falls to a height of h=3.0 \mathrm{~mm} below the normal liquid level inside the capillary due to capillary action. Calculate the surface tension of mercury using the given data. If a detergent is added to the mercury, increasing the surface tension to 1.5 times its original value, calculate the new capillary fall.

Option: 1

0.05 \mathrm{~m}


Option: 2

0.0895 \mathrm{~m}


Option: 3

0.06 \mathrm{~m}


Option: 4

0.002 \mathrm{~m}


Answers (1)

best_answer

Given:

  • Radius of capillary tube r=0.2 \mathrm{~mm}=0.2 \times 10^{-3} \mathrm{~m}
  • Capillary fall h=3.0 \mathrm{~mm}=3.0 \times 10^{-3} \mathrm{~m}
  • Increased surface tension due to detergent T^{\prime}=1.5 \times T, where T is the original surface tension.

Step 1: Calculate the original surface tension T using the capillary fall formula:

            \begin{gathered} T=\frac{4 \times h \times r}{\pi r^2} \\ T=\frac{4 \times 3.0 \times 10^{-3} \times 0.2 \times 10^{-3}}{\pi \times\left(0.2 \times 10^{-3}\right)^2} \\ T=4.772 \times 10^{-2} \mathrm{~N} / \mathrm{m} \end{gathered}

Step 2: Calculate the new surface tension T^{\prime} due to the detergent:

            T^{\prime}=1.5 \times 4.772 \times 10^{-2} \mathrm{~N} / \mathrm{m}

            T^{\prime}=7.158 \times 10^{-2} \mathrm{~N} / \mathrm{m

Step 3: Calculate the new capillary fall h^{\prime} using the increased surface tension T^{\prime} :

            \begin{gathered} h^{\prime}=\frac{T^{\prime}}{4 \times r} \\ h^{\prime}=\frac{7.158 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{4 \times 0.2 \times 10^{-3} \mathrm{~m}} \\ h^{\prime}=0.0895 \mathrm{~m} \end{gathered}

The surface tension of mercury (T) is 4.772 \times 10^{-2} \mathrm{~N} / \mathrm{m}. After adding the detergent, the new capillary fall \left(h^{\prime}\right) is 0.0895 \mathrm{~m}.

So, option B is correct

Posted by

jitender.kumar

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