Question

The following observations were taken for determining surface tension $T$ of water by the capillary method diameter of the capillary, $D=1·25\times {10}^{-2}m$ rise of water, $h=1.45\times {10}^{-2}m$. Using $g=9.80m/s^2$ and the simplified relation $T=\frac{rhg}{2}\times {10}^{3}N{m}^{-1}$ the possible error in surface tension is closest to:

• A. $0·15\%$
• B. $1·5\%$
• C. $2·4\%$
• D. $10\%$

Answer 1

The correct option is B

$1·5\%$

Step 1: Given data

Diameter of capillary, $D=1·25\times {10}^{-2}m$

Rise of water,$h=1.45\times {10}^{-2}m$

Acceleration due to gravity $g=9.80m/s^2$

Relation between mentioned parameter $T=\frac{rhg}{2}\times {10}^{3}N{m}^{-1}$

Step 2: Modify the relation

The relationship equation of surface tension can be written as follows,

$T=\frac{rhg}{2}\times {10}^{3}N{m}^{-1}$

$$\begin{gathered} multiplyanddivideitby2 \\ T=\frac{2rhg}{4}\times {10}^{3}N{m}^{-1} \end{gathered}$$

We know that $D=2r$. So,

$T=\frac{Dhg}{4}\times {10}^3----\left(1\right)$

Step 3: Determine the percentage error

Take the log of equation $\left(1\right)$,

$$\begin{gathered} \log T=\log \frac{Dhg}{4}\times {10}^3 \\ \log T=\log D+\log h+\log g-\log 4+\log 3 \end{gathered}$$

It is understood that,

$\frac{∆T}{T}=\frac{∆D}{D}+\frac{∆h}{h}$. So,

The least count in the measurement of $D$ would be $∆D=0.01\times {10}^{-2}m$

Similarly, The least count in the measurement of $h$ would be $∆h=0.01\times {10}^{-2}m$

$$\begin{gathered} \frac{∆T}{T}=\frac{0·01\times {10}^{-2}}{1·25\times {10}^{-2}}+\frac{0·01\times {10}^{-2}}{1·45\times {10}^{-2}} \\ \frac{∆T}{T}=0·015 \\ \frac{∆T}{T}\times 100=1·5\% \end{gathered}$$

Therefore, the maximum error would be $1·5\%$

Hence, option B is the correct answer.