Question

The following observations were taken for determining surface tension T of water by the capillary method: diameter of the capillary, $\mathrm{D}=1.25\mathrm{x}{10}^{-2}\mathrm{m}$ rise of water, $\mathrm{h}=1.45\mathrm{x}{10}^{-2}\mathrm{m}$. Using $\mathrm{g}=9.80\mathrm{m}/{\mathrm{s}}^2$and the simplified relation $\mathrm{T}=\frac{\mathrm{rhg}}{2}\mathrm{x}{10}^3\mathrm{N}/\mathrm{m},$ 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, $h=1.45\times {10}^{-2}m$
The acceleration due to gravity, $g=9.8m/s^2$

Given, the formula for surface tension is,

$\mathrm{T}=\frac{\mathrm{rhg}}{2}\times {10}^3\mathrm{N}/\mathrm{m}$

Step 2: Formula for surface tension:

The formula for surface tension can be given as:

$\mathrm{T}=\frac{\mathrm{rhg}}{2}\times {10}^3\mathrm{N}/\mathrm{m}$
Also,$\mathrm{r}=\frac{\mathrm{D}}{2}$
where $r$ stands for the radius of the capillary tube and $D$ stands for the diameter of the capillary tube.

Step 3: Calculation of surface tension:
Substituting this value of $r$ in the given equation we can calculate surface tension as:
$\mathrm{T}=\frac{\mathrm{Dhg}}{4}\times {10}^3\mathrm{N}/\mathrm{m}$

Step 4: Expression of the possible error in surface tension:
Taking logarithm on both sides of the above equation, we get
$\log T=\log D+\log h+\log g-\log 4+\log \left({10}^3\right)$
Here, $\log \left({10}^3\right),\log 4$, and $\log g$ have constant value; hence, there will be no error in them, so we can neglect them in the expression.
Hence, the expression will become,
$\log T=\log D+\log h$ …..(1)
Thus, here the error in $T$ can be only due to the error in $D$and $h$

Step 5: Expressing the final expression:
By differentiating, the equation (1) can be written as,
$\mathrm{d}\left(\log \mathrm{T}\right)=\mathrm{d}\left(\log \mathrm{D}\right)+\mathrm{d}\left(\log \mathrm{h}\right)$

$$\begin{gathered} (\because \frac{d\left(\mathrm{logx}\right)}{\mathrm{dx}}=\frac{1}{\mathrm{x}} \\ \Rightarrow d\left(\mathrm{logx}\right)=\frac{\mathrm{dx}}{\mathrm{x}}) \end{gathered}$$
${\displaystyle \frac{\mathrm{\Delta }T}{T}}={\displaystyle \frac{\mathrm{\Delta }D}{D}}+{\displaystyle \frac{\mathrm{\Delta }h}{h}}$
The maximum error in $D$ can be written as,

As the number of significant figures in the given value of $D$ is 3 therefore,
$\mathrm{\Delta }D=0.01\times {10}^{-2}m$
The maximum error in $h$ can be written as,

Since the number of significant figures in the given value of $h$ is 3 therefore,
$\mathrm{\Delta }h=0.01\times {10}^{-2}m$
Step 6: Calculation of the error in surface tension:
Substituting the values in the given expression, we can calculate the surface tension as:
$\frac{\mathrm{\Delta T}}{\mathrm{T}}=\frac{0.01\times {10}^{-2}}{1.25\times {10}^{-2}}+\frac{0.01\times {10}^{-2}}{1.45\times {10}^{-2}}=0.015$

Step 7: Calculation of the percentage error in surface tension:
The percentage error of surface tension can be written as,
$$\begin{gathered} \frac{\mathrm{\Delta T}}{\mathrm{T}}\times 100 \\ =0.015\times 100 \\ =1.5\mathrm{\%} \end{gathered}$$
Thus, the possible error in surface tension is the closest to $1.5\%$.

Hence, option B is correct.