Question

The following observations were taken for determining surface tension $T$ of water by the capillary method:
Diameter of capillary, $D = 1.25 \times {10^{ - 2}}m$
Rise of water, $h = 1.45 \times {10^{ - 2}}m$
Using $g = 9.8m/{s^2}$and the simplified relation, $T = \dfrac{{rhg}}{2} \times {10^3}N/m$ the possible error in surface tension is closest to:
A.$10\%$
B.$0.15\%$
C.$1.5\%$
D.$2.4\%$

Answer 1

In the question, we are given the observations for determining the surface tension of water by the capillary method. The diameter of the capillary tube, the rise of water, and the acceleration due to gravity are given. The expression for surface tension is also given. We have to find the possible error in the surface tension using the given values.

Complete answer:
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,$g = 9.8m/{s^2}$
The relation for surface tension is given by, $T = \dfrac{{rhg}}{2} \times {10^3}N/m$
We know that $r = \dfrac{D}{2}$
where $r$ stands for the radius of the capillary tube and $D$ stands for the diameter of the capillary tube.
Substituting this value of $r$ in the equation for surface tension, we get
$T = \dfrac{{Dhg}}{4} \times {10^3}N/m$
Taking logarithm on both sides, we get
$\log T = \log D + \log h + \log g - \log 4$
Since $\log g$and $\log 4$have constant values that are not experimentally measured, there will not be any error in them.
Hence we can neglect them in the expression.
Now the expression will become,
$\log T = \log D + \log h$
Therefore, the error in $T$ can be only due to the error in $D$and $h$
The error in $T$,$D$and $h$can be written as,
$\dfrac{{\Delta T}}{T} = \dfrac{{\Delta D}}{D} + \dfrac{{\Delta h}}{h}$
The maximum error in $D$ can be written as,
$\Delta D = 0.01$
The maximum error in $h$ can be written as,
$\Delta h = 0.01$
Diameter of capillary, $D = 1.25 \times {10^{ - 2}}m$
Rise of water, $h = 1.45 \times {10^{ - 2}}m$
Substituting the values in the expression for surface tension,
$\dfrac{{\Delta T}}{T} = \dfrac{{0.01}}{{1.25}} + \dfrac{{0.01}}{{1.45}} = 0.015$
The percentage error can be written as,
$\dfrac{{\Delta T}}{T} \times 100 = 0.015 \times 100 = 1.5\%$

The answer is Option (C): $1.5\%$

Note:
The magnitude of the difference between the true value and the measured value of a physical quantity is called the absolute error of the measurement. The relative error or fractional error is defined as the ratio of mean absolute error to the true or mean value. The relative error expressed in percentage is called the percentage error.