The following observations were taken for determining surface tension $$T$$ of water by capillary method:
diameter of capillary, $$D = 1.25 \times 10^{-2}$$ m
rise of water, $$h = 1.45 \times 10^{-2}$$ m
Using $$g = 9.80$$ m 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:

We have been given the simplified working formula

$$T=\frac{r\,h\,g}{2}\times 10^{3}\;{\rm N\,m^{-1}}$$

and the two quantities that have to be measured experimentally are the radius $$r$$ (through the diameter $$D$$) and the height of capillary rise $$h$$. The acceleration due to gravity $$g$$ and the numerical factor $$10^{3}$$ are constants, so any error in $$T$$ can arise only from the errors in $$D$$ and $$h$$.

The theory of errors tells us that, if a result

$$Q=A^{\alpha }B^{\beta }\dots$$

depends on measured quantities $$A,B,\dots$$, then the relative (or percentage) error in $$Q$$ is obtained from

$$\frac{\Delta Q}{Q}=\left|\alpha\right|\frac{\Delta A}{A}+\left|\beta\right|\frac{\Delta B}{B}+\dots$$

In the present experiment we first express the surface-tension formula only through the directly measured quantities. Since

$$r=\frac{D}{2},$$

we may rewrite

$$T=\frac{\left(\dfrac{D}{2}\right)h\,g}{2}\times 10^{3} =\frac{D\,h\,g}{4}\times 10^{3}.$$

Thus $$T$$ is proportional to the first power of $$D$$ and to the first power of $$h$$. Therefore

$$\frac{\Delta T}{T}= \frac{\Delta D}{D}+\frac{\Delta h}{h}.$$

Now we must find the possible (maximum) errors in $$D$$ and $$h$$. The two readings have been quoted to three significant figures:

$$D = 1.25\times 10^{-2}\ {\rm m},\qquad h = 1.45\times 10^{-2}\ {\rm m}.$$

When a quantity is written with three significant figures the last digit can be wrong by at most one unit. Hence the possible absolute error in each reading equals one unit in the last place, i.e.

$$\Delta D = 0.01\times 10^{-2}\ {\rm m}=1.0\times 10^{-4}\ {\rm m},$$ $$\Delta h = 0.01\times 10^{-2}\ {\rm m}=1.0\times 10^{-4}\ {\rm m}.$$

We now compute the relative errors one by one.

For the diameter,

$$\frac{\Delta D}{D}= \frac{1.0\times 10^{-4}} {1.25\times 10^{-2}} =\frac{1.0}{1.25}\times 10^{-2} =0.008 =0.8\%.$$

For the height,

$$\frac{\Delta h}{h}= \frac{1.0\times 10^{-4}} {1.45\times 10^{-2}} =\frac{1.0}{1.45}\times 10^{-2} \approx 0.00690 \approx 0.69\%.$$

Adding these two contributions, we obtain the percentage error in surface tension:

$$\frac{\Delta T}{T}\times 100\% =(0.8\%+0.69\%) =1.49\% \approx 1.5\%.$$

Hence, the correct answer is Option C.