1. Given data

For capillary rise method, $T=\frac{r h g \rho}{2}$ Here the simplified relation is given, so for error calculation we only need proportionality: $T \propto r h$ Since diameter $D$ is measured, and $r=\frac{D}{2}$ so $T \propto D h$

Given observations:

• $D = 1.25 \times 10^{-2}\,\text{m}$
• $h = 1.45 \times 10^{-2}\,\text{m}$
• $g = 9.80\,\text{m/s}^2$

2. Find the possible errors in measured quantities

The possible (maximum) error is taken as half of the least count in the last written digit.

• For $D = 1.25 \times 10^{-2}\,\text{m}$, the last digit corresponds to $0.01 \times 10^{-2}$. So possible absolute error: $\Delta D = 0.01 \times 10^{-2}\,\text{m}$ Hence fractional error: $\frac{\Delta D}{D} = \frac{0.01}{1.25} = 0.008 = 0.8\%$

• For $h = 1.45 \times 10^{-2}\,\text{m}$, similarly, $\Delta h = 0.01 \times 10^{-2}\,\text{m}$ Hence fractional error: $\frac{\Delta h}{h} = \frac{0.01}{1.45} \approx 0.00690 = 0.69\%$

3. Error in surface tension

Since $T \propto D h$ maximum fractional error is $\frac{\Delta T}{T} = \frac{\Delta D}{D} + \frac{\Delta h}{h}$

So, $\frac{\Delta T}{T} = \frac{0.01}{1.25} + \frac{0.01}{1.45}$ $= 0.008 + 0.00690$ $= 0.01490$

Therefore percentage error: $0.01490 \times 100 \approx 1.49\%$

4. Closest option

$1.49\% \approx 1.5\%$

So the correct option is: C: $1.5\%$

5. Comparison with stored answer

Stored correct answer is C. Our derived answer is also C.