The following observations were taken for determining surface tension T of water by capillary method

Q: Why did we change $r$ to $d/2$?

The measurements provided in the question are for the diameter $d$, not the radius $r$. To correctly apply the relative error formula, the variables in the equation must match the measured quantities.

Q: How did we know the absolute error is $0.01 \times 10^{-2}$?

When the error is not explicitly given, it is taken as the least count of the measuring instrument. Since the values $1.25$ and $1.45$ have two decimal places, the smallest measurable value (least count) is $0.01$.

Q: Why is the error in $g$ ignored?

While $g = 9.80$ technically has an uncertainty of $0.01$, $g$ is usually considered a standard constant in such laboratory experiments. Its relative error is extremely small and negligible compared to the errors in measuring $h$ and $d$.

Q: Why do we add the relative errors instead of subtracting them?

In error analysis, we always calculate the maximum possible error. Therefore, whether quantities are multiplied or divided, their relative errors are always added.

Q: What happens to the constants $4$ and $10^3$ in the error formula?

Exact constants have zero uncertainty. Since their absolute error is zero, their relative error is also zero, and they simply drop out of the relative error equation.

Question

The following observations were taken for determining surface tension $T$ of water by capillary method:
Diameter of capillary, $d = 1.25 	imes 10^{-2} 	ext{ m}$
rise of water, $h = 1.45 	imes 10^{-2} 	ext{ m}$.
Using $g = 9.80 	ext{ m/s}^2$ and the simplified relation $T = \frac{rhg}{2} 	imes 10^3 	ext{ N/m}$, the possible error in surface tension is closest to

Options:

(A) $1.5\%$

(B) $2.4\%$

(C) $10\%$

(D) $0.15\%$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a capillary tube dipped in a beaker of water. The water rises to a specific height $h$ due to surface tension, and the tube itself has a diameter $d$. Our goal is to determine the maximum possible percentage error in calculating the surface tension $T$ using the given experimental observations.

### The Master Equation

The problem provides us with a simplified formula for surface tension:
$$T = \frac{rhg}{2} 	imes 10^3 	ext{ N/m}$$

However, our measurements are given in terms of the diameter $d$, not the radius $r$. To avoid any confusion, we must first substitute $r = \frac{d}{2}$ into our equation:
$$T = \frac{dhg}{4} 	imes 10^3 	ext{ N/m}$$

### Formulating the Error Equation

In error analysis, constants do not contribute to the relative error. Here, the numbers $4$ and $10^3$ are exact constants. The acceleration due to gravity $g$ is also treated as a constant in this context because its error is negligible compared to the experimental measurements of $d$ and $h$.

Therefore, the relative error in $T$ is simply the sum of the relative errors in $d$ and $h$:
$$\frac{\Delta T}{T} = \frac{\Delta d}{d} + \frac{\Delta h}{h}$$

### Uncovering the Hidden Errors

Here is the crucial catch in this problem: the absolute errors $\Delta d$ and $\Delta h$ are not explicitly given. We must deduce them from the significant figures of the measured values.

The measured values are $d = 1.25 	imes 10^{-2} 	ext{ m}$ and $h = 1.45 	imes 10^{-2} 	ext{ m}$. Both values are recorded to two decimal places. This implies that the least count of the measuring instrument is $0.01$.

Thus, the absolute error for both measurements is:
$$\Delta d = 0.01 	imes 10^{-2} 	ext{ m}$$
$$\Delta h = 0.01 	imes 10^{-2} 	ext{ m}$$

### Final Calculation

Now, we substitute these values into our error equation to find the percentage error:
$$\frac{\Delta T}{T} 	imes 100 = \left( \frac{0.01 	imes 10^{-2}}{1.25 	imes 10^{-2}} + \frac{0.01 	imes 10^{-2}}{1.45 	imes 10^{-2}} ight) 	imes 100$$

Notice how the $10^{-2}$ terms cancel out perfectly in both fractions:
$$\frac{\Delta T}{T} 	imes 100 = \left( \frac{0.01}{1.25} + \frac{0.01}{1.45} ight) 	imes 100$$

Multiplying the $100$ inside the bracket makes the calculation much easier:
$$\frac{\Delta T}{T} 	imes 100 = \frac{1}{1.25} + \frac{1}{1.45} = \frac{100}{125} + \frac{100}{145}$$

Solving these fractions:
$$\frac{100}{125} = 0.8$$
$$\frac{100}{145} \approx 0.689$$

Adding them together gives:
$$	ext{Percentage Error} = 0.8 + 0.689 = 1.489\%$$

This value is closest to **$1.5\%$**, which is our final answer.

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The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Master Equation

Formulating the Error Equation

Uncovering the Hidden Errors

Final Calculation

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The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Master Equation

Formulating the Error Equation

Uncovering the Hidden Errors

Final Calculation

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