NCERT Solutions

We need to estimate the diameter of a thin thread using a metre scale by measuring a wound coil.

Concept: Averaging method

The averaging method is used to measure dimensions smaller than the least count of the instrument. By winding the thread into a coil of $n$ turns, the total length $L$ becomes large enough to measure accurately.

$$d = \frac{L}{n}$$

Given:

• Instrument: Metre scale (least count = $1\,\text{mm}$).
• Material: A long thread.

Solving:

1. Wind the thread tightly around the metre scale (or a cylindrical rod) such that the turns are adjacent to each other with no gaps.
2. Count the total number of turns, denoted as $n$.
3. Measure the total width of the coil, $L$, using the metre scale.
4. Calculate the diameter $d$ of the thread by dividing the total length by the number of turns:

$$d = \frac{L}{n}$$

Answer

$$d = \frac{L}{n}$$

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We need to determine if increasing the number of circular scale divisions can indefinitely improve a screw gauge's accuracy.

Concept: Least count

The least count ($LC$) of a screw gauge is the smallest distance it can measure, defined by the ratio of pitch to the number of divisions $N$.

$$LC = \frac{\text{Pitch}}{N}$$

Given:

• $\text{Pitch} = 1.0\,\text{mm}$
• $N = 200$

Solving:

1. Calculate the current least count:

$$\begin{aligned} LC &= \frac{1.0\,\text{mm}}{200} \\ &= 0.005\,\text{mm} \end{aligned}$$

2. Mathematically, increasing $N$ decreases $LC$, which suggests higher precision.
3. However, practical limits exist. If $N$ is too large, the divisions become so close that the human eye cannot distinguish them (resolving power limit).
4. Mechanical imperfections, such as backlash error and non-uniformity of the screw thread, eventually exceed the theoretical $LC$.

Answer

$$\text{No, practical limits like resolution and mechanical errors prevent arbitrary accuracy.}$$

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We need to explain why taking 100 measurements of a rod's diameter is superior to taking just a few.

Concept: Random errors

Measurements are subject to random errors, which fluctuate unpredictably. According to the law of errors, the uncertainty in the mean of $n$ measurements is inversely proportional to the square root of $n$.

$$\text{Mean Error} \propto \frac{1}{\sqrt{n}}$$

Given:

• Number of measurements ($n$) = $100$.

Solving:

1. Every individual measurement contains a random error component.
2. By taking the arithmetic mean of $n$ measurements, the random fluctuations tend to cancel each other out.
3. If the error in a single measurement is $\pm \Delta x$, the error in the average of 100 measurements is:

$$\begin{aligned} \text{Error}_{\text{mean}} &= \frac{\Delta x}{\sqrt{100}} \\ &= \frac{\Delta x}{10} \end{aligned}$$

4. Thus, 100 measurements reduce the random error by a factor of $10$ compared to a single measurement, providing a much more reliable and precise value.

Answer

$$\text{To minimize random errors and increase the reliability of the mean value.}$$