NCERT Solutions

We need to identify the most precise instrument by comparing the Least Count of the three given measuring devices.

Concept: Precision

The precision of a measuring instrument is determined by its Least Count (LC), which is the smallest measurement that can be taken accurately. An instrument with a smaller least count is considered more precise.

• For a Vernier callipers:

  $$LC = \frac{1 \text{ Main Scale Division (MSD)}}{\text{Number of divisions on Vernier scale}}$$

• For a Screw gauge:

  $$LC = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$$

• For an Optical instrument:

  $$LC \approx \text{Wavelength of light } (\lambda)$$

Given:

• (a) Vernier callipers: $n = 20$ divisions on the sliding scale.
• (b) Screw gauge: $\text{Pitch} = 1 \text{ mm}$, $N = 100$ divisions on the circular scale.
• (c) Optical instrument: Precision $\approx \lambda \approx 600 \text{ nm} = 6 \times 10^{-7} \text{ m}$.

Solving:

1. For the Vernier callipers, assuming a standard main scale division of $1 \text{ mm}$:

$$\begin{aligned} LC_{a} &= \frac{1 \text{ mm}}{20} \\ &= 0.05 \text{ mm} \\ &= 5 \times 10^{-5} \text{ m} \end{aligned}$$

2. For the screw gauge:

$$\begin{aligned} LC_{b} &= \frac{1 \text{ mm}}{100} \\ &= 0.01 \text{ mm} \\ &= 1 \times 10^{-5} \text{ m} \end{aligned}$$

3. For the optical instrument, the precision is within a Wavelength of light. Taking the average wavelength of visible light as $\lambda \approx 6000 \text{ \AA}$:

$$\begin{aligned} LC_{c} &\approx 6000 \times 10^{-10} \text{ m} \\ &= 6 \times 10^{-7} \text{ m} \end{aligned}$$

4. Comparing the three values:

• $LC_{a} = 500 \times 10^{-7} \text{ m}$
• $LC_{b} = 100 \times 10^{-7} \text{ m}$
• $LC_{c} = 6 \times 10^{-7} \text{ m}$

Since $LC_{c} < LC_{b} < LC_{a}$, the optical instrument has the smallest least count.

Answer

$$(c) \ \text{An optical instrument that can measure length to within a wavelength of light}$$