A vernier callipers has its main scale 10 cm equally divided into 250 parts its vernier scale is of 50 division and coincide with 15mm of the main scale the least count of the instrument is

Options

A. 0.01 cm
B. 0.02 cm
C. 0.001 cm
D. 0.002 cm

Explanation

To find the least count of the vernier callipers, we can follow these steps:

### Step 1: Calculate the value of one main scale division (msd).
The main scale length is given as 10 cm, which is divided into 250 parts.

\[ \text{1 msd} = \frac{\text{Total length of main scale}}{\text{Number of divisions}} = \frac{10 \text{ cm}}{250} = \frac{100 \text{ mm}}{250} = 0.4 \text{ mm} \]

### Step 2: Calculate the value of one vernier scale division (vsd).
The vernier scale has 50 divisions that coincide with 15 mm of the main scale.

\[ \text{1 vsd} = \frac{\text{Length of vernier scale}}{\text{Number of divisions}} = \frac{15 \text{ mm}}{50} = 0.3 \text{ mm} \]

### Step 3: Calculate the least count (LC) of the vernier callipers.
The least count is given by the formula:

\[ \text{LC} = \text{1 msd} - \text{1 vsd} \]

Substituting the values we calculated:

\[ \text{LC} = 0.4 \text{ mm} - 0.3 \text{ mm} = 0.1 \text{ mm} \]

### Step 4: Convert the least count to centimeters.
Since the options are in centimeters, we convert 0.1 mm to centimeters:

\[ 0.1 \text{ mm} = 0.1 \times 10^{-1} \text{ cm} = 0.01 \text{ cm} \]

### Final Answer:
The least count of the instrument is **0.01 cm**.

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