Least count of a vernier callipers is 0.01 cm. When the two jaws are in touch with each other, the `5^(th)` division of vernier scale coinciding with the main scale lies to the left of the zero of the main scale. Further more, while measuring the diameter of the sphere, the zero mark of the vernier lies in between 2.4 cm and 2.5 cm and ther `6^(th)` vernier division coincides with a main scale division.

Options

A. The instrument has a negative error
B. The instrument has a positive correction
C. The true reading of the diamter is 2.51 cm
D. The true reading of the diameter is 2.41 cm

Answer

To solve the problem step by step, we will first identify the key information given and then calculate the true reading of the diameter of the sphere using the vernier calipers.

### Step 1: Identify the Least Count and Zero Error
- The least count of the vernier calipers is given as 0.01 cm.
- It is mentioned that the 5th division of the vernier scale coincides with the main scale to the left of the zero of the main scale. This indicates a negative zero error.

### Step 2: Calculate the Negative Zero Error
- The negative zero error can be calculated as:
\[ \text{Negative Zero Error} = \text{Number of divisions} \times \text{Least Count} \]
- Here, the number of divisions is 5 (since the 5th division coincides with the main scale).
- Therefore:
\[ \text{Negative Zero Error} = 5 \times 0.01 \, \text{cm} = 0.05 \, \text{cm} \]

### Step 3: Determine the Main Scale Reading
- While measuring the diameter of the sphere, it is stated that the zero mark of the vernier lies between 2.4 cm and 2.5 cm.
- Hence, the main scale reading (MSR) is:
\[ \text{MSR} = 2.4 \, \text{cm} \]

### Step 4: Determine the Vernier Scale Reading
- The 6th division of the vernier scale coincides with a main scale division.
- Therefore, the vernier scale reading (VSR) is:
\[ \text{VSR} = 6 \times \text{Least Count} = 6 \times 0.01 \, \text{cm} = 0.06 \, \text{cm} \]

### Step 5: Calculate the True Reading
- The true reading (TR) can be calculated using the formula:
\[ \text{TR} = \text{MSR} + \text{VSR} + \text{Zero Error} \]
- Since we have a negative zero error, we will add it to the total reading:
\[ \text{TR} = 2.4 \, \text{cm} + 0.06 \, \text{cm} + 0.05 \, \text{cm} \]
- Performing the calculation:
\[ \text{TR} = 2.4 + 0.06 + 0.05 = 2.51 \, \text{cm} \]

### Final Answer
The true reading of the diameter of the sphere is **2.51 cm**.

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