Diameter of a steel ball is measured using a Vernier callipers which has divisions of 0.1 cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurement for a ball are given as :
`{:("S.No.","MS (cm)","VS divisions"),(1.,0.5,8),(2.,0.5,4),(3.,0.5,6):}`
It the zero error is -0.03 cm, then mean corrected diameter is :-

Options

A. 0.53 cm
B. 0.56 cm
C. 0.59 cm
D. 0.52 cm

Answer

To solve the problem of finding the mean corrected diameter of a steel ball measured using a Vernier caliper, we can follow these steps:

### Step 1: Understand the Vernier Caliper Specifications
- The main scale (MS) has divisions of 0.1 cm.
- 10 divisions of the Vernier scale (VS) match with 9 divisions of the main scale.

### Step 2: Calculate the Value of One Vernier Scale Division
To find the value of one Vernier scale division (VSD), we can use the formula:
\[ \text{1 VSD} = \left(\frac{9 \text{ MS divisions}}{10 \text{ VS divisions}}\right) \times \text{1 MS division} \]
Substituting the value of 1 MS division:
\[ \text{1 VSD} = \left(\frac{9}{10}\right) \times 0.1 \text{ cm} = 0.09 \text{ cm} \]

### Step 3: Calculate the Least Count of the Vernier Caliper
The least count (LC) can be calculated as:
\[ \text{LC} = \text{1 MS division} - \text{1 VSD} \]
Substituting the values:
\[ \text{LC} = 0.1 \text{ cm} - 0.09 \text{ cm} = 0.01 \text{ cm} \]

### Step 4: Calculate the Readings for Each Measurement
Using the formula for the reading:
\[ \text{Reading} = \text{MS reading} + (\text{VS divisions} \times \text{LC}) \]
For each measurement:
1. **First Measurement**:
\[ R_1 = 0.5 \text{ cm} + (8 \times 0.01 \text{ cm}) = 0.5 \text{ cm} + 0.08 \text{ cm} = 0.58 \text{ cm} \]
2. **Second Measurement**:
\[ R_2 = 0.5 \text{ cm} + (4 \times 0.01 \text{ cm}) = 0.5 \text{ cm} + 0.04 \text{ cm} = 0.54 \text{ cm} \]
3. **Third Measurement**:
\[ R_3 = 0.5 \text{ cm} + (6 \times 0.01 \text{ cm}) = 0.5 \text{ cm} + 0.06 \text{ cm} = 0.56 \text{ cm} \]

### Step 5: Calculate the Mean of the Readings
The mean reading (\(R_{avg}\)) is calculated as:
\[ R_{avg} = \frac{R_1 + R_2 + R_3}{3} = \frac{0.58 \text{ cm} + 0.54 \text{ cm} + 0.56 \text{ cm}}{3} = \frac{1.68 \text{ cm}}{3} = 0.56 \text{ cm} \]

### Step 6: Correct for Zero Error
The zero error is given as -0.03 cm. The mean corrected reading (\(R_{corrected}\)) is calculated as:
\[ R_{corrected} = R_{avg} - \text{Zero Error} \]
Substituting the values:
\[ R_{corrected} = 0.56 \text{ cm} - (-0.03 \text{ cm}) = 0.56 \text{ cm} + 0.03 \text{ cm} = 0.59 \text{ cm} \]

### Final Answer
The mean corrected diameter of the steel ball is:
\[ \text{Mean Corrected Diameter} = 0.59 \text{ cm} \]

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