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9. The jaws of a vernier caliper touch the inner wall of a calorimeter without pressure. The position of the zero mark of the vernier scale on the main scale reads [tex]$3.4 \text{ cm}$[/tex]. The [tex]$6^{\text{th}}$[/tex] vernier scale division coincides with a main scale division. The vernier constant of the caliper is [tex]$0.01 \text{ cm}$[/tex]. Find the actual internal diameter of the calorimeter, given that the vernier scale has a zero error of [tex]$-0.03 \text{ cm}$[/tex].

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To find the actual internal diameter of the calorimeter using the vernier calipers, we need to follow these steps:

### Step 1: Main Scale Reading

From the problem, we know the reading on the main scale:
- Main Scale Reading (MSR) = 3.4 cm

### Step 2: Vernier Scale Reading

Next, we look at the vernier scale:
- The coinciding division is the 6th division of the vernier scale.
- Vernier Scale Division (VSD) that coincides = 6th division.

### Step 3: Vernier Constant

The vernier constant (least count) is given:
- Vernier Constant (VC) = 0.01 cm

### Step 4: Vernier Scale Reading Correction

To correct the reading from the vernier scale, we need to calculate the vernier scale reading correction. This is done by multiplying the vernier constant by the coinciding vernier scale division:
[tex]\[ \text{Vernier Scale Correction} = \text{Vernier Scale Division} \times \text{Vernier Constant} \][/tex]

Substituting the given values:
[tex]\[ \text{Vernier Scale Correction} = 6 \times 0.01 \, \text{cm} = 0.06 \, \text{cm} \][/tex]

### Step 5: Zero Error

We are given a zero error, which needs to be accounted for in our calculation:
- Zero Error = -0.03 cm

### Step 6: Actual Internal Diameter Calculation

To find the actual internal diameter of the calorimeter, we sum the main scale reading, vernier scale correction, and the zero error:
[tex]\[ \text{Actual Internal Diameter} = \text{Main Scale Reading} + \text{Vernier Scale Correction} + \text{Zero Error} \][/tex]

Substituting the given values:
[tex]\[ \text{Actual Internal Diameter} = 3.4 \, \text{cm} + 0.06 \, \text{cm} - 0.03 \, \text{cm} \][/tex]
[tex]\[ \text{Actual Internal Diameter} = 3.4 \, \text{cm} + 0.03 \, \text{cm} \][/tex]
[tex]\[ \text{Actual Internal Diameter} = 3.43 \, \text{cm} \][/tex]

### Result

Thus, the actual internal diameter of the calorimeter, considering the zero error, is:
[tex]\[ \boxed{3.43 \, \text{cm}} \][/tex]
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