1. Find the least count of the Vernier callipers

Given:

• $1$ main scale division $(\text{MSD}) = 1\text{ mm}$
• $10$ Vernier scale divisions $(\text{VSD}) = 9$ main scale divisions

So, $10\,\text{VSD} = 9\,\text{mm}$ $1\,\text{VSD} = 0.9\,\text{mm}$

Least count: $\text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = 1.0 - 0.9 = 0.1\,\text{mm}$

Thus, $\boxed{\text{LC} = 0.1\text{ mm} = 0.01\text{ cm}}$

2. Find the zero error

It is given that when the jaws are closed:

• Vernier zero lies to the right of main scale zero
• $4^{\text{th}}$ Vernier division coincides with a main scale division

This is a positive zero error.

For a direct vernier, if the $n^{\text{th}}$ Vernier division coincides, then zero error is: $\text{Zero error} = n \times \text{LC}$

So, $\text{Zero error} = 4 \times 0.1\text{ mm} = 0.4\text{ mm} = 0.04\text{ cm}$

Hence zero correction is: $\text{Zero correction} = -0.04\text{ cm}$

3. Find the observed reading for the bob

Given:

• Vernier zero lies between $4.1\text{ cm}$ and $4.2\text{ cm}$
• $6^{\text{th}}$ Vernier division coincides

Main scale reading: $\text{MSR} = 4.1\text{ cm}$

Vernier reading: $\text{VR} = 6 \times \text{LC} = 6 \times 0.01 = 0.06\text{ cm}$

Observed reading: $\text{Observed reading} = 4.1 + 0.06 = 4.16\text{ cm}$

4. Apply zero correction

True reading: $\text{True reading} = \text{Observed reading} + \text{Zero correction}$ $= 4.16 - 0.04$ $= 4.12\text{ cm}$

Thus diameter of the bob is $\boxed{4.12\text{ cm}}$

5. Match with the required format

The question asks for: $\underline{\hspace{2cm}} \times 10^{-2}\text{ cm}$

Since $4.12\text{ cm} = 412 \times 10^{-2}\text{ cm}$

So the required integer is: $\boxed{412}$

6. Comparison with stored answer

Derived answer = $412$

Stored correct answer = $412$

So, they agree.