1. Idea: The measurement is written as $3.50\,\text{cm}$, so the instrument must have a least count of at least $0.01\,\text{cm}$, because the value is recorded up to two decimal places in cm.

2. Check each option:

Option A: Screw gauge

Given:

• Pitch $= 1\,\text{mm}$
• Circular scale divisions $= 100$

Least count:

Convert to cm:

So this instrument measures up to $0.001\,\text{cm}$, and readings would typically be written like $3.500\,\text{cm}$, not $3.50\,\text{cm}$.

So, A is not appropriate.

Option B: Screw gauge

Given:

• Pitch $= 1\,\text{mm}$
• Circular scale divisions $= 50$

Least count:

Convert to cm:

This also measures more precisely than $0.01\,\text{cm}$, so the reading would generally be recorded to three decimal places in cm.

So, B is not appropriate.

Option C: Meter scale

A meter scale usually has least count:

So readings are typically written as $3.5\,\text{cm}$ or $3.6\,\text{cm}$, not $3.50\,\text{cm}$.

So, C is not appropriate.

Option D: Vernier calliper

Given:

• $10$ vernier scale divisions $= 9$ main scale divisions
• Main scale has $10$ divisions in $1\,\text{cm}$

So, one main scale division:

Also,

Least count of vernier caliper:

Thus,

This matches the precision implied by $3.50\,\text{cm}$.

So, D is correct.

3. Final Answer: The student used a vernier calliper.

4. Comparison with stored correct answer: Stored correct answer is D, which matches the derived answer.