A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

We begin by recalling that the number of digits written in a measurement tells us up to which place-value the observer was certain.

If a reading is written as $$3.50\ \text{cm}$$, the last digit “0” is significant; hence the observer was sure up to $$0.01\ \text{cm}$$ (one-hundredth of a centimetre).

So, the measuring instrument used must have a least count (i.e. the smallest length it can resolve) of $$0.01\ \text{cm}$$.

Now we examine the least count of each given instrument one by one.

Option A : Meter scale
A common meter scale is ruled millimetre by millimetre.

One main scale division $$=1\ \text{mm}=0.1\ \text{cm}$$.

Hence its least count $$=0.1\ \text{cm}$$.

This is ten times coarser than $$0.01\ \text{cm}$$, so a meter scale cannot justify writing “3.50 cm”.

Option B : Vernier calliper
We are told:

• Main scale has 10 divisions in $$1\ \text{cm}$$, so

$$\text{One MSD}= \frac{1\ \text{cm}}{10}=0.1\ \text{cm}$$.

• Ten vernier divisions coincide with nine main scale divisions, thus

$$10\ \text{VSD}=9\ \text{MSD}=9(0.1\ \text{cm})=0.9\ \text{cm},$$

hence

$$\text{One VSD}= \frac{0.9\ \text{cm}}{10}=0.09\ \text{cm}.$$

The formula for least count of a vernier calliper is

$$\text{Least count} = 1\ \text{MSD} - 1\ \text{VSD}.$$

Substituting, we get

$$\text{Least count}=0.1\ \text{cm}-0.09\ \text{cm}=0.01\ \text{cm}.$$

This exactly matches the required$$0.01\ \text{cm}$$, so such a vernier calliper can legitimately lead to the notation “3.50 cm”.

Option C : Screw gauge (100 divisions, pitch $$1\ \text{mm}$$)
First we convert the pitch:

$$1\ \text{mm}=0.1\ \text{cm}.$$

The formula for least count of a screw gauge is

$$\text{Least count}= \frac{\text{Pitch}}{\text{Number of circular scale divisions}}.$$

Hence

$$\text{Least count}= \frac{0.1\ \text{cm}}{100}=0.001\ \text{cm}\;(10^{-3}\ \text{cm}).$$

With such fine resolution the reading would normally be recorded to three decimal places, e.g. $$3.500\ \text{cm}$$, not $$3.50\ \text{cm}$$.

Option D : Screw gauge (50 divisions, pitch $$1\ \text{mm}$$)
Again converting the pitch gives $$0.1\ \text{cm}$$.

Therefore

$$\text{Least count}= \frac{0.1\ \text{cm}}{50}=0.002\ \text{cm}.$$

This device resolves to the third decimal place (about $$2\times10^{-3}\ \text{cm}$$), so one would also write $$3.502\ \text{cm}$$, $$3.504\ \text{cm}$$ etc., but not simply $$3.50\ \text{cm}$$.

Comparing all four instruments, only the vernier calliper in Option B has the least count $$0.01\ \text{cm}$$ that agrees with the notation used by the student.

Hence, the correct answer is Option B.