Animated Solution for Physics - Physics and Measurement: A student measured the length of a rod and wrote it as $3.50\text{ cm}$. Which instrument did he use to measure it?

In the world of physics, a number is never just a number; it is a story about the instrument that produced it. When a student measures the length of a rod and writes it down as $3.50\text{ cm}$, they are communicating two distinct pieces of information. The first is the magnitude of the length itself. The second, and arguably more profound, is the precision of the instrument used to make that measurement.

You might wonder, why write $3.50\text{ cm}$ instead of just $3.5\text{ cm}$? Mathematically, they represent the same value. However, in experimental physics, the trailing zero is a significant figure. It explicitly states that the measurement is reliable up to the second decimal place in centimeters.

This implies that the instrument is capable of detecting changes as small as $0.01\text{ cm}$. This smallest measurable value is known as the Least Count (LC) of the instrument. Therefore, our mission is to find an instrument among the options that has a least count of exactly $0.01\text{ cm}$.

A Good Start, But Not Enough

Let's evaluate the first option: a standard meter scale. If you look at a typical ruler, the smallest markings are $1\text{ mm}$ apart.

This means the least count of a meter scale is $1\text{ mm}$, which converts to $0.1\text{ cm}$. If the student had used a meter scale, they could only confidently report the measurement as $3.5\text{ cm}$. They would have no way of knowing if the true length was $3.51\text{ cm}$ or $3.49\text{ cm}$. Since $0.1\text{ cm} eq 0.01\text{ cm}$, the meter scale is not our mystery instrument.

Engineering Precision

Now, let's examine the vernier calliper described in option (b). A vernier calliper achieves higher precision by using two scales: a main scale and a sliding vernier scale.

The problem states that the main scale has $10$ divisions in $1\text{ cm}$. This means the length of one Main Scale Division (MSD) is:

It also states that $10$ divisions on the vernier scale match exactly with $9$ divisions on the main scale. We can write this relationship as:

The least count of a vernier calliper is the difference between one main scale division and one vernier scale division:

This matches our required precision perfectly! The vernier calliper is capable of measuring down to the hundredth of a centimeter, making $3.50\text{ cm}$ a valid reading.

Too Much of a Good Thing

Just to be absolutely rigorous, let's check the screw gauges in options (c) and (d). A screw gauge measures by translating rotational motion into linear motion. Its least count is given by the pitch divided by the number of circular divisions.

For option (c):

For option (d):

Both of these instruments are too precise. If the student had used the screw gauge from option (c), they would have reported the measurement with three decimal places, such as $3.500\text{ cm}$.

By analyzing the significant figures in the measurement, we deduced the required least count. We then systematically calculated the least count for each instrument. Only the vernier calliper in option (b) possessed the exact precision of $0.01\text{ cm}$ required to produce the reading $3.50\text{ cm}$.