Animated Solution for Physics - Physics and Measurement: For the four sets of three measured physical quantities as given below. Which of the following options is correct ? (i) $A_1=24.36,B_1=0.0724,C_1=256.2$ (ii) $A_2=24.44,B_2=16.082,C_3=240.2$ (iii) $A_3=25.2,B_3=19.2812,C_3=236.183$ (iv) $A_4=25,B_4=236.191,C_4=19.5$

When we plug numbers into a calculator, it happily spits out a result with as many decimal places as its screen can hold. But in physics, numbers represent real-world measurements, and every measurement has a limit to its precision.

If you measure the length of a table with a standard ruler and the thickness of a piece of paper with a highly precise micrometer, adding those two values together doesn't magically make your ruler measurement more precise. Your final answer is only as reliable as your weakest link. This is the core philosophy behind the rules of significant figures.

The rule for adding or subtracting physical quantities is beautifully simple, yet often misunderstood. When adding or subtracting, the final result must be rounded to the least number of decimal places present in the given values.

Notice that we are looking at decimal places, not the total number of significant figures. This is because addition is about absolute uncertainty (which decimal column the doubt begins in), whereas multiplication is about relative uncertainty.

Let's apply this golden rule to the four sets provided in the problem.

Set 1: We are given $A_1=24.36$ (2 decimal places), $B_1=0.0724$ (4 decimal places), and $C_1=256.2$ (1 decimal place).

The least precise measurement here is $C_1$, which dictates that our final answer can only have one decimal place.

The raw mathematical sum is:

Cutting off after the first decimal place gives us $280.6|324$. Since the first dropped digit is $3$ (which is less than $5$), we simply drop the rest. Thus, $S_1=280.6$.

Set 2: The values are $A_2=24.44$ (2 decimal places), $B_2=16.082$ (3 decimal places), and $C_2=240.2$ (1 decimal place). (Note: The question paper had a typo writing $C_3$ instead of $C_2$, but the intent is clear).

Again, the limiting precision is one decimal place.

The raw sum is:

Cutting off after the first decimal place gives $280.7|22$. The first dropped digit is $2$, so we round down. Thus, $S_2=280.7$.

Set 3: Here we have $A_3=25.2$ (1 decimal place), $B_3=19.2812$ (4 decimal places), and $C_3=236.183$ (3 decimal places).

The limiting precision is once again one decimal place.

The raw sum is:

We cut off after the first decimal place: $280.6|642$. This time, the first dropped digit is $6$, which is greater than $5$. Therefore, we must round up the preceding digit. Thus, $S_3=280.7$.

Set 4: Finally, we look at $A_4=25$ (0 decimal places), $B_4=236.191$ (3 decimal places), and $C_4=19.5$ (1 decimal place).

This is the trap! The number $25$ has zero decimal places. It is the least precise measurement, meaning our final answer must be rounded to the nearest whole integer.

The raw sum is:

We cut off exactly at the decimal point: $280|.691$. The first dropped digit is $6$, so we round up the integer part. Thus, $S_4=281$.

Let's compile our correctly rounded sums: - $S_1=280.6$ - $S_2=280.7$ - $S_3=280.7$ - $S_4=281$

Comparing these values, we get the strict mathematical relationship:

$$S_1<S_2=S_3<S_4$$

If you carefully examine the four options provided in the original JEE paper, you will find that absolutely none of them match this relationship. Because of this, the question was flawed and likely treated as a bonus. However, the rigorous application of significant figures we just performed is exactly what the examiners intended to test!