JEE Physics 2026 Question with Solution

Given: The quantities are $52.01 \, \text{m}$, $153.2 \, \text{m}$ and $0.123 \, \text{m}$.

Find: The correct sum keeping significant figures in view.

When physical quantities are added, the result must be reported to the same number of decimal places as the quantity with the fewest decimal places.

Here:

• $52.01 \, \text{m}$ has $2$ decimal places
• $153.2 \, \text{m}$ has $1$ decimal place
• $0.123 \, \text{m}$ has $3$ decimal places

So the final answer must be rounded to one decimal place.

First, add the given quantities:

$$52.01 + 153.2 + 0.123 = 205.333 \, \text{m}$$

Now round $205.333 \, \text{m}$ to one decimal place:

$$205.333 \, \text{m} \approx 205.3 \, \text{m}$$

Therefore, the sum with significant figures is $205.3 \, \text{m}$. The correct option is D.