JEE Physics 2025 Question with Solution

Given: The measured masses are $435.42 \, \text{g}$, $226.3 \, \text{g}$, and $0.125 \, \text{g}$.

Find: The sum of the masses using the rule of significant figures for addition.

For addition, the result must be rounded to the least number of decimal places present among the given measurements.

• $435.42 \, \text{g}$ has $2$ decimal places.
• $226.3 \, \text{g}$ has $1$ decimal place.
• $0.125 \, \text{g}$ has $3$ decimal places.

So, the final answer must be rounded to $1$ decimal place.

Adding the masses:

$$435.42 + 226.3 + 0.125 = 661.845$$

Thus,

$$661.845 \, \text{g} \approx 661.8 \, \text{g}$$

Therefore, the sum of the masses is $661.8 \, \text{g}$. The correct option is C.

Given: $m_1 = 435.42 \, \text{g}$, $m_2 = 226.3 \, \text{g}$, and $m_3 = 0.125 \, \text{g}$.

Find: The correctly rounded total mass.

A quick way is to first identify the least number of decimal places. Since $226.3 \, \text{g}$ has only $1$ decimal place, the final sum must also have $1$ decimal place.

Now add:

$$m_1 + m_2 + m_3 = 435.42 + 226.3 + 0.125 = 661.845$$

Rounding to one decimal place gives $661.8 \, \text{g}$.

Therefore, the correct option is C.