JEE Physics 2025 Question with Solution

Given: $y = \frac{32.3 \times 1125}{27.4}$ and all digits are significant.

Find: The correctly reported value of $y$ using significant figure rules.

For multiplication and division, the final result must have the same number of significant figures as the quantity with the least significant figures.

Here:

• $32.3$ has $3$ significant figures
• $1125$ has $4$ significant figures
• $27.4$ has $3$ significant figures

So, the reported value of $y$ must have $3$ significant figures.

Now evaluate the expression:

$$y = \frac{32.3 \times 1125}{27.4}$$ $$32.3 \times 1125 = 36337.5$$ $$y = \frac{36337.5}{27.4} \approx 1325.9\text{ to }1330.05$$

Rounding the result to $3$ significant figures gives $1330$.

Therefore, the value of $y$ should be reported as $1330$. The correct option is D.

Given: $y = \frac{32.3 \times 1125}{27.4}$

Find: How to write the final value with the correct number of significant figures.

First identify the significant figures in each number:

1. $32.3$ has $3$ significant figures.
2. $1125$ has $4$ significant figures.
3. $27.4$ has $3$ significant figures.

Thus, the final answer must contain $3$ significant figures.

Now compute:

$$32.3 \times 1125 = 36337.5$$

Then divide by $27.4$:

$$y = \frac{36337.5}{27.4} \approx 1325.9124\ldots$$

or approximately $1330.05$ as shown in the solution approach.

In either case, after rounding to $3$ significant figures, the reported value is:

$$y \approx 1330$$

Therefore, the correct option is D.