JEE MAIN - Physics (2025 - 24th January Morning Shift - No. 20)
Explanation
$$ 32.3 \quad (3\ \text{sig figs}), \quad 1125 \quad (4\ \text{sig figs}), \quad 27.4 \quad (3\ \text{sig figs}) $$
In multiplication and division, the final answer must be reported with the same number of significant figures as the factor with the fewest significant figures, which here is 3.
First, calculate the unrounded result:
$$ y = \frac{32.3 \times 1125}{27.4} \approx \frac{36337.5}{27.4} \approx 1326.186 $$
Expressing $1326.186$ in scientific notation:
$$ 1326.186 \approx 1.326186 \times 10^3 $$
Rounding to 3 significant figures:
The first three significant digits are 1, 3, and 2.
Considering the fourth digit (6) causes the last significant digit to round up.
Thus,
$$ 1.326186 \times 10^3 \approx 1.33 \times 10^3 $$
In standard decimal form, this is written as:
$$ 1330 $$
Hence, the final reported value is:
$$ y = 1330 $$
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