Uncertainty in Measurement — IISER Chemistry MCQs with Solutions

Solution: Measurements clustered tightly around one another are precise, but if they all sit far from the true value they are not accurate — so A is true. R correctly defines precision (mutual closeness) and accuracy (closeness to the true value), which is exactly why precision without accuracy is possible. Hence both are true and R explains A.

Solution: Count significant figures: 0.002050 has 4 (digits 2, 0, 5 and the trailing 0 after the decimal); 25000 has 2 (trailing zeros with no decimal are not significant); 2.5 × 10³ has 2; and 0.025 has 2. So 0.002050 has the most, with 4 significant figures.

Solution: The digit to be removed is exactly 5 with nothing after it. The rounding rule for a trailing 5 keeps the preceding digit unchanged if it is even and raises it by one if it is odd. Here the preceding digit 4 is even, so it stays: 2.45 rounds to 2.4, not 2.5.

Solution: For subtraction, make the exponents equal first: 3.0 × 10⁻³ = 0.30 × 10⁻². Then subtract the digit terms: (5.0 − 0.30) × 10⁻² = 4.7 × 10⁻². Subtracting without aligning exponents (5.0 − 3.0) wrongly gives 2.0 × 10⁻².

Solution: Convert both units: 1 g L⁻¹ = (10⁻³ kg) ÷ (10⁻³ m³) = 1 kg m⁻³, since the two factors of 10⁻³ cancel. Therefore 1.25 g L⁻¹ = 1.25 kg m⁻³. (The g cm⁻³ → kg m⁻³ conversion multiplies by 1000, but for g L⁻¹ the factor is 1 — a common trap giving 1250.)

Solution: In addition, the result cannot have more digits after the decimal point than the term with the fewest. The exact sum is 25.5 + 2.55 + 0.255 = 28.305, but 25.5 has only one decimal place, so the answer is reported to one decimal place: 28.3.

Solution: Multiply the unit factors in series: 3 day × (24 h ÷ 1 day) × (60 min ÷ 1 h) × (60 s ÷ 1 min) = 3 × 24 × 60 × 60 = 259200 s. Omitting the final ×60 gives 4320 s, and 172800 s is the answer for 2 days, not 3.