Uncertainty in Measurement — JEE Main Chemistry MCQs with Solutions

Solution: In 0.200 g the leading zero is not significant, but the two zeros after the decimal (to the right of 2) are, giving 3 significant figures. In scientific notation all digits count, so 6.022 × 10²³ has four; counting numbers are exact with infinite significant figures; and zeros before the first non-zero digit are never significant.

Solution: Multiply the digit terms and add the exponents: (3.5 × 4.0) × 10^(4 + (−7)) = 14 × 10⁻³. Writing 14 in proper form as 1.4 × 10¹ raises the exponent by one, giving 14 × 10⁻³ = 1.4 × 10⁻². Forgetting to raise the exponent during this normalization step wrongly leaves 1.4 × 10⁻³.

Solution: Divide the digit terms and subtract the exponents: (6.4 ÷ 8.0) × 10^(−3 − 5) = 0.8 × 10⁻⁸. To make the digit term lie between 1 and 10, write 0.8 as 8.0 × 10⁻¹, which lowers the exponent by one: 0.8 × 10⁻⁸ = 8.0 × 10⁻⁹. Dividing 8.0 ÷ 6.4 the wrong way gives the 1.25 × 10⁻⁸ trap.

Solution: For addition, first make the exponents equal: 5.0 × 10⁻⁴ = 0.50 × 10⁻³. Then add the digit terms: (4.5 + 0.50) × 10⁻³ = 5.0 × 10⁻³. Adding the digit terms without aligning exponents (4.5 + 5.0) wrongly gives 9.5 × 10⁻³.

Solution: In multiplication, the result carries no more significant figures than the factor with the fewest. 6.4 × 2.51 = 16.064, but 6.4 has only two significant figures, so the answer is rounded to two significant figures: 16.

Solution: By the unit-factor method: 72 (km/h) × (1000 m ÷ 1 km) × (1 h ÷ 3600 s) = 72 × 1000 ÷ 3600 = 72000 ÷ 3600 = 20 m s⁻¹. Multiplying by 3.6 instead of dividing gives the wrong 259.2 m s⁻¹.

Solution: First 3.6 L = 3.6 × 1000 = 3600 cm³. Then convert using 1 cm³ = 10⁻⁶ m³: 3600 × 10⁻⁶ = 3.6 × 10³ × 10⁻⁶ = 3.6 × 10⁻³ m³. Multiplying instead of dividing gives the wrong 3.6 × 10³ m³.