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NCERT Exemplar · Q7

Q.Measure of two quantities along with the precision of respective measuring instrument is A=2.5A = 2.5 m s−1±0.5^{-1} \pm 0.5 m s−1^{-1}, B=0.10B = 0.10 s ±0.01\pm 0.01 s. The value of ABAB will be

(a) (0.25±0.08)(0.25 \pm 0.08) m
(b) (0.25±0.5)(0.25 \pm 0.5) m
(c) (0.25±0.05)(0.25 \pm 0.05) m
(d) (0.25±0.135)(0.25 \pm 0.135) m
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When multiplying measured quantities, the relative errors add. Compute AB=2.5×0.10=0.25AB = 2.5 \times 0.10 = 0.25 m, then find the fractional uncertainty in each factor, sum them to get the fractional uncertainty in the product, and convert back to absolute error. The answer is (A).


Why relative errors add in multiplication

When you multiply two measured quantities, each carrying its own uncertainty, the percentage "wiggle room" in each factor compounds. If AA could be 20% off and BB could be 10% off, the product ABAB inherits both sources of uncertainty. The cleanest way to track this is through relative (fractional) errors:

Δ(AB)AB=ΔAA+ΔBB.\frac{\Delta(AB)}{AB} = \frac{\Delta A}{A} + \frac{\Delta B}{B}.

This formula captures the intuition that errors propagate proportionally when quantities are multiplied or divided.


Step-by-step calculation

  1. Compute the central value of the product.

AB=(2.5 m s−1)×(0.10 s)=0.25 m.AB = (2.5 \,\text{m s}^{-1}) \times (0.10 \,\text{s}) = 0.25 \,\text{m}.

  1. Find the relative error in AA.

ΔAA=0.52.5=0.2=20%.\frac{\Delta A}{A} = \frac{0.5}{2.5} = 0.2 = 20\%.

  1. Find the relative error in BB.

ΔBB=0.010.10=0.1=10%.\frac{\Delta B}{B} = \frac{0.01}{0.10} = 0.1 = 10\%.

  1. Add the relative errors to get the relative error in ABAB.

Δ(AB)AB=0.2+0.1=0.3=30%.\frac{\Delta(AB)}{AB} = 0.2 + 0.1 = 0.3 = 30\%.

  1. Convert back to absolute error. …

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