COMEDK · Physics · 1. Mathematics in Physics
The relative errors in the measurement of two lengths \(1.02 \mathrm{~cm} \pm 0.01 \mathrm{~cm}\) and \(9.89 \mathrm{~cm} \pm 0.01 \mathrm{~cm}\) is
- A \(\pm 1 \%\) and \(\pm 0.1 \%\)
- B \(\pm 1 \%\) and \(\pm 0.2 \%\)
- C \(\pm 1 \%\) and \(\pm 1 \%\)
- D \(0.1 \%\) and \(\pm 1 \%\)
Answer & Solution
Correct Answer
(A) \(\pm 1 \%\) and \(\pm 0.1 \%\)
Step-by-step Solution
Detailed explanation
The relative error in a measurement \(x \pm \Delta x\) is given by \(\dfrac{\Delta x}{x}\). The percentage error is given by \(\dfrac{\Delta x}{x} \times 100 \%\). For the first measurement \(1.02 \pm 0.01 \text{ cm}\), the relative error is \(\dfrac{0.01}{1.02}\). The…
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