Q176 / 180 · COMEDK 2021 · 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
- Correct answerA$\pm 1 \%$ and $\pm 0.1 \%$
- B$\pm 1 \%$ and $\pm 0.2 \%$
- C$\pm 1 \%$ and $\pm 1 \%$
- D$0.1 \%$ and $\pm 1 \%$
Solution
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 percentage error is $\dfrac{0.01}{1.02} \times 100 \% \approx 0.98 \% \approx 1 \%$.
For the second measurement $9.89 \pm 0.01 \text{ cm}$, the relative error is $\dfrac{0.01}{9.89}$. The percentage error is $\dfrac{0.01}{9.89} \times 100 \% \approx 0.101 \% \approx 0.1 \%$.
Thus, the relative errors are approximately $\pm 1 \%$ and $\pm 0.1 \%$.
Answer: $\pm 1 \%$ and $\pm 0.1 \%$
For the first measurement $1.02 \pm 0.01 \text{ cm}$, the relative error is $\dfrac{0.01}{1.02}$. The percentage error is $\dfrac{0.01}{1.02} \times 100 \% \approx 0.98 \% \approx 1 \%$.
For the second measurement $9.89 \pm 0.01 \text{ cm}$, the relative error is $\dfrac{0.01}{9.89}$. The percentage error is $\dfrac{0.01}{9.89} \times 100 \% \approx 0.101 \% \approx 0.1 \%$.
Thus, the relative errors are approximately $\pm 1 \%$ and $\pm 0.1 \%$.
Answer: $\pm 1 \%$ and $\pm 0.1 \%$
Hence, the correct answer is option: 'A'.