Question

The relative errors in the measurement of two lengths \(1.02 \, \text{cm} \pm 0.01 \, \text{cm}\) and \(9.89 \, \text{cm} \pm 0.01 \, \text{cm}\) is

• A. \(\pm 1\% \text{ and } \pm 0.1\%\)
• B. \(\pm 1\% \text{ and } \pm 0.2\%\)
• C. \(\pm 1\% \text{ and } \pm 1\%\)
• D. \(0.1\% \text{ and } 1\%\)

Hint:

Relative error gives the accuracy of measurement. Always use \( \frac{\Delta x}{x} \times 100 \% \) to convert absolute error to percentage error.

Answer 1

The Correct Option is A

Relative error is calculated as: \[ \text{Relative Error} = \left( \frac{\Delta x}{x} \right) \times 100\% \] For the first length: \[ x_1 = 1.02 \text{ cm}, \quad \Delta x_1 = 0.01 \text{ cm} \Rightarrow \text{Relative error} = \frac{0.01}{1.02} \times 100 \approx 0.980\% \approx 1\% \] For the second length: \[ x_2 = 9.89 \text{ cm}, \quad \Delta x_2 = 0.01 \text{ cm} \Rightarrow \text{Relative error} = \frac{0.01}{9.89} \times 100 \approx 0.10\% \]