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Physics · UNITS AND MEASUREMENT
हिन्दीA student measures two different masses using a balance with a least count of \( 0.01 \text{ g} \). The measured masses are \( m_1 = 1.02 \text{ g} \) and \( m_2 = 9.89 \text{ g} \). The ratio of the relative error in the measurement of \( m_1 \) to that in \( m_2 \) is approximately:
Correct answer · Option 1
9.7
Explanation
Relative error in \( m_1 = \frac{\Delta m}{m_1} = \frac{0.01}{1.02} \). Relative error in \( m_2 = \frac{\Delta m}{m_2} = \frac{0.01}{9.89} \). Ratio = \( \frac{0.01/1.02}{0.01/9.89} = \frac{9.89}{1.02} \approx 9.7 \).
