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In an experiment with common balance the mass of a body is found to 2.52g, 2.53g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate1. The mean value of the body2. Mean absolute error3. Percentage error |
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Answer» 1. Mean value, Mmean = \(\frac{2.52+2.53+2.51+2.49+2.54}{5}\) = 2.5g 2. Absolute error, Absolute error ∆m1 = |2.52 – 2.52| = 0 ∆m2 = |2.52 – 2.53| = 0.01 ∆m3 = |2.52 – 2.51| = 0.01 ∆m4 = |2.52 – 2.49| = 0.03 ∆m5 = |2.52 – 2.54| = 0.02 ∴ Mean absolute error \(\frac{0+0.01+0.01+0.03+0.02}{5}\) ∆mmean = 0.014g 3. Percentage error = \(\frac{Δm_{mean}}{m_{mean}}\) × 100 = \(\frac{ 0.014}{2.52}\) × 100 = 0.556. |
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