InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The diameter of a wire as measured by a screw gauge was found to be 0.026 cm, 0.028 cm, 0.029 cm, 0.027cm, 0.024cm and 0.027 cm. Calculate (i) mean value of diameter (ii) mean absoulte error (iii) relative error (iv) percentage error. Also express the result in terms of absolute error and percentage error. |
|
Answer» Correct Answer - (i) 0.027cm 9ii) 0.001 cm(iii) 0.037 (iv) 3.7 % (0.027 +- 0.001) cm 0.027 cm +- 3.7 %` (i) Mean value of diameter, `D = (0.026 + 0.028 + 0.029 +0.027 +0.024 +0.027)/(6)` `D = (0.161)/(6) = 0.268 =0.027cm` Now, `Delta D_1 =0.027 - 0.026 =0.001` `Delta D_2 =0.027 - 0.028 =0.002` `Delta D_3 =0.027 - 0.029 =0.002` `Delta D_4 =0.027 - 0.027 =0.000` `Delta D_5 =0.027 - 0.024 =0.003` `Delta D_6 =0.027 - 0.027 =0.000` (ii) Mean absolute arror `Delta D_1 =0.027 - 0.026 =0.001` `=(| DeltaD_1 |+| Delta D_2 | + |Delta D_3 + |/Delta D_4| + | DeltaD_5|+|DeltaD_6|)/(6)` `Delta D_1 =0.027 - 0.026 =0.001` `= (0.001+ 0.001 + 0.002 + 0.000 +0.003 +0.000)/(6)` `Delta D_1 =0.027 - 0.026 =0.001` `=(0.007)/(6) = 0.001cm` (iii) Relative error `= (0.001)/(0.027) = 0.037` (iv) Percentage error `= 0.037xx10% = 3.7%` (v) Diameter of wire `= (0.27 +- 0.001)cm` `=0.027cm +- 3.7%` |
|