Error Analysis with Searle's experiment

Classical Mechanics Level 3

In searle's experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm.

1) The maximum percentage error in the measurement of Young's modulus of the material of the wire from these data is 4.89%

2) Measurement of length of the wire contributes least to the total percentage error.

3) Measurement of diameter of the wire contributes highest to the total percentage error.

Which of these statements are correct ?

1 & 3 Only All of them 1 & 2 Only 2 & 3 Only

1 solution

Ayon Ghosh
Sep 10, 2017

According to searle 's experiment,

Y = $\frac{4W}{πD^2}$ * $\frac{L}{\Delta L}$ . So errors will be (note here i have used d(x) for error in some quantity x.it is not a differential.).For the measurements there least counts are the maximum errors.

$\frac{d(Y)}{Y}$ $=$ $\frac{2d(D)}{D}$ + $\frac{d(L)}{L}$ + $\frac{d(\Delta L)}{delta L}$ .

$\frac{d(Y)}{Y}$ $* 100$ $=$ $(2* 0.001/0.050 + 0.1/110 + 0.001/0.125)*100$

$\frac{d(Y)}{Y}$ $=$ $4.89$ %.Statements 2 and 3 become clear from the above calculation.

Error Analysis with Searle's experiment

1 solution

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