Q: 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 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of young’s modulus of the material of the wire from these data.

Sol. Maximum percentage error in Y is given by

$\large Y = \frac{W}{\frac{\pi D^2}{4}} \times \frac{L}{x}$

$\large \frac{\Delta Y}{Y} = 2 \frac{\Delta D}{D} + \frac{\Delta L}{L} + \frac{\Delta x}{x} $

$\large \frac{\Delta Y}{Y} = 2 \frac{0.001}{0.05} + \frac{0.1}{110} + \frac{0.001}{0.125} $

= 0.0489