Question

In a Searle's experiment, the diameter of the wire as measured by a screw gauge with 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.

$y=\frac{W}{A}\times \frac{L}{X}$
Where W is the weight suspended from the wire, A is the cross section area, L is the length and X is the extension in the wire.

[IIT JEE 2004]

• A. 4.89 %
• B. 1.26 %
• C. 9.67 %
• D. 15.56 %

Answer 1

The correct option is A

4.89 %

$\frac{△y}{y}=\frac{2△D}{D}+\frac{△x}{x}+\frac{△l}{l}$

(As the Diameter is squared, relative error is added twice)

$\frac{△y}{y}=2(\frac{0.001}{0.05})+(\frac{0.001}{0.125})+(\frac{0.1}{110})=0.0489$

$\therefore$percentage error = 4.89%