Least count of a screw gauge is 0.001 cm. the diameter of a wire measuredby this instrument is 0.025 cm. the percentage error in the measurement is

4 %

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25%

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8 %

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40 %

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Solution

The correct option is A
4 %

d = 0.025 cm; $Δ$ d = 0.001 cm

% error = $\frac{Δ d}{d} \times 100$ = 4%

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Q. The least count of a screw gauge is 0.001 cm.The diameter of a wire measured by it is

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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.

$y = \frac{W}{A} \times \frac{L}{X}$