Diameter of a wire measured by a screw guage in four different readings is found to be 2.20 mm, 2.10 mm, 2.40 mm and 1.90 mm.

Number of significant figures in all readings is two.

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Error in

1^{s t}

reading is negative.

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Mean error of readings is zero.

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Error in

4^{t h}

reading is 0.25 mm.

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Solution

The correct option is D Error in $4^{t h}$ reading is 0.25 mm.
True diameter is, $D = \frac{D_{1} + D_{2} + D_{3} + D_{4}}{4} = \frac{2.2 + 2.1 + 2.4 + 1.9}{4}$
D =2.15 mm
$Δ D_{1} = D - D_{1} = 2.15 - 2.20 = - 0.05$
$Δ D_{4} = D - D_{4} = 2.15 - 1.90 = 0.25$
$Δ D_{m e a n} = \frac{| Δ D_{1} | + | Δ D_{2} | + | Δ D_{3} | + | Δ D_{4} |}{4} = 0.15$
Number of significant figures is 3 in each of the cases.

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