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JEE Main 2022 (28 Jun, Shift 2) · Least Count, Vernier Callipers & Screw Gauge

Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2022 (28 Jun, Shift 2)

Q35JEE Main 2022 · 28 JUN · SHIFT 2Medium

Least Count, Vernier Callipers & Screw Gauge

A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is $\frac{x}{121}\%$ . The value of x is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 150

Mean $= \dfrac{1.22+1.23+1.19+1.20}{4} = 1.21\text{ mm}$
Absolute errors: $0.01, 0.02, 0.02, 0.01$ ; mean absolute error $= 0.015\text{ mm}$
Percentage error $= \dfrac{0.015}{1.21}\times 100 = \dfrac{150}{121}\%$
So $x = 150$ .

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