A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings $5.50 \mathrm{~mm}, 5.55 \mathrm{~mm}, 5.45 \mathrm{~mm} ; 5.65$ mm . The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm . The average diameter of the pencil should therefore be recorded as:

(D) $(5.54 \pm 0.07) \mathrm{mm}$

(A) $(5.5375 \pm 0.0739) \mathrm{mm}$

(B) $(5.5375 \pm 0.0740) \mathrm{mm}$

(C) $(5.538 \pm 0.074) \mathrm{mm}$

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Units & Measurement

A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings

5.50 mm, 5.55 mm, 5.45 mm; 5.65

mm . The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm . The average diameter of the pencil should therefore be recorded as:

(A) $(5.5375 \pm 0.0739) mm$
(B) $(5.5375 \pm 0.0740) mm$
(C) $(5.538 \pm 0.074) mm$
(D) $(5.54 \pm 0.07) mm$ ✓ Correct

Answer: (D)

Solution

A result should never be quoted to more significant figures than the readings justify, and here each reading is recorded to three significant figures, to the nearest

0.01 mm

. The mean

5.5375 mm

and the standard deviation

0.07395 mm

must therefore be rounded to that same precision, becoming

5.54 mm

and

0.07 mm

. The diameter is properly reported as

(5.54 \pm 0.07) mm

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Units & Measurement

Solution

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