In an experiment to find out the diameter of wire using screw gauge, the following observations were noted: (A) Screw moves 0.5 mm on main scale in one complete rotation (B) Total divisions on circular scale = 50 (C) Main scale reading is 2.5 mm (D) 45 th division of circular scale is in the pitch line (E) Instrument has 0.03 mm negative error Then the diameter of wire is: (a) 2.92 mm (b) 2.54 mm (c) 2.98 mm (d) 3.45 mm

Answer: (c). LC = (0.5 mm)/(50) = 0.01 mm Observed reading = 2.5 + 45× 0.01 = 2.95 mm Negative zero error -0.03 mm, so correct reading = 2.95 - (-0.03) = 2.98 mm So the answer is (c) 2.98 mm.

In an experiment to find out the diameter of wire using screw gauge… | JEE Main 2022 (29 Jul, Shift 1)

Q30JEE Main 2022 · 29 JUL · SHIFT 1Easy

Least Count, Vernier Callipers & Screw Gauge

In an experiment to find out the diameter of wire using screw gauge, the following observations were noted:

(A) Screw moves $0.5\text{ mm}$ on main scale in one complete rotation

(B) Total divisions on circular scale $= 50$

(D) $45$ th division of circular scale is in the pitch line

(E) Instrument has 0.03 mm negative error

Then the diameter of wire is:

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Answer: (c)

LC $= \dfrac{0.5\text{ mm}}{50} = 0.01$ mm
Observed reading $= 2.5 + 45\times 0.01 = 2.95$ mm
Negative zero error $-0.03$ mm, so correct reading $= 2.95 - (-0.03) = 2.98$ mm
So the answer is (c) $2.98$ mm.

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Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2022 (29 Jul, Shift 1)

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