JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Units & Measurement

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

(a) Screw moves 0.5 mm on main scale in one complete rotation
(b) Total divisions on circular scale

= 50

4 5^{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✓ Correct
(D)3.45 mm

Answer: (C)

Solution

MSR = 2.5 mm. Least count =

\frac{0.5}{50} = 0.01 mm

. CSR =

45 \times 0.01 = 0.45 mm

. Zero error = −0.03 mm (negative). Diameter = MSR + CSR − (zero error) = 2.5 + 0.45 − (−0.03) = 2.98 mm.

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option

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